position-time and velocity-time graphs uniform motion...

Position-time and velocity-time graphs

Uniform motion problems – algebra

Acceleration and displacement

The kinematics of motion in one dimension: graphing and calculations

Problem-solving strategies

Free fall

Topics:

Sample question:

Horses can run much much faster than humans, but if the length

of the course is right, a human can beat a horse in a race. When,

and why, can a man outrun a horse?

Speed, Velocity, & Acceleration

Speed (v) – how fast you go (scalar)

Velocity (v) – how fast and which way; the rate at which position changes (vector)

Average speed ( v ) – distance / time

Acceleration (a) – how fast you speed up, slow down, or change direction; the rate at which velocity changes

During your 8 mi. trip, which took 15 min., your speedometer displays your instantaneous speed, which varies throughout the trip.

Your average speed is 32 mi/hr.

Your average velocity is 32 mi/hr in a SE direction.

At any point in time, your velocity vector points tangent to your path.

The faster you go, the longer your velocity vector.

Acceleration Acceleration – how fast you speed up, slow down, or

change direction; it’s the rate at which velocity changes. Two examples:

t (s) v (mph)

0 55

1 57

2 59

3 61

t (s) v (m/s)

0 34

1 31

2 28

3 25

a = +2 mph / s a = -3 m / s

s = -3 m / s

2

Velocity & Acceleration Sign Chart

V E L O C I T Y

A

C

C

E

L

E

R

A

T

I

O

N

+ -

+ Moving forward;

Speeding up

Moving backward;

Slowing down

- Moving forward;

Slowing down

Moving backward;

Speeding up

Displacement is

a change of position

in a certain direction,

not

the total distance traveled

Motion Takes place over time

Depends upon the frame of reference

What is frame of reference?

How do we choose one?

Once we decide upon a frame of

reference, it remains fixed for the problem

Keep in mind… Displacement is not always equal to the

distance traveled!

Displacement can be positive or negative!

For motion along x or y axis, the displacement is determined by the x or y coordinate of its final position. Example: Consider a car that travels 8 m, E then 12 m, W.

Net displacement Δx, is

from the origin to the final

position:

What is the

distance traveled?

d = 20 m !!

12 m,W

Δx

Δx = 4 m, W

x 8 m,E

x = +8 x = -4

Displacement When things start moving, the length of a

straight line drawn from the object’s initial position to it’s final position is it’s displacement

In one dimension…

if xxx

Change in position along x-axis = (final position on x-axis) – (initial position on x-axis)

Motion in One Dimension Same concepts apply to any axis!

if yyy

Change in position along y-axis = (final position on y-axis) – (initial position on y-axis)

POSITIVE AND NEGATIVE QUANTITIES

ARE JUST DIRECTIONS IN MOTION PROBLEMS!

Position-time graphs

the relationship

between the

shape of a p-t

graph and the

motion of the

object.

consider a car moving with a constant,

rightward (+) velocity of +10 m/s.

If the position-time data for

the car were graphed, it

would look like the graph at

the right.

If the position-time data for such

a car were graphed, the graph

would look like the graph at the

right.

Now consider a car moving with a rightward (+),

changing velocity - that is, a car that is moving

rightward but speeding up or accelerating.

Constant Velocity Positive Velocity

Positive Velocity Changing Velocity

(acceleration)

.

Position-time graphs for the two types of

motion - constant velocity and changing

velocity (acceleration).

http://www.physicsclassroom.com/Class/1DKin/U1L1e.cfm

Provide a reasonable explanation for the motion diagram.

Draw a position-time graph.

Motion diagram (student walking to school)

(time on the x-axis)

Motion diagram (student walking to school)

Table of data Graph

Interpreting position-time graphs Starting at t = 0:

• Position away from the origin in + direction has + velocity

• Position away from the origin in – direction has a – velocity

• No change in position – 0 velocity

x

t

A

B

C

A … Starts at home (origin) and goes forward slowly

B … Not moving (position remains constant as time

progresses)

C … Turns around and goes in the other direction

quickly, passing up home

Describe this motion.

p-t graph

THS

x E

W

A

B

C

D

Practice problems 1-3, pg. 85

Uniform motion (constant velocity)

Equal displacements occur during

successive equal time intervals.

the slope of the line on a position-time

graph reveals useful information about

the velocity of the object.

