chapter 1 kinematics: the description of...

Etkina/Gentile/Van Heuvelen Process Physics 1/e, Chapter 1 1-1

Chapter 1 Kinematics: The Description of Motion

! Is it possible to safely catch a flying bullet with your hand?

! What is the safe following distance between you and the car in front of you?

! Can you be moving and not moving at the same time?

Make sure you know how to:

1. Define what a point-like object is. (Intro)

2. Use significant digits in calculations (Intro)

While flying their aircraft, World War I pilots were known to catch bullets fired from

enemy soldiers on the ground. Magicians who tried this trick were often seriously injured, or

worse. The bullets traveled about 120 miles per hour (mph). How were the pilots able to catch

bullets with their bare hands, but the magicians could not? By the time you finish working

through this chapter, you’ll know the answer.

Scientists often ask questions about things that most people accept as being “just the way

it is.” In the northern hemisphere, we have more hours of daylight during the summer than in the

winter. In the southern hemisphere, it’s just the opposite. Most people simply accept this fact.

However, scientists want explanations for even the simplest mysteries surrounding us.

We will begin our study of physics with a phenomenon that surrounds us from the

moment we are born—motion. For example, each dawn, we see the Sun in the east, and at dusk it

is in the west. For more than 1000 years, scientists believed that the Sun circled Earth every 24

hours. However now they are convinced that we live in a heliocentric (Sun-centered) solar system

in which Earth and all planets move in orbits around the Sun. What made scientists change their

understanding of Earth’s position and motion in the Solar system? The start of understanding

everyday phenomena such as this is a careful observation and description of the phenomena. In

this chapter, we learn to describe motion.

1.1 What is motion?

Understanding motion quantitatively challenged the minds of scientists for thousands

of years, so let’s start by first defining what motion is. Consider the following experiments

that investigate how the description of motion depends on the observer.

ALG

1.1.1-

1.1.5


Observational Experiment Table 1.1 Describing motion through the observer.

Observational experiments Analysis

Jan focuses on a ball in her

hands as she walks across the

room. Tim, sitting at a desk,

also focuses on the ball.

Jan reaches the other side of

the room continually

focusing on the ball; her head

did not turn. Tim’s head has

turned in order to follow the

ball.

The two observers (Jan and Tim) see the ball

differently. With respect to Jan, the ball’s

position does not change. With respect to

Tim, it does change.

Ted on train A observes

Sue on train B. He does

not turn his head to keep

his eyes on Sue. Joan

standing on the station

platform turns her head to

follow Sue.

Ted sees Sue as stationary whereas Joan sees

her as moving.

Pattern Different observers can describe the same process differently, including whether or not motion is occurring.

The pattern that we discovered by doing these observational experiments was that

different observers describe the same process differently. One person sees the ball moving while

another does not. They are both correct from their perspectives. In order to describe the motion of

something, we need to identify the observer.

Definition of motion Motion is a change in an object’s position relative to a given

observer during a certain change in time. Without identifying the observer, it is impossible

to say whether motion occurred. Physicists say motion is relative, meaning that motion

depends on the point of view of the observer.

Testing your ideas

Are you moving as you read this book? According to our definition of motion, we should

be able to find an observer who says you are moving. Consider a friend walking past you while

you sit reading. Your friend first saw you several feet in front of her, then she saw you next to

her, and finally she saw you several feet behind her. Though you were sitting in a chair, you

definitely were moving with respect to your friend!


We just performed an experiment to test the idea that motion is relative. To test an idea in

science means to use the idea to predict the outcome of a new experiment. If the outcome matches

the prediction, the idea has not been disproved (we would need an infinite number of testing

experiments to prove that the idea is correct in absolutely all cases, a finite number of

experiments cannot do it). If the idea does not match the prediction, then the idea has been

disproved and you need to revise it (if possible).

Now let’s return to you sitting in a chair reading this book. You obviously are not moving

with respect to the chair or with respect to the floor, but you are moving with respect to the Sun

or with respect to a bird flying outside. What makes the idea of relative motion confusing at first

is that people intuitively use the Earth as the object of reference. If an object does not move with

respect to the Earth, many people would say that the object is not moving. That is why it took

scientists thousands of years to understand the reason for days and nights on Earth. An observer

on Earth sees the Sun moving in an arc across the sky (Figure 1.1a). An observer on a spaceship

sees Earth rotating on its axis so that different parts of its surface are facing the Sun at different

times (Figure 1.1b).

Think back to the World War I pilots who could catch bullets with their bare hands. The

speed of airplanes at that time was about 120 mph. If the plane and the bullet were both traveling

relative to Earth at about 120 mph in the same direction, then the bullet was approximately at rest

with respect to the pilot; the pilot could catch the bullet without injuring himself. Does that seem

counterintuitive? The bullet was moving fast from the point of view of the soldier who fired it

from the ground. At the same time, the bullet was almost at rest from the point of view of the

pilot.

Figure 1.1 (a)(b) Two observers see different reasons for the appearance of the Sun relative to Earth

Review question 1.1

Are you moving while sitting on a train that is leaving a station?


1.2 Point-like objects and linear motion

In the Introduction Chapter, we learned about some tools that physicists use to study the

physical world. One tool is models—simplified assumptions about objects, interactions, and

processes when analyzing complicated situations.

Model of a point-like object

One way that physicists model objects is to represent them as point-like objects. We use

the point-like object model to represent a real object when the size and internal structure of the

real object are not relevant for understanding the situation, and/or when all parts of the object

move together without changing its structure. When you represent an object as a point, you

basically think of it as a dot, a small point-like entity of zero volume. Later, we will construct

more sophisticated models of objects (for example, rigid bodies, elastic bodies, ideal gases, and

incompressible fluids) as we build more physics ideas and analyze more complicated situations.

Models of processes

Since we are investigating the phenomenon of motion, we will need a different category

of model—a model of processes. Motion is one of many processes you will study in this book.

Examples of other processes include changes in an object’s color, the transition from a solid

object to liquid, and the drying of a liquid. Just as we made a simplifying assumption about an

object (representing a real object as a point-like object), we can also make simplifying

assumptions about a process. What is the simplest way an object can move?

Imagine that you haven’t ridden a bike in a while. You would probably start by riding in

a straight line to get comfortable. This kind of motion is called linear motion. Eventually you get

comfortable enough that you try changing direction. This kind of motion is called curvilinear

motion. Since linear motion is easier to describe, let’s start with that.

Linear motion A process model of motion using the simplified assumption that an

object, considered as a point-like object, moves along a straight line.

Let’s say we want to construct a model for a car’s motion along a straight stretch of

highway. Since the car is small compared to the length of the highway, we can model it as a

point-like object (a model of an object). Since the highway is long and straight, we can model the

motion as linear motion (a model of a process).

If you think there is more going on, such as the effect the air has on the car, or the effect

that the road surface has on its tires, we agree! The road and the air definitely have important

effects on the motion of the car. They interact with the car. We will start constructing models of

interactions in Chapter 2. For now, we are interested only in the motion of the car and not why the

motion of the car is what it is.

Reference frames

At the beginning of the chapter, we constructed an explanation for how World War I

pilots were able to catch bullets. Specifying the observer before describing the motion is


extremely important. To describe motion quantitatively, physicists use a reference frame. A

reference frame includes an observer, a coordinate system, a scale for measuring distances, and a

clock to measure time. If the object of reference is large (for example, Earth) and cannot be

considered a point, it is important to specify where on the object of reference the origin of the

coordinate system is placed.

Reference frame A reference frame includes three essential components:

(a) An object of reference with a specific point of reference on it.

(b) A coordinate system, which includes coordinate axes, for example, x, y, z, and an origin

located at the point of reference. The coordinate system also includes a unit of measurement for

specifying distances along the axes.

(c) A clock, which includes an origin in time called 0t " and a unit of measurement for

specifying times and time intervals.

Note that the coordinate axes are mutually perpendicular to each other. Each axis has a positive

direction, which runs one way along the axis, and a negative direction, which runs in the opposite

direction. The coordinates of the origin are ( , , ) (0,0,0)x y z " .

For the example of the World War I pilot catching the bullet, we could choose the object

of reference to be the pilot and the point of reference to be the palm of his hand. The coordinate

system could have the positive x-axis pointing in the direction of the front the plane, the positive

y-axis pointing to the pilot’s left, and the positive z-axis pointing straight up. The origin in time

could be the moment the bullet is beside the pilot. In this reference frame, the pilot is at rest, the

bullet is approximately at rest, and the bullet is located at his side.

Review Question 1.2

Describe the point-like object model of a real object. Explain when we can use this model to

describe the motion of a real object.

1.3 A conceptual description of motion

How can we describe linear motion more precisely? We could start by using equations to

represent motion, but let’s first devise a visual representation that does not use mathematical

equations. Later we will construct a mathematical description.

Motion diagrams

We start our investigation with observational experiments involving a bowling ball

rolling on a smooth linoleum floor.

ALG

1.1.6-

1.1.8


Observational Experiment Table 1.2 Using dots to represent motion.

Observational experiment Analysis

(1) You gently push the bowling ball once and

let it roll on a smooth linoleum floor. Place

beanbags each second beside the bowling ball.

The beanbags are evenly spaced.

The diagram below represents with dots the locations of

the bags each second for the slow moving bowling ball.

(2) Experiment (1) is repeated except you

push the ball harder before you let it roll. The

beanbags are farther apart but are still evenly

spaced.

In this diagram the dots are still evenly spaced but now

the distance between them is bigger.

(3) You push the bowling ball and let it roll on

a carpeted floor instead of a linoleum floor.

The distance between the beanbags decreases

as the ball rolls.

In this dot diagram there is a decreasing distance between

the dots.

(4) You roll the ball on the linoleum floor and

gently and continually push on it with a board.

The beanbag separation spreads farther apart

as the pushed ball rolls.

The dots on the diagram spread farther apart.

Pattern

! The spacing of the dots allows us to visualize motion.

! When the object travels without speeding up or slowing down, the dots are evenly spaced.

! When it slows down, the dots get closer together.

! When it moves faster and faster, the dots separate farther.

In the experiments above, the beanbags were an approximate record of where the ball

was located as time passed (Fig. 1.2a). The diagrams with dots allowed us to visualize the motion

of the ball. We can represent motion in even more detail by adding arrows to each dot that

indicate which way the object is moving and how fast it is moving as it passes a particular

position. Such a diagram for the first experiment in Observational Experiment Table 1.2 is shown

in Fig. 1.2b. The dots in the diagram indicate the approximate position of the moving object at

equal time intervals (for example, each consecutive second). The arrows represent the direction of

motion and how fast the object is moving. They are called velocity arrows. The longer the arrow,

the faster is the motion. The tiny arrow above the letter v indicates that this characteristic of

motion has a direction as well as a magnitude—called a vector quantity. For the motion in Fig.

1.2b, the dots are evenly spaced, and the velocity arrows all have the same length and point in the


same direction. This means that the ball was moving equally fast in the same direction at each

point. Similar diagrams with velocity arrows for the other three experiments in Table 1.2 are

shown in Fig. 1.2c-e.

Figure 1.2 (a)Person dropping bags as ball rolls (b)(c) Motion diagrams for steady motion (d)(e) and

changing motion

Velocity change arrows

The ball in Experiment 4 described by the motion diagram in Fig. 1.2e was moving

increasingly fast while being pushed. The velocity arrows were getting increasingly longer. We

can represent this change in the velocity arrows by a velocity change arrow v#!

. The# here means

a change in whatever quantity follows the# , a change in v!

in this case. To illustrate how to

determine qualitatively the velocity change between any two adjacent points in the diagram, we

redraw the diagram shown in Fig. 1.2e again in Fig. 1.3a. For illustration purposes only, the v!

arrows for each position are now numbered consecutively. How for example do we determine

qualitatively the velocity change arrow as the ball moves from position 2 to position 3 in Fig.

1.3a? Place the second arrow 3v!

directly above the first arrow 2v!

as shown in Fig. 1.3b. The


object was moving faster at position 3 than at position 2. Arrow 2v!

can be turned into 3v!

by

placing the tail of a velocity change arrow v#!

at the head of 2v!

so that the head of v#!

makes

the combination of 2 23v v$ #! !

the same length as 3v!

. This is called graphical vector addition:

2 23 3v v v$ # "! ! !

, as shown in Fig. 1.3c. Note that if we move 2v!

to the other side of the equation,

then:

23 3 2v v v# " %! ! !

Thus, 23v#!

is the difference of the third velocity arrow and the second velocity arrow—the

change in velocity between position 2 and position 3. (Appendix 1 has a detailed description of

vectors.)

Figure 1.3 Determining velocity change arrows

Making a complete motion diagram

We now place in our diagrams the v#!

arrows above and between the dots where the

change occurred (see Fig. 1.4a). These new diagrams are called motion diagrams. A v#!

arrow

points in the same direction as the v!

arrows when the object is speeding up; the v#!

arrow

points opposite the v!

arrows when the object is slowing down. In most of our problems, the v!

arrows change by the same amount during each consecutive time interval. Then, the v#!

arrows

for each time interval are the same lengths. In such cases only one v#!

arrow is needed for the

entire motion diagram (see Fig. 1.4b).

Figure 1.4 A motion diagram that includes v#!

arrows


The following Reasoning Skill box (Constructing a Motion Diagram) summarizes the

procedure for constructing a motion diagram. Notice that in the experiment represented with this

diagram, the object is slowing down.

