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Notes/Cues Here Unit: Kinematics (Motion)

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AP Physics 1: Algebra-Based Mr. Bigler

Linear Motion, Speed & Velocity Unit: Kinematics (Motion)

NGSS Standards: N/A

MA Curriculum Frameworks (2006): 1.1, 1.2

AP Physics 1 Learning Objectives: 3.A.1.1, 3.A.1.3

Knowledge/Understanding Goals:

understand terms relating to position, speed & velocity

understand the difference between speed and velocity

Language Objectives:

Understand and correctly use the terms “position,” “distance,” “displacement,” “speed,” and “velocity.”

Accurately describe and apply the concepts described in this section using appropriate academic language.

Labs, Activities & Demonstrations:

Walk in the positive and negative directions (with positive or negative velocity).

Walk and change direction to show distance vs. displacement.

Notes:

coördinate system: a framework for describing an object’s position (location), based on its distance (in one or more directions) from a specifically-defined point (the origin). (You should remember these terms from math.)

direction: which way an object is oriented or moving within its coördinate system. Note that direction can be positive or negative.

position (x): the location of an object relative to the origin (zero point) of its coördinate system. We will consider position to be a zero-dimensional vector, which means it can be positive or negative with respect to the chosen coördinate system.

distance (d ): [scalar] how far an object has moved.





displacement ( d

or x ): [vector] how far an object’s current position is from its starting position (“initial position”). Displacement can be positive or negative (or zero), depending on the chosen coördinate system.

rate: the change in a quantity over a specific period of time.

motion: when an object’s position is changing over time.

speed: [scalar] the rate at which an object is moving at an instant in time. Speed does not depend on direction, and is always nonnegative.

velocity: )(v

[vector] an object’s displacement over a given period of time.

Because velocity is a vector, it has a direction as well as a magnitude. Velocity can be positive, negative, or zero.

uniform motion: motion at a constant velocity (i.e., with constant speed and direction)

An object that is moving has a positive speed, but its velocity may be positive, negative, or zero, depending on its position.





Variables Used to Describe Linear Motion

Variable Quantity MKS Unit

x position m

d, Δx distance m

d

, x displacement m

h height m

v

velocity sm

v

average velocity sm

The average velocity of an object is its displacement divided by the time, or its change in position divided by the (change in) time:

ttt

xx

to

ΔxΔxdv

(Note that elapsed time is always a difference )( t , though we usually use t

rather than t as the variable.)

We can use calculus to turn v into v by taking the limit as Δt approaches zero:

dt

dx

t

xLimvt

0

i.e., velocity is the first derivative of displacement with respect to time.

We can rearrange this formula to show that displacement is average velocity times time:

tvd

Position is the object’s starting position plus its displacement:

txxx oo vd

where x 0* means “position at time = 0”. This formula is often expressed as:

txx o vd

* x o is pronounced “x -zero” or “x -naught”.





Note that t

x

is the slope of a graph of position (x ) vs. time (t ). Because

vt

x

, this means that the slope of a graph of position vs. time is equal to the

velocity.

In fact, on any graph, the quantity you get when you divide the quantity on the x-axis by the quantity on the y-axis is, by definition, the slope. I.e., the slope is

x

y

, which means the quantity defined by

axis-

axis-

x

y will always be the slope.

Recall that velocity is a vector, which means it can be positive, negative, or zero.

On the graph below, the velocity is + sm4 from 0 s to 2 s, zero from 2 s to 4 s, and

sm2 from 4 s to 8 s.





Sample problems:

Q: A car travels 1200 m in 60 seconds. What is its average velocity?

A:

sm20

s06

m1200

v

t

dv

Q: A person walks 320 m at an average velocity of sm25.1 . How long did it take?

A: “How long” means what length of time.

stt

t

dv

256

2031.25

It took 256 seconds for the person to walk 320 m.

Add Important Linear Acceleration Page: 141




Linear Acceleration Unit: Kinematics (Motion)

NGSS Standards: N/A

MA Curriculum Frameworks (2006): 1.1, 1.2

AP Physics 1 Learning Objectives: 3.A.1.1, 3.A.1.3

Knowledge/Understanding Goals:

what linear acceleration means

what positive vs. negative acceleration means

Skills:

calculate position, velocity and acceleration for problems that involve movement in one direction

Language Objectives:

Understand and correctly use the term “acceleration.”

Accurately describe and apply the concepts described in this section using appropriate academic language.

Labs, Activities & Demonstrations:

Walk with different combinations of positive/negative velocity and positive/negative acceleration.

Drop a dollar bill or meter stick and have someone try to catch it.

Drop two strings of beads, one spaced at equal distances and the other spaced at equal times.

Drop a bottle of water with a hole near the bottom or bucket of ping-pong balls.

Notes:

acceleration: a change in velocity over a period of time.

uniform acceleration: when an object’s rate of acceleration (i.e., the rate at which its velocity changes) is constant.





If an object’s velocity is increasing, we say it has positive acceleration.

If an object’s velocity is decreasing, we say it has negative acceleration.

time

velo

cit

y

time

velo

cit

y

time

velo

cit

y

positive acceleration acceleration = zero negative acceleration

Note that if the object’s velocity is negative, then increasing velocity (positive acceleration) would mean that the velocity is getting less negative, i.e., the object would be slowing down in the negative direction.