The slope of the line on a position-time graph reveals

useful information about the velocity of the object.

The importance of Slope

Whatever characteristics the velocity has, the

slope will exhibit the same (and vice versa). If the

velocity is constant, then the slope is constant (i.e.,

a straight line). If the velocity is changing, then the

slope is changing (i.e., a curved line). If the

velocity is positive, then the slope is positive (i.e.,

moving upwards and to the right). This very

principle can be extended to any motion

The object represented by the graph on the right is traveling

faster than the object represented by the graph on the left.

Slow, Rightward(+) Constant Velocity

Fast, Rightward(+) Constant Velocity

Negative (-) Velocity Slow to Fast

Leftward (-) Velocity Fast to Slow

Consider the two graphs below. Both graphs show plotted points forming a curved line.

The curved line of changing slope is a sign of accelerated motion (i.e., changing velocity).

Applying the principle of slope to the graph on the left, the object depicted by the graph is

moving with a negative velocity (since the slope is negative ). The object is starting with a

small velocity (small slope) and finishes with a large velocity. This object is moving in the

negative direction and speeding up.

The graph on the right also depicts negative velocity (since there is a negative

slope). The object begins with a high velocity (the slope is initially large) and

finishes with a small velocity (since the slope becomes smaller). So this object

is moving in the negative direction and slowing down.

Applying algebra to uniform motion

The slope of the line on a position-time graph

is equal to the velocity of the object.

If the object is moving with a velocity of +2 m/s, then the slope of the

line will be +2 m/s.

Slope = rise/run

For p-t graph rise = ∆d run = ∆t

slope of the line = ave. velocity

V ave = ∆d = d1 – d0

∆t t1 – t0

ave. velocity is ∆d NOT d*

∆t t

*instantaneous speed

book, pg.86

0

-2

2

4

-4

0.2 0.4 0.6 0.8 s

m

A

A va = 2 - (-2) = 4 m = 10 m/s

0.4 – 0 .4

B

B va = -4 - (-2) = -2 m = -5 m/s

0.4 – 0 .4

For Uniform motion the derivative:

Vave = ∆d = d1 – d0

∆t t1 – t0

When t0 = 0, it can be eliminated from the problem

Vave = d1 – d0

t1

d1 = Vave (t1) + d0

Vave = v because constant velocity – so no need for ave.

d = vt + d0

d0 represents position at t = 0

An airplane is moving at a uniform velocity of 75 m/s from a

starting position of 40 m. Find the position of the airplane after 2.5

seconds.

d = vt + d0

d = 40 m + (75 m/s)(2.5 s)

d = 230 m

practice problems 9-12, pg. 89

http://4.bp.blogspot.com/_jUAaH9qqHrE/TPSB4wP_J6I/AAAAAAAAA9c/VWr9mdo0y98/s1600/cartoon-airplane.jpg

The Cheetah: A cat that is built for speed. Its

strength and agility allow it to sustain a top speed of

over 100 km/h. Such speeds can only be maintained

for about ten seconds.

graphing velocity

Definition of Speed Speed is the distance traveled per unit of

time (a scalar quantity).

vs = = d

t

20 m

4 s

s = 5 m/s

Not direction dependent!

A

B d = 20 m

Time t = 4 s

Definition of Velocity Velocity is the displacement per unit

of time. (A vector quantity.)

Direction required!

A

B s = 20 m

Time t = 4 s

Δx=12 m

20o = 3 m/s at 200 N of E

Average Speed and Instantaneous Velocity

The instantaneous

velocity is the magnitude

and direction of the

velocity at a particular

instant. (v at point C)

The average speed depends ONLY on

the distance traveled and the time

required.

A

B s = 20 m

Time t = 4 s

C

Example: A runner runs 200 m, east, then changes direction and runs 300 m, west. If the entire trip takes 60 s, what is the average speed and what is the average velocity?