Reasoning Skill: Constructing a Motion Diagram

Tip! When drawing a motion diagram always specify the observer with respect to whom a

particular motion occurs.

Below is a conceptual exercise—a practice example that involves the skill that you have

just learned. When working on a conceptual exercise, first read it several times. Next, visualize

the situation and draw a sketch of what is happening. Then, construct a physics representation (in

this case, a motion diagram) for the process.

Conceptual Exercise 1.1 Driving in the city A car at a traffic light initially at rest speeds up

when the light turns green. The car reaches the speed limit in 4 seconds, continues at the speed

limit for 3 seconds, then slows down and stops in 2 seconds while approaching the next stoplight.

Finally, the car is at rest for 1 second until the light turns green. Meanwhile, a cyclist approaching

the first green light keeps moving without slowing down or speeding up. She reaches the second

stoplight just as it turns green. Draw a motion diagram for the car, and another for the bicycle as

seen by the observer on the ground. If you place one diagram below the other, it will be easier to

compare them.

Sketch and Translate Visualize the motion for the car and the bicycle as seen by the observer on

the ground. The car trip has four distinct parts:

(1) starting at rest and moving faster and faster for 4 seconds;

(2) moving at a constant rate for 3 seconds;

(3) slowing down to a stop for 2 seconds; and

(4) sitting at rest for 1 second.

""!"v "

!"v "

!"v

"#!"v

3. Draw a velocity change "#!"v arrow to indicate how the

""!"v arrows are changing between adjacent positions.

2. Point velocity "!"v arrows in the direction of motion

and draw their relative lengths to indicate

approximately how fast the object is moving.

1. Draw dots to represent the position of

the object at equal time intervals.


The bicycle moves at a constant rate with respect to the ground for the entire time and is at the

same initial location and same final location as the car—only moving when at these locations.

Simplify and Diagram We model the car and the bicycle as point-like objects (dots). In each

motion diagram, there will be 11 dots, one for each second of time (including one for time zero).

The last two dots for the car will be on top of each other since the car was at rest from time = 9 s

to time = 10 s. The dots for the bicycle are evenly spaced. The motion diagrams are shown in Fig.

1.5.

Figure 1.5 Car and bicycle trips between stop lights

Try It Yourself: Two bowling balls are rolling along a linoleum floor. One of them is moving

twice as fast as the other. At time zero, they are next to each other on the floor. Construct motion

diagrams for each ball’s motion during a time of four seconds, as seen by an observer on the

ground. Indicate on the diagrams the locations at which the balls were next to each other at the

same time. Indicate possible mistakes that a student can make answering the question above.

Answer: See Fig. 1.6. They are side-by-side only at time zero—the first dot for each ball. It looks

like they are side-by-side when at the 2-m position, but the slow ball is at the 2-m position at 2 s

and the faster ball is there at 1 s. Similar reasoning applies for the 4-m positions - they are not

side-by-side when at the 4-m position as they reach that point at different times.

Figure 1.6 Two bowling ball motion diagrams

Review Question 1.3

What information about a moving object can we extract from a motion diagram?


1.4 Quantities for describing motion

So far we learned how to represent motion qualitatively using a motion diagram.

However, to analyze situations more precisely, for example, to determine how far a car will travel

after the brakes are applied to avoid a person crossing a street, we need to investigate how to

describe motion quantitatively. As we learned in the Introduction Chapter, physicists describe

processes using physical quantities. In this section, we will construct some of the quantities we

need to describe linear motion. In most situations, we are interested in the positions of objects at

specific times; thus, the quantities we use in the remainder of this section involve time and

position.

Time and time interval

The names of many physical quantities have been borrowed from everyday language,

where they often have imprecise meanings. In physics, the name given to a physical quantity has

a very specific meaning, and it’s important to define these quantities carefully to avoid confusion

with the everyday terms.

Time is an example of such a quantity. People use the word ‘time’ to talk about different

things. One meaning is the reading on a clock when some event happened. Another is how long a

process takes. The latter use of the word ‘time’ is the difference in time between the start of the

process and the end of the process (it involves two times). To make the distinction clear,

physicists use two different words—time (a clock reading) and time interval (a difference in clock

readings).

Time and time interval Time (clock reading) t is the reading on a clock or some other time

measuring instrument. Time interval ( 2 1–t t ) or t# is the difference of two times. In the SI

system (metric units), the unit of time and of time interval is the second. Time and time interval

are both scalar quantities. They can be measured in other units such as minutes, hours, days, and

years.

Position, displacement, distance, and path length

Along with a precise definition for time and time interval, we need to precisely define what we

mean by position. There are four quantities related to this idea: position, displacement, distance,

and path length.

Position, displacement, distance, and path length The position of an object is its

location with respect to a particular coordinate system (usually indicated by x or y). The

displacement of an object, usually indicated by d!

, is an arrow (a vector) from an object’s initial

position to its final position. The magnitude (length) of the displacement vector is called distance.

The path length is how far the object moved as it traveled from its initial position to its final

position. Imagine laying a string along the path the object took. The length of string is the path

length.


Figure 1.7a shows a car’s initial position i

x at initial time i

t . It moves toward the origin

of the coordinate system at 0x " , stops, and then moves in the positive x direction to its final

position f

x . Notice that the initial position and the origin of a coordinate system are not

necessarily the same points! The path length is the distance from i

x to 0 plus the distance from 0

to f

x (Fig. 1.7b). The displacement for the whole trip is a vector that points from the starting

position at i

x to the final position at f

x (Fig. 1.7c). The distance for the trip is the magnitude of

the displacement (always a positive value); note that it does not equal the path length.

Tip! The subscripts 1, 2, 3, or i (initial) and f (final) for times and the corresponding positions

sometimes communicate a sequence of different and distinguishable stages in a process involving

the motion of an object. For us the process starts at the beginning of observations of motion and

stops at the end; however, the actual motion occurs before we start observing it and continues

after we stop.

Figure 1.7 Position, path length, displacement and distance for short car trip.

Scalar component of displacement for motion along one axis

One or more of these four quantities (position, displacement, distance, and path length)

can be used to describe linear motion quantitatively. To do so, we need to specify a reference

frame. For linear motion the simplest description is when one coordinate axis points either

parallel or anti-parallel (opposite in direction) to the object’s direction of motion. In this case, we

need only one coordinate axis to describe the object’s changing position. Imagine that at time i

t

(initial) a car is at position & 'ix t and that at time

ft (final) the car is at & 'f

x t (Fig. 1. 8). The


notation ( )x t indicates that the car’s position x is a function of the quantity time t – here time is

an independent variable and position is a dependent variable. The car’s position can change as

time progresses.

Figure 1.8 Using function x(t) to identify position

We will use the notation ( )x t when dealing with the functional relationship. However,

when we need to note a specific value of position x at a specific clock reading 1t , instead of

writing 1( )x t we will write 1x . The same applies to i

x , f

x , etc. The displacement vector points

from the initial position i

x to the final position f

x .

The quantity that we determine through the operation f i

x x% is called the x scalar

component of the displacement vector and is abbreviated x

d (usually we will drop the term

‘scalar’ and just call this the x-component of the displacement). Figure 1.9a shows that the initial

position of person A is A 3.0 mi

x " $ and the final position is A 5.0 mf

x " $ ; thus the x-

component of person A’s displacement is A A A– ( 5.0 m) ( 3.0 m) = +2.0 m

x f id x x" " $ % $ ;

the displacement is positive since the object moved in the positive x-direction. In Fig. 1.9b, person

B moved in the negative direction from initial position of +5.0 m to the final position of +3.0 m;

thus the x-component of displacement of person B is

B, B B – ( 3.0 m) – (+5.0 m) = – 2.0 mx f i

d x x" " $ and is negative. To learn more about the

vectors and their components, see Appendix 1.

Figure 1.9 Component of displacement (a) positive and (b) negative


Distance is always positive, as it equals the absolute value of the displacement 2 2x x% .

In the example above, the displacements for A and B are different but the distances are both

+2.0 m (always positive).

If we choose a different reference frame for the same trip, the values for f

x and i

x may

be different as would be the displacement and the distance. Suppose for example that we choose a

reference frame whose object of reference is person C, and person C is walking next to person A.

We choose the positive x-axis in this new reference frame as pointing in the same direction as A’s

motion relative to the sidewalk. In person C’s reference frame, person A is not moving. The

initial and final positions of person A are the same. Thus the displacement for person A and the

distance traveled relative to person C’s reference frame are both zero.

Tip! The quantities used to describe motion have no meaning unless the reference frame is

specified.

Using the position-related quantities to describe a two-dimensional trip

To understand better how vector quantities work, let’s use the quantities position,

displacement, distance, and path length to describe a trip from St. Louis to Indianapolis to

Louisville. (Fig.1.10a) We use Earth as the object of reference and place the origin of the

coordinate system at Indianapolis. The positive x-axis points towards St. Louis, and the positive

y-axis points toward Louisville. Notice that the orientations of the axes are different from the

traditional orientation used in mathematics. In physics we often choose axes by convenience and

not necessarily by tradition. As long as the x and y-axes are perpendicular to each other, and we

move in a counterclockwise direction from the x-axis to the y-axis, we have a Cartesian

coordinate system. Because the three cities form a right triangle (approximately), we could line

up the coordinate system with the three cities as shown. We use the scale of the map to determine

quantitatively the positions of the three cities.

Although all three cities have considerable size, these sizes are relatively small compared to

the distances between them. Using the map’s scale, we can estimate the distance between St.

Louis and Indianapolis is between 230 and 250 miles, depending on which parts of the cities we

travel between. Since the map does not allow us to be very precise in determining this distance,

we need to specify not just the distance but also the uncertainty in that distance. In this case, we

choose the uncertainty to be ±10 mi ; thus we record the St. Louis and Indianapolis distance as

240 10 mi( . Using this same idea, we measure the distance between Indianapolis to Louisville

as 110 10 mi( .

Since the origin of the coordinate system is at Indianapolis, the position coordinates of

Indianapolis are I 0 mix " , I 0 miy " ; the position coordinates of St. Louis are S 240 mix " $ ,

S 0 miy " ; and the position of Louisville is L 0 mix " , L 110 miy " $ .


The displacement vectors for the two parts of the trip are shown in Fig. 1.10b. The

displacement vector from St. Louis to Indianapolis ( S Id )

!) represents the first part of the trip. The

displacement vector from Indianapolis to Louisville ( I Ld )

!) represents the second part of the trip.

The magnitude of each displacement vector is the distance for each part of the trip. The x and y

components of the displacement vector for the first part of the trip S Id )

! are:

I, I S 0 – (240 mi) = 240 miS x

d x x) " % " %

S I, 0 0 0 miy

d ) " % " .

The magnitude of the first displacement vector is 240 mi. The x and y components of the

displacement vector for the second part of the trip I Ld )

!are:

I L, 0 0 0 mix

d ) " % "

I L, L I (110 mi) – 0 = +110 mi.y

d y y) " % " .

The magnitude of the second displacement vector is 110 mi. The total path length for the trip is

240 mi 110 mi 350 mi$ " .

Figure 1.10c shows the displacement vector for the whole trip. Thus, according to the rules

of vector addition (see appendix), it is the vector sum of the displacement vectors S Id )

! and

I Ld )

!. We can find the length of the total displacement vector using the Pythagorean theorem,

and the lengths of the two perpendicular displacement vectors:

2 2

L S (240mi) (110mi) 260 mi.d ) " $ "

Using a calculator, you would get 264.00757 miles. However, because we designated an

uncertainty of ±10 miles, there is no reason to write more than two significant digits. In fact, the

roads are not straight, so the uncertainty is possibly greater than this. Notice that the length of the

total displacement vector is less than the path length traveled. If you returned back to St. Louis

from Louisville along Route 64, the total path length for the whole trip would be about

240 mi 110 mi + 260 mi 610 mi$ " , while the total displacement would be zero (you’re back

where you started).


Figure 1.10 A two dimensional trip

Review Question 1.4

Which of the following (quantities initial position, final position, distance, displacement, and path

length) have values that depend on the choice of coordinate system?

1.5 Representing motion with

data tables and graphs

So far, we have learned how to

represent motion with motion diagrams,

and developed precise definitions for

position, time, and other related concepts.

In this section, we learn to represent

constant linear motion with data tables

and graphs.

Imagine walking across the front

Table 1.3 Time-position data for linear motion.

Clock reading (time) Position

t0 = 0.0 s 0x = 1.00 m

t1 = 1.0 s 1x = 2.42 m

t2 = 2.0 s 2x = 4.13 m

t3 = 3.0 s 3x = 5.52 m

t4 = 4.0 s 4x = 7.26 m

t5 = 5.0 s 5x = 8.41 m

t6 = 6.0 s 6x = 10.00 m

ALG 2.1.1;

2.1.2


of your classroom. In order to record your position as a function of clock reading, you drop a

beanbag on the floor each second (Fig. 1.11a). We choose the floor as the object of reference. The

origin of the coordinate system is 1.00 m from the first beanbag and the position axis points in the

direction you are walking. Each bag’s positions and the clock reading that it was dropped are

shown in Table 1.3. It is difficult to drop the bags exactly each second and the bags have to fall a

certain distance from your hand to the floor. The position data recorded in Table 1.3 includes an

uncertainty of ±0.01 m. The uncertainty in the time data is more complicated as it involves

human reaction time; however, we will designate an uncertainty of about ±0.1 s.