Variables Used to Describe Acceleration

Variable Quantity MKS Units

a

acceleration 2s

m

g

acceleration due to gravity 2s

m

By convention, physicists use the variable g

to mean acceleration due to gravity,

and a

to mean acceleration caused by something other than gravity.





Because acceleration is a change in velocity over a period of time, the formula for acceleration is:

t

v

t

v

t

vva o

and, from calculus:

dt

dv

t

va

t

0Lim

The units must match the formula, which means the units for acceleration must be velocity (distance/time) divided by time, which equals distance divided by time squared.

Because dt

dxv , this means that acceleration is the second derivative of position

with respect to time: 2

2

)(dt

xd

dt

dx

dt

dv

dt

d

dt

dva

However, in an algebra-based physics course, we will limit ourselves to problems in which acceleration is constant.

We can rearrange this formula to show that the change in velocity is acceleration times time:

atvvv o

Note that when an object’s velocity is changing, the final velocity, v , is not the same as the average velocity, v . (This is a common mistake that first-year physics students make.)





t

v

is the slope of a graph of velocity (v ) vs. time (t ). Because a

t

v

, this

means that acceleration is the slope of a graph of velocity vs. time:

Note the relationship between velocity-time graphs and position-time graphs.

positive acceleration acceleration = zero negative acceleration

time

velo

cit

y

time

velo

cit

y

time

velo

cit

y

concave up linear concave down





Note also that tv is the area under a graph (i.e., the area between the curve and

the x-axis) of velocity (v ) vs. time (t ). Because dtv , this means the area under a graph of velocity vs. time is the displacement (Δx). Note that this works both for constant velocity (the graph on the left) and changing velocity (as shown in the graph on the right).

In fact, on any graph, the quantity you get when you multiply the quantities on the x- and y-axes is, by definition, the area under the graph.

In calculus, the area under a curve is the integral of the equation for the curve. This means:

t

dtvd0

where v can be any function of t .





In the graph below, between 0 s and 4 s the object is accelerating at a rate of

2sm2.5 .

Between 4 s and 6 s the object is moving at a constant velocity (of sm01 ), so

the acceleration is zero.

a = 2s

m2.5

m5)2)(5.2( 221 d

m5)5)(2(21 A

a = 2sm2.5

m20)4)(5.2( 221 d

m20)10)(4(21 A

a = 0 m20)2)(10( tvd

m20)10)(2( A

In each case, the area under the velocity-time graph equals the total distance traveled.





To show the relationship between v and ,v we can combine the formula for

average velocity with the formula for acceleration in order to get a formula for the position of an object that is accelerating.

atv

tvd

However, the problem is that v in the formula atv is the velocity at the end,

which is not the same as the average velocity v .

If the velocity of an object is changing (i.e., the object is accelerating), the average velocity, v (the line over the v means “average”), is given by the formula:

2

vvv o

If the object starts at rest (not moving, which means 0ov ) and it accelerates at

a constant rate, the average velocity is therefore the average of the initial velocity and the final velocity:

vvvvv

v o21

22

0

2

Combining all of these gives, for an object starting from rest:

221

21

21 )( attatvttvd

If an object was moving before it started to accelerate, it had an initial velocity,

or a velocity at time = 0. We will represent this initial velocity as ov*. Now, the

formula becomes:

221 attvdxx oo

* pronounced “v-zero” or “v-naught”

distance the object would travel at its initial velocity

additional distance the object will travel because it is accelerating





This equation can be combined with the equation for velocity to give the following equation, which relates initial and final velocity and distance:

advv o 222

Finally, when an object is accelerating because of gravity, we say that the object is in “free fall”.

On earth, the average acceleration due to gravity is approximately 2sm807.9 at

sea level (which we will usually round to 2sm10 ). Any time gravity is involved

(and the problem takes place on Earth), assume that 2s

m10ga .

Extensions

Just as a change in velocity is called acceleration, a change in acceleration with

respect to time is called “jerk”: t

aj

.

While questions about jerk have not been seen on the AP exam, some AP problems do require you to understand that the area under a graph of acceleration vs. time would be the change in velocity (Δv), just as the area under a graph of velocity vs. time is the change in position.





Homework Problems: Motion Graphs

1. An object’s motion is described by the following graph of position vs. time:

a. What is the object doing between 2 s and 4 s? What is its velocity during that interval?

b. What is the object doing between 6 s and 7 s? What is its velocity during that interval?

c. What is the object doing between 8 s and 10 s? What is its velocity during that interval?





2. An object’s motion is described by the following graph of velocity vs. time:

a. What is the object doing between 0 s and 2 s? What are its velocity and acceleration during that interval?

b. What is the object doing between 2 s and 4 s? What is its acceleration during that interval?

c. What is the object doing between 6 s and 9 s? What is its acceleration during that interval?



Use this space for summary and/or additional notes:


3. The graph on the left below shows the position of an object vs. time. Sketch a graph of velocity vs. time for the same object on a graph similar to the one on the right.

4. In 1991, Carl Lewis became the first sprinter to break the 10-second barrier for the 100 m dash, completing the event in 9.86 s. The chart below shows his time for each 10 m interval.

distance (m) 0 10 20 30 40 50 60 70 80 90 100

time (s) 0 1.88 2.96 3.88 4.77 5.61 6.45 7.29 8.12 8.97 9.86

Plot Lewis’s displacement vs. time and velocity vs. time on graphs similar to the ones below.

linear motion, speed & velocity page: unit: kinematics ... · add important linear motion,...

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