Recall that average speed is a

function only of total distance

and total time:

500 m

60 s

total pathAverage speed

time

Avg. speed= 8 m/s Direction does not matter!

start

s1 = 200 m s2 = 300 m

Now we find the average velocity, which is the net displacement divided by time. In this case, the direction matters.

xo = 0

t = 60 s

x1= +200 m x = -100 m

xo = 0 m; x = -100 m

Direction of final

displacement is to the left as

shown.

Average velocity:

Note: Average velocity is directed to the west.

Graphing velocity: uniform motion (constant velocity)

is represented by a horizontal line on a v-t graph.

Increasing speed?

A

B

C

D

0

+

V (m/s)

Constant speed?

Increasing speed?

Backwards and increasing

speed?

Backwards at a constant

speed?

Decreasing speed?

? What does the intersection of 2 lines on a v-t graph tell us?

Note: v-t graphs give NO information about position.

Here is a motion diagram of a car moving along a straight stretch

of road:

Which of the following velocity-versus-time graphs matches this

motion diagram?

Checking Understanding

A. B. C. D.

15

A graph of position versus time for a

basketball player moving down the

court appears like so:

Which of the following velocity graphs matches the above

position graph?

A. B. C. D.

A graph of velocity versus time for a

hockey puck shot into a goal appears

like so:

Which of the following position graphs matches the above

velocity graph?

A. B. C. D.

displacement vs. time velocity vs. time acceleration vs. time

slope slope

Area underneath = Δv Area underneath = Δd

displacement vs. time velocity vs. time acceleration vs. time

slope slope

Area underneath = Δv Area underneath = Δd

p-t

time (s)

area: none

slope: m/s = v Rate at which

displacement

changes

v-t

time (s)

m/s x s = m = Δd

m/s/s = a Rate at which

velocity changes

a-t

time (s)

m/s2 x s = m/s = Δv

Acceleration

The rate of change of velocity

Think about this....

What are three ways to change the velocity of a car?

Accelerate

Decelerate

Change direction

average acceleration is the rate of change in velocity

between t0 and t1 a = ∆v

∆t

If a car moves at a

constant velocity, then its

acceleration is zero

acceleration = change in velocity

change in time

acceleration (m/s2) = (vf) - (vi) time

These four motion diagrams show the motion of a particle along

the x-axis. Which motion diagrams correspond to a positive

acceleration? Which motion diagrams correspond to a negative

acceleration?

Slide 2-22


These four motion diagrams show the motion of a particle along

the x-axis. Rank these motion diagrams by the magnitude of the

acceleration. There may be ties.


Slide 2-21

These six motion diagrams show the motion of a particle along

the x-axis. Rank the accelerations corresponding to these motion

diagrams, from most positive to most negative. There may be

ties.


Slide 2-23

©2008 by W.H. Freeman and

Company

Acceleration

There is a difference between negative acceleration and

deceleration:

Negative acceleration is acceleration in the negative direction as

defined by the coordinate system.

Deceleration occurs when the acceleration is opposite in direction to

the velocity.

Acceleration acceleration is the rate of change of velocity.

t

v

tt

vva

12

12

©2008 by W.H. Freeman

and Company

Acceleration

Acceleration is a vector, although in one-

dimensional motion we only need the sign.

here is an example of deceleration:

a = v2 – v1

∆t

Helpful rule:

If the sign of both the acceleration and the

velocity are the same = speeding up

If the signs of the acceleration and

velocity are different = slowing down

THINK ABOUT THIS

Why do

highways

have speed

limits and not

velocity limits?

or deriving for dum dums

Quick calculation: 1. A plane travels in a single direction on a runway. It reaches the end of the runway in 20 seconds and its velocity is 80 m/s. What is its acceleration?

2. A skateboarder is traveling at 8 m/s. He slows and comes to a stop in 4 sec. What was the acceleration?

+ 4 m/s2

- 2 m/s2

3. A sailboat is moving at 12 m/s when a gust of wind changes its velocity to 18 m/s. The wind lasts 10 seconds. For that 10 sec, what was its acceleration?