Do you see a pattern in the table’s data? One way to determine if there is a pattern is to

plot the data on a graph. For simplicity, we will not represent uncertainty in the data points (Fig.

1.11b).

Time (or clock reading) t is the independent variable; so the horizontal axis of the graph

will be the t axis. Position x is the dependent variable; so the vertical axis will be the x-axis. As

we noted earlier in the chapter, in physics we designate axes by convenience rather than tradition.

Most of the time, they are not called x and y.

A row in the data table turns into two points, one on each axis. Each point on the

horizontal axis represents a clock reading. Each point on the vertical axis represents the position

of a beanbag. Lines drawn through these points and perpendicular to the axes intersect at a single

location—a dot on the graph that simultaneously represents a time and the corresponding position

of the object. This dot is not a location in real space but rather a representation of the position of

the beanbag at a specific clock reading.

Is there a trend in the locations of the dots on the graph? We see that the position

increases as the time increases, which is what we expect for this person’s motion. We can draw a

smooth best-fit curve that passes as close as possible to the data points (Fig. 1.11c). It looks like a

straight line in this particular case – the position is linearly dependent on time. The graph we just

made is called a position-versus-time kinematics graph. The word kinematics refers to motion.

Kinematics graphs contain more precise information about an object’s motion than motion

diagrams.

Figure 1.11 Constructing a kinematics Position vs. Time graph


Relating a motion diagram and position-versus-time graph

To understand how graphs and motion diagrams are related, let’s consider the motion

represented by the data in Table 1.3 and in Fig. 1.11c. A modified motion diagram for the data in

Table 1.3 is shown in Fig. 1.12 (the dot times are shown and the v#!

arrows have been removed

for simplicity). The corresponding position-versus-time graph is also shown in Fig. 1.12. Each

position dot on the motion diagram corresponds to a point on the position axis for a particular

clock reading. The graph line combines the information about the position of an object and the

clock reading when this position occurred. Note for example, the 4.0 st " dot on the motion

diagram at position 7.26 mx " is at 7.26 m on the position axis. The corresponding dot on the

graph is at the intersection of the vertical line passing through 4.0 s, and the horizontal line

passing though 7.26 m.

Tip! The quantity that appears on the vertical

axis of the graph can represent the position of

an object whose actual position is changing

along a horizontal axis, along a vertical axis, or

along an inclined axis. The line shown on a

kinematics graph is not a picture of the actual

path that the object takes as it moves.

Figure 1.12

The role of a reference frame

Always keep in mind that representations of motion (motion diagrams, tables, kinematics

graphs, equations, etc.) depend on the reference frame chosen. Let’s look at the representations

of the motion of a cyclist using two different reference frames.


Conceptual Exercise 1.2 Effect of reference frame on

motion description Two observers each use different

reference frames to record the changing position of a

bicycle rider. (Fig. 1.13a) Both reference frames use the

earth as the object of reference but the origins of the

coordinate systems and the directions of the x-axes are

different. The data for the cyclist’s trip is presented in

Table 1.4 for observer 1 and in Table 1.5 for observer 2—

the data depend on the reference frame used. Sketch a

motion diagram and a position-versus-time graph for the

motion when using each reference frame.

Figure 1.13 Describing cyclist in different

reference frames

Sketch and Translate Table 1.4 indicates that the observer in reference frame 1 sees the cyclist at

time 0 0.0 st " at position 0 40.0 mx " and at 4 4.0 st " at position 4 0.0 mx " . Thus, the

cyclist is moving in the negative direction relative to the coordinate axis in reference frame 1.

Meanwhile, the observer in reference frame 2 sees the cyclist at time 0 0.0 st " at position

0 0.0 mx " and at time 4 4.0 st " at position 4 40.0 mx " . Thus, the cyclist is moving in the

positive direction relative to reference frame 2.

Simplify and Diagram Since the size of the cyclist is small compared to the distance he is

traveling, we can represent him as a point-like object. A motion diagram for the cyclist is also

shown in Fig. 1.13a. Using the data in the tables, we plot position-versus-time kinematics graphs

for each observer—see Fig. 1.13b and c. Although the graph in Fig. 1.13b looks very different

from the one in Fig. 1.13c, they represent the same motion. The graphs look different because the

reference frames are different.

Table 1.4 Time–position data for cyclist when using

reference frame 1.


to = 0.0 s 0x = 40.0 m

t1 = 1.0 s 1x = 30.0 m

t2 = 2.0 s 2x = 20.0 m

t3 = 3.0 s 3x = 10.0 m

t4 = 4.0 s 4x = 0.0 m

Table 1.5 Time–position data for cyclist when using

reference frame 2.


to = 0.0 s 0x = 0.0 m

t1 = 1.0 s 1x = 10.0 m

t2 = 2.0 s 2x = 20.0 m

t3 = 3.0 s 3x = 30.0 m

t4 = 4.0 s 4x = 40.0 m


Figure 1.13

Notice that in reference frame 1, the cyclist is initially at 0 40 mx " and is moving left

toward the origin—toward smaller values of x, shown as a downward-sloped line in Fig. 1.13b.

However in reference frame 2, the positive direction of the axis points left and its origin and the

cyclist’s initial position 0 0x " are at the same place. Using this axis, the cyclist is moving to the

left, away from the origin and toward larger values of x. Both graphical descriptions are correct

relative to the reference frames used.

Try It Yourself: A third observer recorded the following values for the time and position of the

same cyclist. Describe the reference frame of this observer.

Answer: The point of reference could be another cyclist moving in the opposite positive direction

from the direction in which the first cyclist is traveling and each covering the same distance

relative to the ground during the same time interval (the point of reference cyclist moves 40 m to

the right relative to the ground while the cyclist being analyzed moves 40 m to the left relative to

the ground).

Review Question 1.5

A position-versus-time graph representing a moving object is shown

in Fig. 1.14. What are the positions of the object at clock readings 2.0

s and 5.0 s?

Figure 1.14


to = 0.0 s 0x = 0.0 m

t1 = 1.0 s 1x = -20.0 m

t2 = 2.0 s 2x = -40.0 m

t3 = 3.0 s 3x = -60.0 m

t4 = 4.0 s 4x = -80.0 m


1.6 Constant velocity motion

In the last section, we used a table of data to construct a graphical representation of motion. Here

we will connect graphs to mathematical representations. Let’s investigate the motion of two

motorized toy cars starting next to each other at the same time and racing toward a finish line. We

collect data marking the locations of the cars every second and analyze the data by constructing a

graph of the positions of the cars as a function of time.

When the cars are released, car B moves faster than car A and reaches the finish line first

(Fig. 1.15). The data that we collect is shown in

Observational Experiment Table 1.6. The reference

frame has Earth as the object of reference. The origin

of the coordinate system (the point of reference) is

1.0 m to the left of the initial position of the cars.

The positive x-direction points in the direction of the

cars’ motions. Now let’s use the provided data to find

some pattern. It is common in science to analyze data

collected by others with the goal of finding a pattern,

explaining, etc. Figure 1.15

Observational Experiment Table 1.6 Graphing the motion of cars

Observational experiment Analysis

Data for car A Data for car B

To analyze the data with the goal of

finding a pattern, we use a graphical

analysis.

0 0.0 st "

1 1.0 st "

2 2.0 st "

3 3.0 st "

4 4.0 st "

5 5.0 st "

0 1.0 mx "

1 1.4 mx "

1 1.9 mx "

3 2.5 mx "

4 2.9 mx "

5 3.5 mx "

0 0.0 st "

1 1.0 st "

2 2.0 st "

3 3.0 st "

4 4.0 st "

5 5.0 st "

0 1.0 mx "

1 1.9 mx "

2 3.0 mx "

3 3.9 mx "

4 5.0 mx "

5 6.0 mx "

Pattern

It looks like a straight line is the simplest reasonable choice for the best-fit curve in both cases (the data

points do not have to be exactly on the line). The slope of the line representing the motion of car B is

greater than the slope of the line representing the motion of car A. What is the physical meaning of this

slope?

When we write a functional relationship in mathematics, usually x is our independent

variable and y is the dependent variable; a function ( ) ( )y x f x" is an operation that one needs to


do to x as an input, to have y as the output. For a straight line, the function y(x) is ( )y x kx b" $ ,

where k is the slope and b is the y intercept—the value of the y when 0x " .

In the case of the cars, the independent variable is time t and the dependent variable is

position x. The equation of a straight line becomes & 'x t kt b" $ , where b is the x-intercept of

the line, and k is the slope of the line. The x-intercept is the x -position when 0t " , also called

the initial position of the car 0x . Both cars started at the same location: A0 B0 1.0 mx x" " .

To find the slope k of a straight line, we can choose any two points on the line and divide

the change in the vertical quantity ( x# in this case) by the change in the horizontal quantity ( t#

in this case): 2 1

2 1

–

–

x x xk

t t t

#" "

#. For example, for car A the slope of the line is

A

3.5 m – 1.0 m0.5 m s

5.0 s – 0.0 sk " " $ . The slope of the line for car B is

B

6.0 m – 1.0 m1.0 m s

5.0 s – 0.0 sk " " $ . Now we have all the information we need to write

mathematical equations that describe the motion of each of the two cars:

Car A: & ' & 'A 0.5 m s 1.0 mx t" $ $

Car B: & ' & 'B 1.0 m s 1.0 mx t" $ $

What is the meaning of the slopes of the lines for the motion of the two cars? Notice that

the units of the slope are meters per second. The slope indicates how the object’s position

changes with respect to time. That sounds like speed. We have learned that some of the terms

used in physics are borrowed from everyday speech, but that they have very specific meanings in

physics. In everyday speech we know that speed is usually measured in miles per hour, which

sounds similar to meters per second. However, the slope of the line contains more information

than just how fast the car is going. It also tells us the direction of motion relative to the coordinate

axis.

Consider the motions represented graphically in Fig. 1.16a. The slope of the position-

versus-time graph representing motion of car C is +20 m/s, but the slope of car D’s position-

versus-time graph is –20 m/s. What is the significance of the minus sign? Car C is moving in the

positive direction but car D is moving in the negative direction (in Fig. 1.16a this motion is

represented by the increasing position for car C with time and decreasing position for car D—also

shown in the motion diagrams in Fig. 1.16b). The magnitudes of the slopes of their position-

versus-time graphs are the same but the signs are different. Thus, the slope contains information

about how fast the car is traveling (which is what we call speed) and in what direction it is

traveling. A combination of this information is called velocity, and is what the slope of a position-

versus-time kinematics graph represents. You are already familiar with the term “velocity arrow”

used on motion diagrams. Now you have a formal definition for this quantity.


Figure 1.16 Sign of slope indicates direction of motion

Velocity and speed for constant velocity linear motion For constant velocity linear motion,

the component of velocity x

v along the axis of motion can be found as the slope of the

position-versus-time graph or the ratio of the component of the displacement of an object

2 1x x% during any time interval 2 1t t% :

2 1

2 1

x

x x xv

t t t

% #" "

% # (1.1)

Examples of units of velocity are: m/s, mi/h, and km/h. Speed is the magnitude of the

velocity and is always a positive number.

Tip! In Eq (1.1) you can use any change in position divided by the time interval during which that

change occurred and obtain the same number—as long as the position-versus-time graph is a

straight line, or the object is moving at constant velocity. Later in the chapter, you will learn how

to modify this equation to find the velocity of an object if the velocity is not constant.

Note that velocity is a vector quantity. If we write the definition in a vector form for the

motion at constant velocity, it is d

vt

"#

!!

. Here we divide a vector by a scalar. According to the

operations with vectors (see more in appendix), the resultant vector has the same direction as the

displacement vector – thus the direction for the velocity vector shows the direction of motion

(same as the direction of the displacement vector) and the magnitude shows the speed. But as it is

difficult to operate mathematically with vectors, we will work with components.

Also note that the velocity and speed of an object depend on the observer. Think of a stop

sign on the road. A bus driver approaching the stop sign will say that the velocity and the speed of

the stop sign are non-zero relative to the driver while a person standing at the stop sign would say

its velocity and speed are zero.


Equation of motion for constant velocity motion

We can rearrange Eq. (1.1) into a form that allows us to determine the position of an

object at time 2t knowing only its position at time 1t and the x-component of its velocity:

& '2 1 2 1xx x v t t" $ % . If we apply this equation for time zero ( 0 0t " ) when the initial position is

0x , then the position x at any later position time t can be written as follows.

Position equation for constant velocity linear motion

0 xx x v t" $ (1.2)

where x is the function ( )x t , position 0x is the position of the object at time 0 0t " with

respect to a particular reference frame, and the (constant) x -component of the velocity of

the object x

v is the slope of the position-versus-time kinematics graph.

Below you see a new type of task—a quantitative exercise. The purpose of quantitative

exercises is to help you apply the mathematical representations you have just learned to analyze a

different physical situation. They have different steps than conceptual exercises. These steps

focus on constructing mathematical representations of the situation, answering the mathematical

question being asked, and evaluating the result.

Quantitative Exercise 1.3 A cyclist In Conceptual Exercise 1.2, you constructed graphs (Figs.