0.6 m/s2

4. A speed skater accelerates at 5 m/s2. What will his speed be 4 seconds later?

20 m/s

At highway speeds, a particular

automobile is capable of an

acceleration of about 1.6 m/s2. At

this rate, how long does it take to

accelerate from 80 km/h to 110

km/h?

.

The time can be found from the average acceleration,

va

t

2 2

1m s30 km h

3.6 km h110 km h 80 km h5.208s 5 s

1.6 m s 1.6 m s

vt

a

One more….

A chicken runs across the road at a speed of 12 m/s. If the road is 36 meters across, how long does it take for the chicken to cross the road?

3 s

A. Displacement with constant acceleration

B. Velocity with constant acceleration

C. Displacement with constant acceleration

D. Final velocity after any displacement

Displacement with constant acceleration

Displacement depends on acceleration, velocity and time.

Vavg = Δd

Δt

v-t

time (s) 0

0

V (cm/s)

vi

vf

vavg

For an object moving with constant

acceleration, the average velocity is equal to

the average of the initial and final velocities.

Δd = Vavg = vi + vf

Δt 2

Δd = ½ (vi + vf) Δt

Multiply both sides by ∆t

Try one:

A race car reaches a speed of 42 m/s. It then begins a

uniform negative acceleration, using its parachute and

breaking system, and comes to rest 5.5 s later. Find how

far the car moves while stopping.

Displacement with Constant Acceleration

A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance the car travels during this time.

20 m

One more:

When Maggie applies the brakes of her car, the car

slows uniformly from 15.00 m/s to 0.00 m/s in 2.50 s.

How many meters before a stop sign must she apply

the brakes in order to stop at the sign?

18.8 m

Velocity with Constant Acceleration

Rearrange the equation for acceleration to find vf

Multiply by ∆t and add the initial velocity to both sides

Final velocity = initial velocity + (acceleration x time interval)

a = vf -vi = vf -vi tf - ti ∆t

vf = vi + a ∆t


Substitute this expression in for vf into the displacement with constant acceleration equation:

Distribute ∆t

vf = vi + a ∆t

∆d = 1 (vf - vi) ∆t 2

∆d = 1 (vi + a∆t +vi) ∆t 2

∆d = 1 [2vi ∆t + a(∆t)2] 2

∆d = vi ∆t + 1 a (∆t)2 2


∆d = vi ∆t + 1 a (∆t)2 2

This equation is useful not only for finding how far an object travels under constant acceleration but also for finding the distance required for an object to reach a certain speed or to come to a stop

Practice Problems:

A boat with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed and the displacement of the boat during this time.

Hint: you need to solve using two equations

28.9 m

35.6 km/hr

displacement/velocity when time is unknown

∆d = 1 (vf + vi) ∆t 2

You can find the final velocity of an accelerating object without knowing how long it has been

accelerating.

2∆d = (vf + vi) ∆t

multiply by 2

2∆d = ∆t

(vf + vi)

You can use this equation to derive another equation for finding final velocity after any displacement.

Final Velocity after any Displacement

substitute the expression for ∆t into the final v equation

vf = vi + a ∆t vf = vi + a ( 2∆d) (vf + vi)

Subtract vi from

both sides vf -vi = a ( 2∆x) (vf + vi)

Multiply both

sides by (vf +

vi) to get

velocities on the

same side

vf2 - vi

2 = 2a∆x

Solve

for vf2

vf2 = vi

2 + 2a∆d

2∆d = ∆t

(vf + vi)

Final Velocity after any Displacement cont.

When using this equation remember you MUST take the square root of the right side to find the final velocity!

practice

A baby sitter pushing a stroller starts from rest and accelerates at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

vf2 = vi

2 + 2a∆d

? vf ? vi = 0 m/s a = 0.500 m/s2 ∆d = 4.75m

x +

vf2 = (0 m/s)2 + 2(0.500 m/s2)(4.75 m)

vf2 = 4.75m2/s2

vf = 4.75m2/s2 = + 2.18 m/s

2. An electron is accelerated from rest in an accelerator at 4.5 X 107 m/s2 over a distance of 95 km. Assuming constant acceleration, what is the electron’s final velocity?

vf2 = vi

2 + 2a∆d

watch your units!