1.13b-c) for the motion of a cyclist using two different reference frames. Now construct

mathematical representations (equations) for the cyclist’s motion for each of the two graphs. Do

the equations indicate the same position for the cyclist at time 6.0 st " ?

Represent Mathematically The cyclist moves at constant velocity; thus the general mathematical

description of her motion is 0 xx x v t" $ , where

2 1

2 1

x

x xv

t t

%"

%.

Solve and Evaluate Using reference frame 1 and the graph in Fig. 1.13b, we see that the biker’s

initial position is 0 40 mx " $ . The velocity along the x-axis (the slope of the graph line) is

0 m - 40 m10 m s

4 s - 0 sx

v " " %

The minus sign in front of the cyclist’s speed indicates that the velocity points in the negative x-

direction (toward the left) relative to that axis. The motion of the bike with respect to reference

frame 1 is described by the equation:

& '0 40 m 10 m sx

x x v t t" $ " $ %

Using reference frame 2 and the graph in Fig. 1.13c, we see that the cyclist’s initial

position is 0 0 mx " . The x-component of the velocity along the axis of motion is


40 m 0 m10m s

4 s 0 sx

v%

" " $%

The positive sign in front of the cyclist’s velocity indicates that it points in the positive x -

direction (toward the left). The motion of the bike relative to reference frame 2 is described by the

equation:

& '0 0 m 10 m sx

x x v t t" $ " $

Let’s see what these two equations say about the position of the cyclist at time 1 6 t s" .

With respect to reference frame 1:

& '& '1 40 m 10 m s 6 s 20 mx " $ % " %

With respect to reference frame 2:

& '& '1 0 m 10 m s 6 s 60 mx " $ $ " $

How can the position of the cyclist be both 20 m% and 60 m$ ? Remember that the

description of motion of an object depends on the reference frame. If you put a dot on coordinate

axis 1 at the 20 m% position and a dot on coordinate axis 2 at to +60 m position, you find that

the dots are in fact at the same location, even though that location is indicated by a different

number in each reference frame (Fig. 1.17). Both descriptions of the motion are correct and

consistent; but each one is meaningful only with respect to the reference frame that was used.

Figure 1.17 Position of cyclist at t = 6 s in two different reference frames

Try it yourself: Use the data for the motion of the cyclist as seen by the third observer in the Try It

Yourself part of Example 1.2 to write the equation of motion. Why is the magnitude of the

cyclist’s velocity different compared to 10 m/s in the example above?

Answer: (0 m) ( 20 m/s)x t" $ % . The observer is moving with respect to the earth at the same

speed in the direction opposite to the cyclist.

Let’s try an example that is a little more complex. We will use the same steps to solve the

rest of the examples.

Example 1.4 You chase sister Your young sister is running at 2.0 m/s toward a mud puddle that

is 6.0 m in front of her. You are 10.0-m behind her running at 5.0 m/s to catch her before she

enters the mud. Will she need a bath?


Sketch and Translate We start our analysis by

drawing a sketch to visualize what is

happening. Next, we choose a reference frame

with the earth as the object of reference. The

point of reference (the origin) will be your

initial position and the positive direction will be

toward the right, in the direction that you both

run (Fig. 1.18a). Figure 1.18(a)

Why choose this reference frame rather than some other one? If we use this reference

frame, the positions of you and your sister will always be positive, as will the components of your

velocities. Also, your initial position will be zero. We could have chosen any reference frame, but

often the description of the motion of object(s) will be simplest in one particular reference frame.

With this choice of reference frame, we can mathematically describe the positions and

velocities of you and your sister at the beginning of the process. The initial time is zero at the

moment that you are at the origin. Note that you were both running before time zero; this just

happened to be the time when we started analyzing the process. We want to know the time when

you and your sister are at the same position. This will be the position where you catch your sister.

Simplify and Diagram We assume that you and your sister

are point objects. To sketch graphs of the motions, find

the sister’s position at 1 second by multiplying her speed

by 1 second and adding it to her initial position. Do it for 2

seconds and for 3 seconds as well. Plot these values on a

graph for the corresponding clock readings (1, 2, 3 s, etc.)

and draw a line that extends through these points. Repeat

this for yourself (Fig. 1.18b.) Figure 1.18(b) Position vs. Time graph for

you and your sister’s motion

Represent Mathematically Use Eq. (1.2) to construct mathematical representations for your and

your sister’s motion. The form of the equation is the same for both ( 0 xx x v t" $ ); however, the

values for the initial positions and the components of the velocities along the axis are different.

Sister: & ' & 'S 10.0 m 2.0 m sx t" $

You: & ' & 'Y 0.0 m 5.0 m sx t" $

From the graphs we see that the distance between you and your sister is shrinking with time; do

the equations tell the same story? For example, at time 2.0 st " , your sister is at position

& ' & '& 'S (2 s) 10.0 m 2.0 m s 2.0 s 14.0 mx " $ " and you are at

& ' & '& 'Y (2 s) 0.0 m 5.0 m s 2.0 s 10.0 mx " $ " . You are catching up to your sister.


Solve and Evaluate The time t at which the two of you are at the same position can be found by

setting S Y( ) ( )x t x t" :

& ' & ' & ' & '10.0 m 2.0 m s 0.0 m 5.0 m st t$ " $

Rearrange the above to determine the time t when you are both at the same position:

& ' & ' & ' & '2.0 m s 5.0 m s 0.0 m 10.0 mt t% " %

& ' & '3.0 m s 10.0 mt% " %

3.33333333 st "

The 3.33333333 s number produced by our calculator has many more significant digits

than the givens. Should we round it to have the same number of significant digits as the givens?

The rule of thumb is that if it is the final result, you need to round this number to 3.3, as the

answer cannot be more precise than the given information. However, if the number is instead a

result of an intermediate calculation, do not round it; use it as is to calculate the next value, and

then round the final result.

Sister: & ' & 'S ( ) 10.0 m 2.0 m s (3.33333333 s) 16.7 mx t " $ "

You: & ' & 'Y ( ) 0.0 m 5.0 m s (3.33333333 s) 16.7 mx t " $ "

Note that if you used the rounded number 3.3 s, you would get 16.5 m for your sister and 16.6 m

for you. These would be slightly less than the result calculated above. However, for our purposes

it does not matter, as the goal of this example was to decide if you could catch your sister before

she reaches the puddle. Since you caught her at a position of about 16.7 m, with the uncertainty of

about 0.1 m, this position is greater than 16.0 m distance to the puddle. Therefore your sister

reaches the puddle before you. This answer seems consistent with the graphical representation of

their motion shown in Fig. 1.18b.

Try It Yourself: Describe the problem situation using a reference frame with the sister (not the

earth) as the object and point of reference, and the positive direction pointing toward the puddle.

Answer: With respect to this reference frame, the sister is at position 0 and at rest; you are

initially at –10.0 m and moving toward your sister with velocity 3.0 m s$ ; and the mud puddle

is initially at +6.0 m and moving toward your sister with velocity 2.0 m s% (Fig. 1.18c).

Figure 1.18(c)


Unit Conversions

Suppose that in the last example, you were told the distance to the puddle in meters, and

the speeds of the mother and daughter in miles per hour. To analyze the situation, you would first

need to convert the speeds into meters per second. When physical quantities are substituted into

equations to determine other quantities, the units of the quantities must be consistent—for

example, all SI metric units. How is a quantity expressed in one set of units converted into

another set? Suppose your speed is reported as 11 mph. To find your speed in m/s, use the

conversions on the inside of the front cover of this book to convert miles into m (meters) and

hours into s (seconds):

1609 m11 miles

11 miles 1 mile 4.9 m/s3600 s1 hour

1 hour 1 hour

" "

What we have actually done is multiplied the numerator and denominator by the equivalent of the

number 1 (since 1609 m = 1 mile, then 1609 m/1 mile = 1; and since 3600 s = 1 hour, then 3600

s/1 hour = 1). Notice that in these fractions, the unit that we are converting from (miles or hours)

appears in the denominator, and the unit we are converting to (m or s) appears in the numerator.

Graphing velocity

So far, we constructed only a single type of kinematics graph (the position-versus-time

graph). We could also construct a graph of an object’s velocity as a function of time. What would

this graph look like for you and your sister in the last example (using the earth as the object of

reference)? You are moving at a constant velocity whose x component is 5 m/sx

v " $ , and your

sister at +2.0 m/s. Put clock readings on the horizontal axis

and the x-component of your and your sister’s velocities on

the vertical axis; then plot points for these velocities at each

time (see Fig. 1.19a). The best-fit curve for each person is a

horizontal line, which makes sense since the velocity is not

changing. For you the equation of the best-fit line is

& 'Y 5.0 m sx

v t " $ and for your sister & 'S 2.0 m sx

v t " $ ,

where ( )x

v t represents the x component of velocity as a

function of time. Figure 1.19(a) V vs t graph lines

with Earth as object of reference

Do you think these graph lines will be different if instead we choose your sister as the

object of reference? Her velocity with respect to herself is zero; so the best-fit curve will again be

a horizontal line, but it will be at a value of 0.0 m s instead of +2.0 m/s. With her as the object

of reference, your velocity is +3.0 m/s (see Fig. 1.19b) and the mud velocity is –2.0 m/s. The

ALG

2.1.3


minus sign indicates that from your sister’s point of view, the

mud is moving in the negative direction toward her at speed

2.0 m/s.

Figure 1.19(b) V vs t graph lines

with sister as object of reference

Tip! Notice that a horizontal line on a position-vs-time graph means that the object is at rest (the

position is constant with time). The same horizontal line on the velocity-vs-time graph means that

the object is moving at constant velocity (it does not change with time).

Finding displacement from a velocity graph

We just learned to construct a new type of kinematics graph, a velocity-versus-time

graph. Can we get anything more out of such graphs besides being able to represent velocity

graphically? Earlier you learned that for constant velocity linear motion, the position of an object

changes with time according to 0 xx x v t" $ . This equation connects an object’s position x at

time t to its initial position x0 at time zero and its velocity vx. Rearranging this equation a bit we

get 0 xx x v t% " . The left side is the displacement of the object from time zero to time t . Now

look at the right side; vx is the vertical height of the velocity-versus-time graph line and t is the

horizontal width from time zero to time t (see Fig. 1.20). This means we can interpret the right

side as the shaded area between the velocity-

versus-time graph line and the time axis. In the

previous equation, this area (the right side of the

equation) equals the displacement of the object

from time zero to time t on the left side of the

equation. Here the displacement is a positive

number as we chose an example with positive

velocity. If we wish to extend our reasoning to all

cases, we need to take the direction into account.

Figure 1.20 Using x vs. t to find displacement

Displacement is the area between a velocity-versus-time graph line and the time axis For

motion with constant velocity, the displacement 2 1x x% of an object during a time interval from

t1 to t2 is the area between a velocity-versus-time graph line and the time axis between those two

clock readings. The displacement is positive for areas above the time axis and negative for areas

below the time axis.


Quantitative Exercise 1.5 Displacement of you and your sister Use the velocity vs. time

graphs shown in Fig. 1.19a for you and your sister (see Example 1.4) to find your

displacements with respect to the earth for the time interval from 0 to 3.0 s. The velocity

components relative to earth are +2.0 m/s for your sister and +5.0 m/s for you.

Represent Mathematically For constant velocity motion, the object’s displacement 0x x% is the

area of a rectangle whose vertical side equals the object’s velocity x

v and the horizontal side

equals the time interval 0 0t t t% " % during which the motion occurred (Fig. 1.20).

Solve and Evaluate The displacement of your sister heading toward the mud puddle during the

3.0-s time interval is the area of a rectangle whose vertical side is the velocity and whose

horizontal side is the time interval:

S 0 S( ) ( 2.0 m/s)(3.0 s) 6.0 md x x" % " $ " $ .

Your displacement during that same 3.0-s time interval is the product of your constant velocity

and the time interval:

Y 0 Y( ) ( 5.0 m/s)(3.0 s) 15.0 md x x" % " $ " $ .

Try It Yourself: Determine the magnitudes of the displacements of you and your sister from time

zero to time 2.0 s and your positions at that time. Your initial position is zero and your sister’s

is10 m.

Answer: The sister’s values are S 4.0 md " and S 14.0 mx " ; and your values are Y 10.0 md "

and Y 10.0 mx " . She is 4.0 m ahead of you at that time.

The first person to use this graphical method to find displacement was Nicolas Oresme at

the University of Paris early in the fourteenth century (1320-1382). He was a chaplain to King

Charles V. Historians of science are interested in his work in kinematics but also because before

Copernicus he proposed a Sun centered model for the universe (as opposed to an Earth centered

model).

Review Question 1.6

The position-versus-time graph for a moving object is a straight line with a slope equal to –15.0

m/s. Explain what this means about the motion of this object.

1.7 Motion at constant acceleration

In the previous section, we learned to describe linear motion at constant velocity. In that kind of

motion, the speed and the direction of travel do not change; therefore it is a very restrictive kind

of motion. Since the goal of this chapter is to construct models of motion, we should try to make

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2.1.4


models that are more widely applicable. The goal of this section is to construct a model of linear

motion at changing velocity.

Instantaneous velocity

In the last section, the function vx (t) was a

horizontal line on the velocity-versus-time graph because

the velocity was constant. How would the graph look if the

velocity were changing? One example of such a graph is

shown in Fig. 1.21. The velocity of the object is positive all

the time; however its magnitude changes with time. A point

on the curve indicates the velocity of the object shown on

the vertical axis at a particular time (clock reading) shown

on the horizontal axis. Figure 1.21

The velocity of an object at a particular time is called the instantaneous velocity of the

object. Figure 1.21 shows velocity-versus-time graph for the motion when the instantaneous

velocity of the object is continually changing and is different just before and just after a particular

clock reading. A car’s speedometer measures a car’s instantaneous speed (the magnitude of its

velocity) relative to the road surface. But, the speedometer does not indicate the direction of the

car’s motion; so it does not measure the car’s instantaneous velocity. However, some cars are

equipped with a compass. The compass reading together with the speedometer reading does

provide enough information to specify the car’s instantaneous velocity—its speed and direction.

When an object’s velocity is changing, we cannot use Eq. (1.1) to determine its

instantaneous velocity, as the ratio 2 1

2 1

x

x x xv

t t t

% #" "

% #

is not the same for different time intervals

the way it was when the object was moving at constant velocity. However, we can still use the

same equation to determine a new quantity called the average velocity, which is the ratio of the

change in position and the time interval during which this change occurred. For motion at

constant velocity, the instantaneous and average velocity equal each other; for the motion at

changing velocity they are not.

When an object moves with changing velocity, its velocity can change quickly or slowly.

To characterize the rate at which the velocity of an object is changing, we need a new physical

quantity.

Acceleration

We’ll analyze motion with changing velocity in a way that is similar to how we

analyzed constant velocity motion. We start by looking for the simplest type of linear motion with

changing velocity. This occurs when the velocity of the object increases or decreases at a

constant rate, that is, by the same amount during the same time interval. Imagine that a cyclist is

speeding up so that his velocity is increasing at a constant rate with respect to an observer on the

ground. Graphically, this process is represented as the velocity-versus-time graph shown in Fig.

1.22a. The velocity-versus-time graph is a straight line that is not horizontal. Now imagine that a

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2.1.5


car next to the cyclist is also speeding up, but at a greater rate than the biker. Its velocity-versus-

time graph is shown in Fig. 1.22b. The larger slope indicates that the car’s velocity increases at a

faster rate. This physical quantity that characterizes the change in velocity during a particular

time interval, equal to the slope of the velocity-versus-time graph, is called acceleration a!

.

Similar to velocity, the acceleration of an object moving in a straight line is defined as:

2 1

2 1

–

–

x x x

x

v v va

t t t

#" "

#

The acceleration can be either positive or negative. In the examples above, if the cyclist

or the car had been slowing down, their velocity-versus-time graph would instead have a negative

slope, which corresponds to a decreasing speed and a negative acceleration, as 2xv is smaller than

1xv (Fig. 1.22c). However, an object can have a negative acceleration and speed up! This happens

when the object is moving in the negative direction, has a negative component of velocity, but its

speed in the negative direction is increasing in magnitude (see Fig. 1.22d).

Figure 1.22

Because velocity is a vector quantity and the acceleration shows how quickly the velocity

changes as time progresses, acceleration is also a vector quantity. We can define acceleration in a

more general way. The average acceleration of an object during a time interval is:

2 1

2 1

–

–

v v va

t t t

#" "

#

! ! !!

You can see that to determine the acceleration, we need to determine the velocity change

vector v#!

. Remember that the capital delta # means “final value minus initial value”, or in this

case, the velocity of the object at a later time 2t minus the velocity of the object at an earlier time

1t ( 2 1v v v# " %! ! !

). This equation involves the subtraction of vectors. However, it is possible to


think about the same equation as addition by moving the 1v!

and v#!

to the same side of the

equation ( 1 2v v v$ # "! ! !

). Note that v#!

is the vector that we add to 1v!

to get 2v!

(Fig. 1.23). We

did this when making motion diagrams in Section 1.3. The acceleration vector /a v t" # #! !

is in

the same direction as the velocity change vector v#!

, as the time interval t# is a scalar quantity.

Figure 1.23 The change in velocity

Acceleration An object’s average acceleration during a time interval t# is the change in its

velocity v#!

divided by that time interval:

2 1

2 1

–

–

v v va

t t t

#" "

#

! ! !!

(1.3)

If t# is very small, then the acceleration given by this equation is the instantaneous

acceleration of the object. For one-dimensional motion, the component of the acceleration

along a particular axis (for example, for the x axis) is:

2 1

2 1

–

–

x x x

x

v v va

t t t

#" "

# . (1.4)

The unit of acceleration is (m/s)/s = m/s2.

Note that if an object has an acceleration of 2+6 m/s , it means that its velocity changes by

+6 m/s in 1 s, or by +12 m/s in 2 s [2(+12 m/s)/(2 s) = + 6 m/s ].

Tip! If it is difficult for you to think about velocity and acceleration in abstract terms, try

calculating the acceleration for simple integer velocities, as done above.

Note that is possible for an object to have a zero velocity and a non-zero acceleration –

for example when an object starts moving from rest – zero initial velocity and speeds up as it time


progresses. An object can also have a non-zero velocity and zero acceleration – for example an

object moving at constant velocity. Note also that the acceleration of an object depends on the

observer. For example, a spending up car is accelerating for an observer on the ground but is not

for the driver of that car.

Determining the velocity change from the acceleration

If at 0 0t " for linear motion, the x-component of the velocity of some object is 0xv

and its acceleration x

a is constant, then its velocity x

v at a later time t can be determined by

substituting these quantities into Eq. (1.4):

0–

– 0

x x

x

v va

t"

Rearranging, we get an expression for the changing velocity of the object as a function of

time:

0x x xv v a t" $ . (1.5)

For one-dimensional motion, the directions of the vector components x

a , x

v , and 0xv are

indicated by their signs relative to the axis of motion—positive if in the positive x- direction and

negative if in the negative x-direction.

Example 1.6 Car’s changing velocity Suppose that in a 2.0-s time interval, the velocity of a car

changes from –10 m/s to –14 m/s. Describe the motion of the car and determine its acceleration.

Sketch and Translate The process is represented in Fig. 1.24a. The car is moving in the negative

direction with respect to the chosen reference frame. The components of the car’s velocity along

the axis of motion are negative: 0 –10 m/sx

v " at time 0 0.0 st " and –14 m/sx

v " at

2.0 st " . The speed of the car (the magnitude of its velocity) increases. It is moving faster in the

negative direction.

Simplify and Diagram A motion diagram for the car is shown in Fig. 1.24b. Note that v#!

points

in the negative x-direction.

Figure 1.24 Car speeds up in negative direction

Represent Mathematically We apply Eq. (1.4) to determine the acceleration:

0

0

–

–

x x

x

v va

t t"

Solve and Evaluate Substituting the given velocities and times, we get:


& ' & ' 2

x

14m s 10m s2.0m s

2.0 s – 0.0 sa

% % %" " %

The car’s x-component of velocity at time zero was –10 m/s. It’s velocity was changing by –2 m/s

each second. So 1 second later, its velocity was:

1s 0 = + = (–10 m/s) + (–2 m/s) = –12 m/sx x x

v v v# .

During the second 1-second time interval, the velocity changed by another –2 m/s and was now

(–12 m/s) + (–2 m/s) = (–14 m/s) . In this example, the car was speeding up by 2 m/s each

second in the negative direction.

Try It Yourself: What is the car’s acceleration if its velocity changes from +14 m/s to +10 m/s

in 2.0 seconds?

Answer: 2(–2.0 m/s)/s –2.0 m/s

xa " " . The car’s velocity in the positive x-direction is

decreasing 2.0 m/s each second.

Tip! When an object is speeding up, the acceleration is in the same direction as the velocity, and

the velocity and acceleration have the same sign. When an object slows down, the acceleration is

in the opposite direction relative to the velocity; they have opposite signs. For example, when an

object is moving in the negative direction and slowing down, its acceleration is in the positive

direction and has a positive sign.

Displacement of an object moving at constant acceleration

For motion at constant velocity, we know that the area between the velocity-versus-

time graph line and the time axis between two clock readings equals the magnitude of the

object’s displacement during that time interval. Now we have a different situation—the

velocity is changing. Is the magnitude of an object’s displacement during a time interval still the

area under its velocity-versus-time graph line between those two times?

Consider the displacement of the object during the small shaded time interval shown in

Fig. 1.25a. The velocity is almost constant during that time interval. Note that x

xv

t

#"#

or

• ( ) = x

xv t t x

t

## " # #

#. Thus, the small displacement x# during that small time interval t# is

the small shaded area x

v t#! under that curve (the height times the width of the small rectangle).

We can repeat the same procedure for many successive small time intervals (Fig. 1.25b), building

up the area under the curve as a sum of areas of small rectangles of different heights. This is what

integration is, an important operation in calculus. The total area, shown in Fig. 1.25c, is the total

displacement of the object during the time interval between the initial time 0t and the final time

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2.1.6


t . In general, the area will give you the displacement. A negative area (below the time axis)

corresponds to a negative displacement

Displacement from a v-vs-t graph The displacement 2 1x x% of an object during a time

interval 2 1t t% is the area between the velocity-versus-time curve and the time axis between

those two clock readings. The displacement is negative for areas below the time axis and

positive for areas above.

Figure 1.25

Equation of motion—position as a function of time

We can use the above idea to find an equation for the position x of an object at different

times t . Consider the displacement of an object whose velocity-versus-time graph line is shown

in Fig. 1.26a. Its displacement 0–x x during the time interval from 0 0t " to some arbitrary time

t is the shaded area under the trapezoid during that time interval.

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2.1.7,

2.1.8


Figure 1.26

We can find the area between this curve and the time axis by breaking the area into a

triangle on top and a rectangular below (Fig. 1.26b). The area of a triangle is 1

base height2* * .

The base of the triangle is 0t % and the height is x 0xv v% . So, the area of the triangle is

& '& ' & '& ' 2

triangle x 0

1 1 10

2 2 2x x x

A t v v t a t a t" % % " " ,

where we substituted x 0x xv v a t% " from Eq. (1.5) into the above. Note that vx is the value of the

x- component of the velocity at time t .

The area of the rectangle equals its width times its height:

& 'rectangle 0 0x

A v t" %

The total area under the curve (the displacement 0x x% of the object) is

2

0 rectangle triangle 0

2

0 0

1

2

1

2

x x

x x

x x A A v t a t

x x v t a t

% " $ " $

+ " $ $

(Note that the symbol + means that this equation follows from the previous equation.) Does the

above result make sense? Consider a limiting case, for example when the object is traveling at a

constant velocity (when 0x

a " ). In this case the equation should reduce to the result from our

investigation of linear motion with constant velocity ( 0 0xx x v t" $ ). It does. We can also check

the units of each term in this equation for consistency (when terms in an equation are added or

subtracted, each of those terms must have the same units). Each term has units of meters, so the

units also check.

Position of an object during linear motion with constant acceleration For any initial

position 0x at clock reading 0 0t " , we can determine the position x of an object at any later

time t , provided we also know the initial velocity 0xv of the object and its constant

acceleration x

a :

2

0 0

1

2x x

x x v t a t" $ $ (1.6)


Since 2

t appears in Eq. (1.6), the position-versus-time graph for this motion will be a

parabola (Fig. 1.27a). Unlike the position-versus-time graph for constant velocity motion where

xconst

t

#"

#, the position-versus-time graph line for accelerated motion is not a straight line; it

does not have a constant slope. At different times, the change in position x# has a different value

for the same time interval t# (Fig. 1.27b). The line tangent to the position-versus-time graph line

at a particular time has a slope x

t

##

that equals the velocity x

v of the object at that time. The

slopes of the tangent lines at different times for the graph line in Fig. 1.27b differ—greater when

the position changes more during the same time interval. An object’s velocity at a particular time

can also be determined using Eq. (1.5) 0x x xv v a t" $ , which describes an object’s velocity as a

function of time for constant acceleration motion.

Figure 1.27 Velocity is the slope of x vs. t graph

Example 1.7 Acceleration during a car collision A car traveling at 45 mi/h collides head-on

with a concrete pillar. During the collision, the front of the car compresses 1.0 m. The driver, if

belted securely in the car, also travels about 1.0 m while stopping during the collision. Determine

the acceleration of the driver during the collision.

Figure 1.28(a)


Sketch and Translate The sketch in Fig. 1.28a represents the initial and final situations during the

collision. The car crumples 1.0 m. This crumpling distance is the distance that the driver travels

during the collision. Since the driver comes to a stop over such a short distance, we expect his

acceleration to have a large value.

Simplify and Diagram As the shape of the car changes, we cannot model it as a point like object.

But we can think of a driver as a point-like object that moves 1.0 m during the collision. Another

simplification involves the acceleration. We don’t know the details of the stopping motion so

assume that the acceleration is constant equal to the average acceleration. The motion diagram in

Fig. 1.28b represents the person’s motion while stopping, assuming constant acceleration.