9.2 x 105 m/s

2 part acceleration problems

Dinner at a Distance, Part I

Chameleons catch insects with their tongues, which they can

extend to great lengths at great speeds. A chameleon is aiming for

an insect at a distance of 18 cm. The insect will sense the attack

and move away 50 ms after it begins. In the first 50 ms, the

chameleon’s tongue accelerates at 250 m/s2 for 20 ms, then

travels at constant speed for the remaining 30 ms. Does its tongue

reach the 18 cm extension needed to catch the insect during this

time?

Slide 2-24

Dinner at a Distance, Part II

Cheetahs can run at incredible speeds, but they can’t keep up

these speeds for long. Suppose a cheetah has spotted a gazelle.

In five long strides, the cheetah has reached its top speed of 27

m/s. At this instant, the gazelle, at a distance of 140 m from the

running cheetah, notices the danger and heads directly away.

The gazelle accelerates at 7.0 m/s² for 3.0 s, then continues

running at a constant speed that is much less than the cheetah’s

speed. But the cheetah can only keep running for 15 s before it

must break off the chase. Does the cheetah catch the gazelle, or

does the gazelle escape?

Slide 2-25

What is my acceleration right now?

According to Galileo

With no air resistance, all freely falling objects have the same constant acceleration.

The Mechanical Universe

The magnitude of acceleration of a falling object:

g= -9.81 m/s/s

Aug. 2, 1971, astronaut David Scott

Galileo and Apollo

©2008 by W.H. Freeman and Company

Falling Objects

Near the surface of the Earth, all objects experience

approximately the same acceleration due to gravity.

This is one of the most

common examples of

motion with constant

acceleration.

Consider the position of the free-falling object at regular time intervals, every 1 second.

The fact that the distance which the ball travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward.

It is accelerating!

Falling Objects

The acceleration due to

gravity at the Earth’s

surface is approximately

9.80 m/s2.

Misconceptions about falling objects.

Because acceleration due to gravity is directed downward to the center of the earth, acceleration of a falling object is always negative - even when the object is thrown up into the air!

+

+

Based on the usual axis, downward

direction is -.

-9.81 m/s2

A curved line on a position vs. time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration of g = -9.81 m/s/s, you would expect that its position-time graph would be curved.

A closer look at the position-time graph - the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object, the initial small slope indicates a small initial velocity and the final large slope indicates a large final velocity.

Last, but not least, the negative slope of the line indicates a negative (i.e., downward) velocity.

the velocity-time graph shows the object starts with a zero velocity (from rest) and finishes with a large, negative velocity (the object is moving in the negative direction and speeding up). An object moving in the negative direction and speeding up has a negative acceleration.

Analysis of the slope of the graph is consistent with this – moving

with a constant acceleration of -9.81 m/s2.

Free Fall

Objects thrown in the air have a + velocity and – acceleration.

This means the object is slowing down . The object moves upward but with smaller and smaller velocity.

At the top of its path, v has decreased and is momentarily 0. Now it moves downward with – v and still – a (this means the object is speeding up!)

If everything accelerates at the same rate, does that mean everything falls at the same rate?

Even if they have a different weight? yes

Even if they are different sizes? yes

Even if they are different shapes? not if you include air resistance

If you put it in a vacuum, then yes

Free Fall Acceleration

Objects moving toward earth move faster and faster as they fall.

Acceleration Due to Gravity Every object on the earth

experiences a common force: the force due to gravity.

This force is always directed toward the center of the earth (downward).

The acceleration due to gravity is relatively constant near the Earth’s surface.

Earth

W g

Gravitational Acceleration In a vacuum, all objects fall

with same acceleration.

Equations for constant acceleration apply as usual.

Near the Earth’s surface:

a = g = -9.80 m/s2 (-32 ft/s2 )

Directed downward (usually negative).

Acceleration due to Gravity

a = -g = -9.8 m/s2 9.8 m/s2

Near the surface of the

Earth, all objects accelerate

at the same rate (ignoring air

resistance).