Figure 1.28(b)

Represent Mathematically The challenge in representing this situation mathematically is that

there are two unknowns: the magnitude of acceleration x

a of the driver (which is what we are

looking for) and the time t at which he comes to rest. However, both Eqs. (1.5) and (1.6)

describe linear motion with constant acceleration and have x

a and t in them. Since we have two

equations and two unknowns, we can handle this challenge by solving Eq. (1.5) for the time t

( 0x x

x

v vt

a

%" ) , and substitute it into rearranged Eq. (1.6):

2

0 0

1

2x x

x x v t a t" $ $

The result is:

2

0 00 0 2

( )( )

2

x x x x x

x

x x

v v a v vx x v

a a

% %% " $ ,

where x and x

v are the position and velocity of the driver at the later time t . Using algebra we

can simplify the above equation:

& ' & ' & '& ' & ' & '& '

& '

2

0 0 0 0

2 2 2

0 0 0 0 0

2 2

0 0

2 2

0

0

2 2

2 2 2 2

2

2

x x x x x x

x x x x x x x x

x x x

x x

x

a x x v v v v v

a x x v v v v v v v

a x x v v

v va

x x

+ % " % $ %

+ % " % $ % $

+ % " %

%+ "

%

Solve and Evaluate Now we can use the above equation to solve for the acceleration. We also

need to convert the initial speed of the car in mi/h to m/s:

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2.1.10


& '

& '

& '

22

2 220

0

mi 1 h 1609 m0 45

h 3600 s 1 mi200 m s

2 2 1.0 m 0

x x

x

v va

x x

, -% * *. /% 0 1" " " %% %

.

This is a huge acceleration; but the human body can withstand acceleration such as this.

Try It Yourself: Using the expression & '

2 2

0

02

x x

x

v va

x x

%"

%, decide how the acceleration would change

if: (a) the stopping distance 0( )x x% doubled, and (b) the initial speed v0x doubled. Note that the

final velocity is 0x

v " .

Answer: (a) ax would be half as great; (b) ax would be four times greater.

In the Represent Mathematically step above, we developed a new equation for the

acceleration by combining Eqs. (1.5) and (1.6). This useful equation is rewritten and described

briefly below.

Alternate equation for linear motion with constant acceleration:

& ' 2 2

0 02 –x x x

a x x v v% " (1.7)

The above equation can be used to determine one of the following quantities if the others are

known: the acceleration ( xa ), the displacement 0( )x x% of an object, its initial velocity

( 0xv ), and its final velocity ( xv ) when at position x . This equation is useful for situations in

which you do not know the time interval during which the changes in position and in velocity

occurred.

Review Question 1.7

(a) Give an example in which an object with negative acceleration and is speeding up. (b) Give an

example in which an object with positive acceleration is slowing down.

1.8 Skills for Analyzing Situations Involving Motion

In this section we will practice translating between these different types of concrete and abstract

ways of representing motion (called representations). We start with linear motion with constant

velocity since that’s the simplest kind of motion we have encountered.

Motion at constant velocity

To help understand physical processes involving motion, we will represent processes in

multiple ways: the words in the problem statement, a sketch, one or more diagrams, possibly a

graph, and a mathematical description. The sketches and diagrams help give meaning to the

mathematical descriptions for such processes.


Example 1.8 Two walking friends You sit on a sidewalk bench and observe two friends walking

on the sidewalk at constant velocity. At time zero Jim is 4.0 m east of you and walking away

from you at speed 2.0 m/s. At time zero, Sarah is 10.0 m east of you and walking toward you at

speed 1.5 m/s. Represent their motions with an initial sketch, motion diagrams, and then

mathematically.

Sketch and Translate We need separate descriptions of Jim and of Sarah since their motions are

different. We choose the earth as the object of reference with your position as the reference

point. The positive direction will point to the east. Jim’s initial position is 0 4.0 mx " $ m and

his constant velocity is 2.0 m/sx

v " $ . Sarah’s initial position is 0 = +10.0 mX m and her

constant velocity is = –1.5 m/sx

V (the velocity is negative since she is moving westward). The

sketch in Fig. 1.29a represents the initial situation.

Simplify and Diagram Both friends can be modeled as point-like objects since the distances they

move are somewhat greater than their own sizes. The motion diagrams in Fig. 1.29b represent

their motions.

Figure 1.29

Represent Mathematically Now construct equations to represent Jim’s and Sarah’s motion:

Jim: 4.0 m +(2.0 m/s)x t" $

Sarah: 10.0 m (–1.5 m/s)X t" $ $

Solve and Evaluate We were not asked to solve for any quantity. We will do it in the following

Try It Yourself exercise.

Try It Yourself: Determine the time when Jim and Sarah are at the same position, that is, when

x X" and where is that position?

Answer: They are at the same position when 1.7 st " and when 7.4 mx X" " .

Tip! Note that in the last example we indicated the positions and velocities of Jim and of Sarah

using different symbols. Using the same x and v for each would be confusing.

Equation Jeopardy Problems

Learning to read the math language of physics with understanding is an important skill.

To help develop this skill, we will do examples similar to the TV game show Jeopardy. You are

given one or more equations and are asked to use them to construct a consistent sketch of a

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2.2.8


process. You then convert the sketch into a diagram of a process that is consistent with the

equations and sketch. Finally, you can invent a word problem that the equations might have been

used to solve. Note that there are often many possible consistent word problems for a particular

mathematical description. Here is our first example of an Equation Jeopardy Problem.

Conceptual Exercise 1.9 Equation Jeopardy The following equation describes an object’s

motion:

& ' & '5.0 m 3.0 m sx t" $ %

Construct a sketch, a motion diagram, kinematics graphs, and a verbal description of a situation

that is consistent with this equation. There are many possible situations that the equation

describes equally well.

Sketch and Translate This equation looks like a specific example of our general equation for the

linear motion of an object with constant velocity: 0 0xx x v t" $ . The minus sign in front of the

3.0 m/s indicates that the object is moving in the negative x-direction. At time zero, the object is

located at position 0 5.0 mx " $ with respect to some chosen object of reference. Let’s imagine

that this chosen object of reference is a running person (the observer) and the equation represents

the motion of a person sitting on a bench (object of interest) as seen by the runner. (You can

probably think of easier possible processes described by the equation.) A labeled sketch of this

possible situation is shown in Fig. 1.30a. The positive axis points from the observer (the runner)

towards the bench and the person sitting on it is 5.0 m in front of the runner and coming closer to

the runner as time elapses.

Figure 1.30(a)

Simplify and Diagram We are modeling the object of interest as a point-like object. A motion

diagram for the situation is shown in Fig. 1.30b. The equal spacing of the dots and the equal

lengths of velocity arrows both indicate that the object of interest is moving with constant

velocity with respect to the observer.

Figure 1.30(b)


Motion at constant non-zero acceleration

Position-versus-time and velocity-versus-time kinematics graphs of the process are

shown in Fig. 1.30c and d. The position-versus-time graph has a constant –3.0 m/s slope and a

+5.0 m intercept with the vertical (position) axis. The velocity-versus-time graph has a constant

value (–3.0 m/s) and a zero slope (the velocity is not changing). The following verbal description

describes this particular process: A jogger sees a bench in the park 5.0 meters in front of her. The

bench is approaching at a speed of 3.0 m/s. The direction pointing from the jogger to the bench is

positive.

Figure 1.30(c)(d)

Try It Yourself: Suppose we switch the observer and object of reference roles in the last example.

Now the person on the bench is the object of reference and observes the runner. We choose to

describe the process by the same equation as in the example:

& ' & '5.0 m 3.0 m sx t" $ %

Construct an initial sketch and a motion diagram that are consistent with the equation and with

the new observer and new object of reference.

Answer: Figure 1.31a is an initial sketch for this process and Fig. 1.31b is a consistent motion

diagram. The kinematics graphs for this process are the same as in the example.


Figure 1.31

We’ve just learned how to analyze linear motion with constant velocity (zero

acceleration). Let’s apply what we’ve learned to analyze linear motion with constant (non-zero)

acceleration.

Example 1.10 Equation Jeopardy Problem A process is represented mathematically by the

following equation:

& ' & ' & '2 260 m 10 m s 1.0 m s ,x t t" % $ $

Use the equation to construct an initial sketch, a motion diagram, and words to describe a process

that is consistent with this equation.

Sketch and Translate The above equation appears to be an application of Eq. (1.6), which we

constructed to describe linear motion with constant acceleration, if we assume that 21.0 m/s in

front of 2

t is the result of dividing 22.0 m/s by 2:

2

0 0

1

2x x

x x v t a t" $ $ .

& ' & ' & '2 2160 m 10 m s 2.0 m s

2x t t" % $ $

It looks like the initial position of the object is 0 = –60 mx , its initial velocity is 0 = +10 m/sx

v ,

and its acceleration is 2 = +2.0 m/s

xa . Let’s imagine that this equation describes the motion of a

car passing a van in which you, the observer, are riding in the van on a straight highway. The car

is 60 m behind you and moving 10 m/s faster than your van. The car speeds up at a rate of

22.0 m/s with respect to the van. The object of reference is you in the van; the positive direction

is the direction in which the car and van are moving. A sketch of the initial situation with the

known information is shown in Fig. 1.32a.

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2.2.2-

2.2.7;

2.2.9


Figure 1.32(a)

Simplify and Diagram The car can be considered as a point-like object—much smaller than the

dimensions of the path it travels. The car’s velocity and acceleration are both positive. Thus, the

car’s speed in the positive x-direction is increasing as at moves toward the van (toward the

origin). A motion diagram for the car’s motion as seen from the van is shown in Fig. 1.32b. The

successive dots in the diagram are spaced increasingly farther apart as the velocity increases; the

velocity arrows are drawn increasingly longer. The velocity arrow (and the acceleration) point in

the positive x-direction, that is, in same direction as the velocity arrows.

Figure 1.32(b)

Represent Mathematically We were given the mathematical representation of the situation at the

start of the Equation Jeopardy Problem.

Solve and Evaluate To evaluate what we have done, we can check the consistency of the different

representations. For example, we can check if the initial position and velocity are consistent in the

equation, the sketch, and the motion diagram. In this case, they are.

Try It Yourself: Describe a different scenario for the same mathematical representation.

Answer: This could be a mathematical representation of the motion of a cyclist moving on a

straight path as seen by a person standing on a sidewalk 60 m in front of the cyclist. The positive

direction is in the direction the cyclist is traveling. When the person starts observing the cyclist,

he is moving at an initial velocity of 0 = +10 m/sx

v and speeding up with acceleration

2 = +2.0 m/sx

a .

Problem solving procedure: kinematics problems

The problem-solving procedure in this text uses a multiple representation strategy that

was found to be extremely useful in helping solve physics problems. The same strategy will be

adapted to the different physics subject areas in the book. Most of the chapters will include one


problem-solving procedure that can be applied to typical problems in that chapter. The complete

strategy for a particular chapter will be presented with an example at the side. The problem-

solving steps for kinematics (motion) problems is presented in the next example on the left side

and illustrated for that problem on the right side.

Example 1.11 Car arriving at a red light A car’s

motion is represented by the velocity-versus-time

graph in Fig. 1.33. What time interval is needed for

the car to stop and how far did it travel while

stopping?

Figure 1.33 A car’s v vs. t graph when stopping

Sketch and Translate:

! Sketch the situation described in the problem.

! Include an object of reference, a coordinate

system and indicate the origin and the positive

direction.

! Label the sketch with relevant information.

From the graph, we see that the car’s velocity at time

zero is 0

14 m/sx

v " $ . The object of reference is the

ground. The plus sign means it is moving in the

positive x-direction. From the graph, we see that the

car’s velocity in the positive x-direction decreases by

2.0 m/s for each second; thus the slope of the graph

xv

t

#

#is (–2.0 m/s)/s . A labeled initial situation

sketch is shown below. The car’s initial position is

unknown—we’ll choose to place it at position 0

0x "

at 0

0t " .

Simplify and diagram:

! Decide how you will model the moving object

(for example, as a point-like object)

! Can you model the motion as constant velocity or

constant acceleration?

! Draw motion diagrams and kinematics graphs if

needed.

We will model the car as a point-like object moving

along a straight line. A motion diagram for the car is

shown below. The velocity arrows get increasingly

smaller since the acceleration arrow points opposite

the velocity arrows.


Represent Mathematically:

! Use the sketch(s), motion diagram(s), and

kinematics graph(s) to help construct a

mathematical representation (equations) of the

process. Be sure to consider the sign of each

quantity.

Rearrange Eq. (1.6) to determine the time at which

the velocity decreases to zero:

– 0x x

x

v v

at "

The position of the car at that time can then be

determined using the position equation of motion:

2 21(–2.0 m/s ) .

20 (14 m/s) tx t" $ $

Solve and Evaluate:

! Solve the equations to find the answer to the

question you are investigating.

! Evaluate the results to see if they are reasonable.

To do this, check the units, decide if the

calculated quantities have reasonable values

(sign, magnitude), and check limiting cases.

Substituting the known information in the first

equation above:

2

(0 m/s) – (14 m/s)7.0 s

–2.0 m/st ""

The car stops after a time interval 0( 7.0 s)t t% " .

The car’s position when it stops is:

2 21(–2.0 m/s ) 49 m.

20 (14 m/s)(7.0 s) (7.0 s)x "" $ $

The units are correct, and the magnitudes are

reasonable.

Try It Yourself: A cyclist is moving in the negative x-direction at a speed of 6.0 m/s (the positive

direction points to the right). The cyclist sees a red light and stops in 3.0 seconds. What is the

cyclist’s acceleration?

Answer: 23.0 m/s

xa " $ . Why is the acceleration positive even though the cyclist’s speed

decreased? Hint: Draw a motion diagram and think about the direction of the velocity change

arrow and the acceleration arrow relative to the chosen positive direction.

Using the quadratic equation

In Example 1.7 we used a mathematical strategy in which two independent mathematical

expressions were used to solve for a situation involving two unknown quantities. In the next

problem we use another important mathematical strategy—the quadratic equation—to solve the

problem.

Example 1.12 Stopping for a cross walk You are driving at a moderate speed of 8.0 m/s (18

mph). Suddenly, you become distracted and fail to notice a crosswalk until you are 3.0 m from

the crosswalk (you are supposed to stop at or before the crosswalk). You hit the brakes and slow

down at a rate of 8.0 m/s each second. What is time interval needed for the car to reach the

crosswalk, and how fast is the car moving at that instant?

Sketch and Translate From the text of the problem we see that the motion is described with

respect to the ground. Let’s choose a coordinate axis that points toward the right, opposite your


direction of motion with the origin at the edge of the crosswalk, as shown in Fig. 1.34a. With this

choice, the following information about the car’s motion relative to that coordinate axis is known:

initial position 0 3.0 mx " $ , initial velocity 0 8.0 m/sx

v " % , and acceleration 28.0 m/s

xa " $ .

We wish to know the time t when the car reaches the crosswalk, and its speed x

v at that time.

Figure 1.34(a)

Simplify and Diagram A motion diagram for the car is shown in Fig. 1.34b. Note that the velocity

arrows pointing in the negative direction are getting shorter as the car moves toward the left.

Recall that earlier in the chapter, we saw that the velocity change arrow and the acceleration

arrow point opposite the direction of the velocity when an object’s speed is decreasing. This

means the acceleration points in the positive direction and is positive.

Figure 1.34(b)

Represent Mathematically We are now ready to construct a mathematical description of the

motion. Since this is linear motion with constant acceleration, we can apply Eqs. (1.6) and (1.5)

to this problem:

2

0 0

1

2x x

x x v t a t" $ $

0x x xv v a t" $

These equations applied to the situation with this car become:

& ' & '2 213.0 m 8.0 m s 8.0 m s

2x t t" $ % $

& ' & '28.0 m s 8.0 m sx

v t" % $

To determine when the car passes the 0x " position, we set 0x " in the first equation above:

& ' & '2 210 3.0 m 8.0 m s 8.0 m s

2t t" $ % $

Solve and Evaluate To determine the time when the position is zero, we need to solve the last

equation for the unknown time t. Notice that it is not a linear equation, but a quadratic equation.


One way to solve a quadratic equation is to simplify it so that it has the form 2 0At Bt C$ $ " .

Recall from algebra that such an equation has two solutions:

2 4

2

B B ACt

A$

% $ %" and

2 4

2

B B ACt

A%

% % %" .

The quadratic equation for the car in the problem (leaving out the units) is:

24.0 – 8.0 3.0 0.t t $ "

With = 4.0, = –8.0, and = 3.0A B C , the first solution is:

22 (–8.0) (–8.0) 4 • 4.0 •3.041.5 s.

2 2 • 4.0

B B ACt

A$

% $ %% $ %" " " $

The second solution is:

22 (–8.0) (–8.0) 4 • 4.0 •3,040.5 s.

2 2 • 4.0

B B ACt

A%

% % %% % %" " " $

Which answer should we use? The velocity of the car with the second solution is:

& ' & '28.0 m s 8.0 m s (0.5 s) –4.0 m/s.x

v " % $ "

The car is moving in the same direction as it was initially moving, but now slower; that makes

sense. The velocity of the car at 1.5 s, the second solution, is:

& ' & '28.0 m s 8.0 m s (1.5 s) 4.0 m/s.x

v " % $ " $

The car must have stopped, and because of the continuing positive acceleration, it is now moving

in the opposite direction. This is not a reasonable solution in terms of real life and will be rejected

even though it is an appropriate math solution.

Try It Yourself: At what time and location in this example did the car stop, and where was it when

its velocity was zero?

Answer: Use the velocity equation to determine the time when the velocity was zero:

20 –8.0 m/s + (8.0 m/s )t" .

The velocity will be zero when 1.0 st " . The car is now at:

& ' & '2 213.0 m 8.0 m s (1.0 s) 8.0 m s (1.0 s) –1.0 m.

2x " $ % $ "

Evidently the car continued to slow down in the negative direction until 1.0 st " and when 1.0

m beyond the start of the crosswalk. Because of the positive acceleration, the car unrealistically

moved backward in the positive direction reaching the cross walk stop sign again at 1.5 s.

Review Question 1.8 A car’s motion with respect to the ground is described by the following

function:

& ' & ' & '2 248 m 12 m s 2.0 m sx t t" % $ $ %


Determine the car’s position and velocity at time zero. Also, determine its acceleration. At what

times (if any) does the car’s velocity become zero?

1.9 Free Fall

We learned about two simple models of motion – linear motion with constant velocity and linear

motion with constant acceleration. Do these models represent commonly occurring motions? For

example, if we drive a car in a straight line on a highway at roughly constant velocity, there

would still be slight variations in the car’s speed and direction of motion. Yet, the constant

velocity motion model applies nicely. What about constant acceleration motion? That is much

harder. Can you drive your car in such a way that the rate at which your velocity increases is

constant? It’s not easy.

There is another common situation that can be modeled reasonably as constant

acceleration motion. Let’s start with the following observational experiment. Tear out a sheet of

paper from your notebook and hold the paper in one hand. Hold a textbook in the other hand and

then drop both side by side. The book lands first. Next, crumple the paper into a tight ball. Now

drop the book and the crumpled paper side-by-side. This time they land at about the same time.

Does the motion depend on how heavy the objects are or on their shapes?

Aristotle believed that heavier objects fall faster than lighter objects. This is consistent

with our first observation with the book and paper, and people believed his explanation for almost

2000 years. But, the second observation with the book and crumpled paper is inconsistent with

this explanation.

In the 1600s, long after Aristotle, Galileo Galilei observed

swinging chandeliers in a church and noticed that the time interval

needed to complete each swing did not depend on how heavy the

chandeliers were. Instead, he noticed that it depended on the length of

the ropes that supported the chandeliers. This pattern made him think

of falling objects, as he suspected that the reason that suspended

chandeliers swung as they did was the same as the reason for the

objects to fall when they were not suspended. That led him to

construct the following hypothesis: the heaviness of objects does not

affect the way they fall. Galileo said that the reason some objects seem to fall differently from

others is because of the effect the air has on them. Put another way, Galileo’s hypothesis says that

in the absence of air, all objects, regardless of mass or shape, would fall in exactly the same way.

There is a legend that Galileo, then a mathematics professor at the University of Pisa,

tested his hypothesis by dropping from the Pisa Tower (Fig. 1.35) two stones of about the same

size but made of different materials—one was much heavier than the other. He recorded the time

that it took them to fall by listening for the thud when they hit the ground. Although he could hear


the heavier stone hit the ground slightly sooner than the light one, the difference in the times that

the two stones fell was tiny compared to how heavy they were.

If Galileo could get rid of air completely, he would observe that less massive and more

massive objects fall exactly the same way. Today scientists can use vacuum pumps to remove

nearly all of the air from a container. When most of the air is pumped out, all objects reach the

bottom at the same time—for example, a feather and a rock.

At this point, we do not know why this is happening (we will learn why in Chapter 2).

For the moment this is just a pattern that we have observed about the motion of falling objects:

when air is not present, all objects fall in exactly the same way. Galileo’s idea that the heaviness

of an object was not important in describing their falling motion is supported by the results of the

testing experiment we just described.

What type of motion should we use to model the motion of these falling objects? We

know from experience that a breakable object is more likely to break when it falls from a table

then from a low shelf. Is it because it is moving faster when falling from the table? If this is the

case, then objects do speed up when falling. But do they speed up at constant acceleration or does

the acceleration change during the fall? Galileo decided that constant acceleration was the

simplest possible motion and he did not see any reason why the motion of falling objects should

be more complicated. This idea was difficult to test. Galileo did not have a video camera or a

watch with a second hand. He measured time with his heartbeat. Objects fell too quickly for his

crude measuring instruments. How could he slow down falling objects so he could collect data to

support or reject his hypothesis?

Galileo decided to reduce the speed of falling objects by rolling balls on a tilted track,

and recording the time interval it takes them to cover a certain distance. His original hypothesis

was that the speed of a falling object should increase in a simple linear fashion as a function of

time – increasing by the same amount every second or in other words, that the objects would fall

with constant acceleration. He extended this hypothesis to the rolling balls – when they travel

down the ramp, their speed increases in linear proportion to time. This hypothesis makes a

specific prediction about the distance the balls should travel as a function of time. If the ball starts

at zero velocity and velocity increases in linear proportion to time x x

v a t" , the distance traveled

will be proportional to time squared 21

2x

x a t" (Eq. (1.5). Thus, if the ball travels a distance of 1

unit during the first second of its motion, its displacement after the first 2 seconds should be 4

units, after the first 3 seconds should be 9 units, etc. These distances are easy to measure! The

outcome of the experiment that Galileo performed was consistent with this prediction and thus

supported his hypothesis (Fig. 1.36). This result convinced Galileo that his hypothesis that in the

absence of air, all objects fall with constant acceleration was correct. He also thought that the

acceleration of freely falling objects would be larger than the acceleration of the balls rolling

down the incline.


Figure 1.36 Distance ball travels while rolling down incline

Let’s test Galileo’s actual hypothesis using a real falling

object. Imagine that we videotape a metal ball that is dropped beside

a ruler (Fig. 1.37). If the hypothesis is correct, the speed of the ball

should increase linearly with time. After recording the fall, we step

through the video frame-by-frame and record the ball’s position

every 0.100 s (Table 1.7). We use the earth as our object of reference

and place the origin of our coordinate axis at the initial location of

the ball. The positive direction points down.

To determine the average velocity during each time interval,

we calculate the displacement of the ball between consecutive times

and then divide by the time interval (see Table 1.7). For example, the

average speed between 0.100 s and 0.200 s is

(0.196 m – 0.049 m)/(0.200 s – 0.100 s) = 1.47 m/s . These

Figure 1.37 Falling ball calculated velocities in the last row are associated with the clock

readings *t at the middle of each time interval (the third row). Finally, we use these velocities and

*t times to construct a velocity-versus-time graph (Fig. 1.38). Notice that the best-fit curve for

this data is a straight line, indicating that the motion of the metal ball can be modeled accurately as

motion with constant acceleration. The slope of the line equals 9.8 m/s2.

Figure 1.38 Velocity-vs-time for falling ball


Table 1.7 Position and time data for a falling block.

t (s) 0.000 0.100 0.200 0.300 0.400 0.500

y (m) 0.000 0.049 0.196 0.441 0.784 1.225

We can represent the motion of a falling ball mathematically using the equations of

motion for constant acceleration [Eqs. (1.5) and (1.6)] with 2= 9.8 m/s

ya g" :

2

0 0= (9.8 m/s ) y y y y

v v a t v t$ " $ (1.8)

2 2

0 0

1(9.8 m/s )

2y

y y v t t" $ $ . (1.9)

where 0 0 and y

y v are the position and instantaneous velocity of the object at the clock reading

0 0t " . These equations apply if the positive y direction is down. If using an upward pointing y-

axis, a minus sign is placed in front of the 9.8 m/s2. The magnitude of the object’s acceleration

while falling without air resistance is given a special symbol 29.8 m/sg " .

If we videotape a small object thrown upwards and then use the data to construct a velocity-

versus-time graph, we find that its acceleration is still the same at all clock readings. Using an

upward pointing axis, its acceleration is 2–9.8 m/s on the way up;

2–9.8 m/s on the way down;

and even 2–9.8 m/s at the instant when the object is momentarily at rest at the highest point of its

motion. At all times during the object’s flight, its velocity is changing at a rate of –9.8 m/s each

second. A motion diagram and the graphs representing the position-versus-time, velocity-versus-

time and acceleration-versus-time are shown in Fig. 1.39. The positive direction is up. Notice that

when the position-versus-time graph is at its maximum (object is at maximum height), the velocity

is instantaneously zero (the slope of the position-versus-time graph is zero). The acceleration is

never zero, even at the moment when the velocity of the object is zero.

t* (s) 0.050 0.150 0.250 0.350 0.450

vav (m/s) 0.49 1.47 2.45 3.43 4.41


Figure 1.39 Representing vertical motion of object thrown upward

Tip: It might be tempting to think that at the instant an object is not moving, its acceleration must

be zero. This is only true for an object that is at rest and remains at rest. In the case of an object

thrown upward, if its acceleration at the top of the flight is zero, it would never descend (it would

remain at test at the very top of the flight).

Another tip! Physicists say that an object is in a state of free fall even when it is thrown upward,

as its acceleration is the same on the way up as on the way down. The acceleration points down in

the vertical y direction. When the y -axis points down, the y-component of the acceleration

29.8 m/sy

g " ; when the y -axis points up 29.8 m/s

yg " % .

Galileo

Galileo Galilee was a founder of the contemporary practice of science—science that is

based on making careful observations, finding patterns, proposing hypotheses/explanations for

these patterns, and testing these explanations by predicting the outcomes of new experiments. He

was the first scientist to write his papers in his native language rather than in Latin. He was the

first to point a telescope at the Moon and see the craters there. He was the first to observer the

moons of Jupiter—four satellites that are now called the Galilean moons of Jupiter. Jupiter has

many more moons than just four, but only the largest were visible to Galileo with his telescope.