Interpretation: Velocity decreases by 9.8 m/s each second,

meaning velocity is becoming less positive or more

negative. Less positive means slowing down while going

up. More negative means speeding up while going down.

This acceleration

vector is the

same on the way

up, at the top,

and on the way

down!

Example: A ball is thrown vertically upward with an initial velocity of 30.0 m/s. What are its position and velocity after 4.00 s, and 7.00 s? Find also the maximum height attained

a = g

+

vo = +30.0 m/s

Given: a = -Δ9.8 m/s2

vo = 30.0 m/s

t = 4.00 s; 7.00 s

Find:

Δy = ? – displacement

v = ? - final velocity

After those three “times”

Δy = ? – maximum height

Given: a = -9.8 m/s2; vo = 30.0 m/s

t = 2.00 s; 4.00 s; 7.00 s

Solutions:

For t = 4.00 s:

For t = 7.00 s:

Given: a = -9.8 m/s2; vo = 30.0 m/s

t = 2.00 s; 4.00 s; 7.00 s

Solutions:

For maximum height, v = 0 (the ball stops at maximum height):

Can use equations for constant acceleration to solve problems.

Example 1: A flower pot falls from a window 25.0 m above the sidewalk a. How fast is it moving when it hits the ground? b. How much time does someone on the ground have to get out of the way?

1. diagram the problem

25 m

2. Define what you know, and what you’re looking for

Δd = 25 m vi = 0 m/s a = -9.8 m/s2

?vf ? ?∆t?

vf2 = vi

2 + 2a∆y

3. Choose the equation(s)

vf = vi

+ a∆t

Example 1: A flower pot falls from a window 25.0 m above the sidewalk a. How fast is it moving when it hits the ground? b. How much time does someone on the ground have to get out of the way?

4. Solve !

25 m

a. vf2 = vi

2 + 2a∆y

vf2 = 0 + 2(-9.8)(25)

vf = √490 = 22 m/s

b. vf = vi

+ a∆t

∆t = vf /a

= 22 m/s

-9.8 m/s2

= 2.25 s

Example 2: Jan sets a volleyball straight up, leaving her hands at 12.0 m/s. a. How high does it go? b. If, when the ball is on the way down, it is spiked at the same height at which it was set, how long was it in flight? c. How fast is it traveling when it is spiked?

1. diagram the problem

floor

0

Place starting point of the ball at the

origin (yi = 0 at t = 0)

2. Define what you know, and what you’re looking for

di = 0 m vi = 12 m/s a = -9.8 m/s2 ?vf ? ?∆t?

floor

0

vi = 12 m/s

a = - 9.81 m/s2

4. solve! a. At the top of its flight, the ball has instantaneous

velocity of 0. vf = 0

so: 0 = 122 + 2(-9.8 m/s) ∆y

= 144 + (-19.6)Δy

Δy = 7.35 m

3. Choose the equation(s)

vf2 = vi

2 + 2a∆y

floor

0

b. To analyze the rest of the problem: remember that the down half of the trip is a mirror image of the up half. So, if, while going up, the ball passes through a particular height at a particular velocity, on its way down it will pass through that height at the same speed (with its velocity directed down). This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its max height, and then double that to get the total time.

OR, you can do the math: df = di + vit + ½ at2

0 = 0 + 12t + ½(-9.8)t2

0 = 12 – 4.9t t = 12/4.9 = 2.45 s

c. The answer has to be 12 m/s down, because of the mirror-image

relationship between the up half of the flight and the down half.

OR, you can do the math: vf = vi

+ a∆t

vf = 12 + -9.8(2.45s)

= -12m/s

Practice!

Reviewing Concepts

Pg. 107-108, 1, 2, 4, 6-9, 12, 17-24

website the Physics Classroom

Practice Problems

pg. 109 –

Problems 27, 28, 34, 35, 40, 43, 45-47

Questions like #49 – 59, 64- 76

Lab Time!!

Determining acceleration

position-time and velocity-time graphs uniform motion...

Documents