He got in trouble for his study of sunspots. At the time it was believed the Sun was not supposed

to have any spots at all. He not only saw sunspots, but he saw that they moved across the face of

the Sun. His hypothesis was that the Sun rotated, which turned out to be correct (read more about

sunspots in Chapter 17). His discoveries were evidence that Earth was just one planet out of


many, and that it wasn’t located at the center of the Universe (which was the accepted belief at

the time).

Example 1.13 Research rocket test flight The European REXUS 3 research rocket is used to test

electronic parts, which will be used on future spaceflights. In April 2006, the engine fired during

takeoff for 25.0 seconds, giving the rocket an average acceleration of about 60 m/s2. After the

rocket’s engine shut off, the rocket continued upward into the atmosphere at decreasing speed (like

an upward thrown ball slowing down after it leaves the thrower’s hand). Determine (a) the

REXUS 3’s height above Earth’s surface at the end of the 25.0 s fuel burn, (b) its speed at that

time, and (c) the height of REXUS 3 above Earth’s surface when it reached its highest altitude.

Sketch and Translate We choose the earth as the object of reference. The point of reference is the

initial position of the rocket. The positive direction is upward. We need to analyze the rocket’s

motion in two parts because there are two different values for acceleration. During the first part of

the rocket’s motion, the engine is firing and the rocket has a constant acceleration of 260 m/s$ .

During the second part of the rocket’s motion, the engine is off and its acceleration is 2–9.8 m/s .

For (a) and (b), we need to find the rocket’s height 1y and speed 1yv at the end of part I of the

motion. For (c), we need to find the highest point of the rocket, or

2y at the end of part II. After this, the rocket’s velocity becomes

negative, and it begins falling back toward Earth. The process is

pictured in Fig. 1.40a.

Simplify and Diagram We make several assumptions: 1) the

rocket is a point-like object, 2) the acceleration of the rocket

during part I is constant at 2+60 m/s , and 3) the acceleration

during part II is constant at 2–9.8 m/s . Assumptions 2) and 3)

are questionable. The rocket will be traveling at increasing speed

and very rapidly; air resistance is likely to have a changing effect

on the rocket motion, making its acceleration vary. But, if we do

not make these assumptions, we will not be able to apply our

model of constant acceleration motion. Figure 1.40b shows a

motion diagram for both parts of the rocket’s motion. Notice the

acceleration vectors. Figure 1.40 Rocket flight

Represent Mathematically The standard kinematics equations for constant accelerated motion will

be adapted for each part of the motion:

0y y yv v a t" $ (1.5)

2

0 0

1

2y y

y y v t a t" $ $ (1.6)

Solve and Evaluate We start by applying the above for the fuel burn.


Part I (the rocket burn): The rocket’s initial position is 0 0y " , its initial speed 0 0y

v " , the

acceleration during fuel burn 2

I 60 m/sy

a " $ , and the time at the end of the burn 25 st " . The

speed of the rocket at the end of the burn is determined using Eq. (1.5):

& '& '2 3

1 0 I 1 0 60 m s 25.0 s 1500 m s 1.5 10 m sy y y

v v a t" $ " $ " " * .

We’ve switched to scientific notation here because when we wrote 1500 m s , it was not possible

to indicate how many significant digits there were. Are there two, three, or four? There are two,

because the acceleration has just two significant figures. Use Eq. (1.6) to determine the position

of the rocket at the end of the fuel burn):

& '& '22 2 4

1 0 0 I

1 10 0 60 m s 25.0 s 2.0 10 m

2 2y y

y y v t a t" $ $ " $ $ " *

Part II (after the rocket burn): The end of part I is the beginning of part II, so the position and

velocity of the rocket at the end of part I is the position and velocity of the rocket at the beginning

of part II. At the end of part II, the rocket is at its highest point and has zero velocity for a brief

instant (it had a positive velocity a moment earlier, and will have a negative velocity a moment

later.) This means that 2 0y

v " . We use Eq. (1.7) to find the height of the rocket at the end of part

II:

& '

& '& ' & '

2 2

2 1 II 2 1

22 32 2

2 1 4

2 1 2II

5

2

2

0 1.5 10 m s2.0 10 m

2 2 9.8 m s

1.3 10 m = 130 km = 81 mi.

y y y

y y

y

v v a y y

v vy y

a

y

" $ %

% *%+ " $ " $ *

%

+ " *

Assuming our assumptions are reasonable, the model for the rocket’s motion tells us that

it will attain a maximum altitude of 81 miles. However, the assumptions must be somewhat

wrong because the rocket actually reached only 20 miles in height. This is due primarily to the

acceleration not being constant, leading to a breakdown in our simplified equations of motion.

Try It Yourself: Determine the time interval for part II of the flight (from the time the fuel stopped

burning until the rocket reached its highest point).

Answer: 150 s .

Review Question 1.9

How do you decide if you should use a positive or negative sign for the acceleration of an object

moving in a vertical direction?


1.10 Putting it all together

In this section, we use the models developed thus far to analyze a situation that

happens on the road every day—tailgating, or when one car follows too closely behind

another. The National Highway Traffic Safety Administration of the US has reported that

rear-end collisions account for 23 percent of all motor vehicle crashes, resulting in

approximately 2000 deaths and 950,000 injuries.

Drivers count on their ability to apply the brakes in time if the car in front of them suddenly

slows. However, this is not always possible. Factors affecting the possibility of a collision due to

tailgating include:

! Car decelerations (‘deceleration’ is a term used to describe acceleration that is opposite

the direction of motion).

! Driver reaction time (the time interval it takes for a driver to respond by applying the

brakes from the time he or she sees the need to slow down).

! The cruising speeds of the cars before they start applying their brakes.

! The distance between the lead and tailgating car.

Let’s look at motion of two vehicles in what appears to be a safe driving situation.

Example 1.14 An accident involving tailgating A sports car follows about two car lengths (10.0

m) behind a van. At the beginning of the process, both vehicles are traveling at a conservative 25

m/s speed (56 mph). The driver of the van suddenly slams on the brakes decelerating at 9.0 m/s2

to avoid an accident. The car’s driver’s reaction time is 0.80 s and the car’s maximum

deceleration is 29.2 m/s . Will the car be able to stop before hitting the van?

Sketch and Translate Figure 1.41a represents this situation for each vehicle. We’ll use capital

symbols to indicate quantities referring to the van and lower case symbols for quantities referring

to the car. We use the coordinate system shown in Fig. 1.41a with the origin of the coordinates at

the initial position of the sports car’s front bumper. The positive direction is in the direction of

motion.

The process starts when the van starts braking (constant negative acceleration). It moves

at constant acceleration throughout the problem situation. The motion of the car is broken into

two parts: 1) the motion before the driver applies the brakes (constant velocity), and 2) its motion

after driver starts braking (constant negative acceleration).

Simplify and Diagram We model each vehicle as a point-like object, but since we are trying to

determine if they collide, we need to be more specific about their locations. The position of the

car will be the position of its front bumper. The position of the van will be the position of its rear

bumper. We look at the motion of each vehicle separately. If the car’s final position is greater

than the van’s final position, then a collision has occurred at some point during their motion.

Assume that the vehicles have constant acceleration so that we can apply our model of motion

ALG

2.2.10-

2.2.18


with constant acceleration. A velocity-versus-time graph line for each vehicle is shown in Fig.

1.41b.

Figure 1.41 Sports car tailgates a van

Represent Mathematically: We use the constant acceleration mathematical principles constructed

in this chapter.

The Van

Equation (1.7) can be used to determine the distance the van travels while stopping:

& ' 2 2

0 0

2 2

00

2 –

2

x x x

x x

x

A X X V V

V VX X

A

% "

%+ " $

The Car Part 1

Since the car is initially traveling at constant velocity, we use Eq. (1.2). The subscript ‘0’

indicates the moment the driver sees the van start slowing down. The subscript ‘1’ indicates the

moment the car driver starts braking. The subscript ‘2’ indicates the moment the car stops

moving.

1 0 0 1xx x v t" $

The Car Part 2

After applying the brake, the car has an acceleration of –9.2 m/s2. We represent this part of the

motion using Eq. (1.7):


& ' 2 2

2 1 2 1

2 2

2 12 1

2 2

2 12 0 0 1

2 –

2

( )2

x x x

x x

x

x x

x

x

a x x v v

v vx x

a

v vx x v t

a

% "

%+ " $

%+ " $ $

The last step came from inserting the result from part 1 for x1.

Solve and Evaluate The van’s initial velocity 0 = +25 m/sV , its final velocity final = 0V , and its

acceleration 2 = –9.0 m/sA . Its initial position is two car lengths in front of the front of the car,

so 0 2 5.0 m 10 mX " * " . The final position of the van is:

& '

& '

222 2

00 2

0 25 m s10 m 45 m.

2 2 9.0 m s

x x

x

V VX X

A

%%" $ " $ "

%

The car’s initial position 0 0x " , its initial velocity is 0 25 m/sx

v " , its final velocity

2 0x

v " , and its acceleration when braking is 2–9.2 m/s

xa " . The car’s final position is:

& '

& ' & '& '222 2

2 12 0 0 1 2

0 25 m s ( ) [0 m 25 m s 0.8 s ] 54 m.

2 2 9.2 m s

x x

x

x

v vx x v t

a

%%" $ $ " $ $ "

%

The car would stop about 9.0 m beyond where the van would stop. There will be a collision

between the two vehicles. Another way to solve this problem is to switch to a reference frame

associated with the van. In this reference frame the van is at rest and car approaches it. If it

reaches the van while moving, the collision occurs.

This analysis illustrates the reason why tailgating is such a big problem. The car traveled

at a 25-m/s constant velocity during the relatively short 0.80 s reaction time. During the same

0.80 s, the van’s velocity decreased by 2(0.80 s)(–9.0 m/s ) = 7.2 m/s from 25 m/s to about 18

m/s. So, the van was moving somewhat slower than the car when the car finally started to brake.

Since they were both decelerating at about the same rate, the tailgating vehicle’s velocity was

always greater than that of the vehicle in front (see Fig. 1.41b)—until they hit.

Try It Yourself: Two cars, one behind the other, are traveling at 30 m/s (13 mph). The front car

hits the brakes and has a 10-m/s2 deceleration—the same as that of the back car. The driver of the

second car has a 1.0 s reaction time. The front car’s speed has decreased to 20 m/s during that 1.0

s. The back car traveling at 30 m/s starts braking. How far behind the front car should the back

car be so it does not hit the front car?

Answer: The back car should be at least 25 m behind the front car.

If a drivers’ reaction time were 0.1 s instead of 0.8 s, then tailgating might not be such a

problem, but such short reaction times are extremely rare. Many tailgating accidents occur when

we are not alert, either while eating, talking on a cell phone, or looking at the GPS unit. As a


result, reaction times can be as high as two seconds. Here’s a rule of thumb for determining a safe

following distance. Choose a marker on the side of the road ahead of you, such as a tree or a road

sign. Two seconds should pass between when the car in front of you passes that marker and when

your car does. This will give you a large enough following distance so that if the car in front does

slam on the brakes, you will be able to avoid a collision.

Review Question 1.10

Why is “following too closely” not a very good explanation for why a tailgating accident occurs?


Summary

Word Description Picture or Diagrammatic

Description

Mathematical Description

A reference frame consists of an object of

reference, a point of reference on that

object, a coordinate system whose origin is

at the point of reference, and a clock.

A pictorial description of motion includes

a sketch of the situation, a coordinate axis,

and the values of known physical

quantities that describe that process.

! Time or clock reading t (a scalar) is the

reading on a clock or on a stopwatch or on

some other time measuring instrument.

! Time interval t# (a scalar) is the

difference of two times.

Time: t

Time interval:

2 1t t t# " %

! Position x (a scalar) is the location of

an object relative to the chosen origin.

! Displacement d!

is a vector drawn from

the initial position of an object to its final

position. The x component of the

displacement x

d is the change in position

of the object along the x -axis.

! Distance d (a scalar) is the magnitude

of the displacement and is always positive.

0 x

x x d" $ or

0 0 x

d x x" % 2 if d!

points

in the positive direction of x

axis.

0 0 x

d x x" % 3 if d!

points

in the negative direction.

0d x x" %

! Velocity v!

(a vector) is the displacement

of an object during a time interval divided

by that time interval. The velocity is

instantaneous if the time interval is very

small and average if the time interval is

longer.

! Speed v (a scalar) is the magnitude of

the velocity.

For linear constant velocity

motion d

v

t

"#

!

2 1

2 1

x

x xxv

t t t

%#" "# %

(1.1)

xv

t

#"#

! Acceleration a!

(a vector) is the change

in an object’s velocity v#!

during a time

interval t divided by the time interval.

The acceleration is instantaneous if the

time interval is very small and average if

the time interval is longer.

va

t

#"#

!!

0

0

–

–

x x x

x

v v v

a

t t t

"#

"# (1.4)

Motion with constant velocity or constant

acceleration can be represented with a

sketch, a motion diagram, with kinematics

graphs, and mathematically.

0x x x

v v a t" $ (1.5)

2

0 0

1

2x x

x x v t a t" $ $(1.6)

2 2

0

0

2

x x

x

v vx x

a

%% "

(1.7)

chapter 1 kinematics: the description of...

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