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CHAPTER 2

Describing Motion: Kinematics in One Dimension

http://www.physicsclassroom.com/Class/1DKin/1DKinTOC.html

• Reference Frames and Displacement • Average Velocity • Instantaneous Velocity • Acceleration • Motion at Constant Acceleration • Solving Problems • Falling Objects • Graphical Analysis of Linear Motion

DISPLACEMENT There are two aspects to any motion. There is the movement itself and there is the issue of what caused the motion or what changed it. This requires that forces be considered. Kinematics deals with the concepts that are needed to describe motion, without any reference of forces. Dynamics deals with the effect that forces have on motion. Together, kinematics and dynamics form the branch of physics known as mechanics.

Quantities in Motion

• Any motion involves three concepts

– Displacement – Velocity – Acceleration

• These concepts can be used to study objects in motion To describe the motion of an object its location must be known at all times. Position

• Defined in terms of a frame of reference

– One dimensional, so generally the x- or y-axis – Defines a starting point for the motion

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Displacement

• Displacement is a vector that points from an object’s initial position to its final position and has a magnitude that equals the shortest distance between the two positions. The SI unit for displacement is the meter (m).

Scalars are quantities that are fully described by a magnitude (or numerical value) alone.

Vectors are quantities that are fully described by both a magnitude and a direction.

Generally denoted by boldfaced type and an arrow over the letter + or – sign is sufficient for this chapter

_

f stands for final and i stands for initial

– May be represented as y if vertical – Units are meters (m) in SI, centimeters (cm) in cgs or feet (ft) in US

Customary

Example of Position, Distance, and Displacement

The distance is the total length of travel; if you drive from your house to the grocery store and back, you have covered a distance of 8.6 mi.

Displacement is

the change in position. If you drive from your house to the grocery store and then to your friend’s house, your displacement is 2.1 mi and the distance you have traveled is 10.7 mi.

Displacement Isn’t Distance

• The displacement of an object is not the same as the distance it travels

– Example: Throw a ball straight up and then catch it at the same point you released it

• The distance is twice the height • The displacement is zero

f ix x x

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Speed

• The average speed of an object is defined as the total distance traveled divided by the total time elapsed. If a motorcycle travels 300 meters in 10 seconds the average speed is 30 m/s. The average speed is the distance traveled divided by the time required to cover the distance traveled divided by the time required to cover the distance.

– Speed is a scalar quantity

• Average speed totally ignores any variations in the object’s actual motion during the trip

• The total distance and the total time are all that is important • SI units are m/s

Velocity

• It takes time for an object to undergo a displacement • The average velocity is rate at which the displacement occurs

generally use a time interval, so let ti = 0

• Direction will be the same as the direction of the displacement (time interval is always positive)

– + or - is sufficient • Units of velocity are m/s (SI), cm/s (cgs) or ft/s (US Cust.)

– Other units may be given in a problem, but generally will need to be converted to these

Speed vs. Velocity

• Cars on both paths have the same average velocity since they had the same displacement in the same time interval

• The car on the blue path will have a greater average speed since the distance it traveled is larger

total distanceAverage speed

total time

dv

t

f iaverage

f i

x xxv

t t t

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Graphical Interpretation of Velocity

• Velocity can be determined from a position-time graph • Average velocity equals the slope of the line joining the initial and final positions • An object moving with a constant velocity will have a graph that is a straight line

Average Velocity, Constant

• The straight line indicates constant velocity • The slope of the line is the value of the

average velocity

Example 1: The position of a runner as a function of time is plotted as moving along the x axis of a coordinate system. During a 3.00 s time interval, the runner’s position

changes from1 50x m to

2 30.5x m . What was the runner’s average velocity?

given find formula solution

3.00t s

(Note: distance (x) is to the left on the x-axis that is why it is negative) Example 2: You step on a nail with your bare feet. A nerve impulse, generated in your foot, travels through your nervous system at an average speed of 110 m/s. How much time does it take for the impulse, which travels a distance of 1.8 m, to reach your brain? given find formula solution

v = 110 m/s t x

tv

1.8

0.016sec110 /

mt

m s

x = 1.8 m Example 3: A plane is sitting on a runway, awaiting takeoff. On an adjacent parallel runway, another plane lands and passes the stationary plane at a speed of 45 m/s. The arriving plane has a length of 36 m. By looking out of a window, a passenger on the stationary plane can see the moving plane. For how long a time is the moving plane visible? given find formula solution

v = 45 m/s t x

tv

36

0.8sec45 /

mt

m s

x = 36 m

1 50.0x m v2 1x x

vt

2 30.5x m

19.56.50 /

3.00

mv m s

s

5

Example 4: How far can a cyclist travel in 2.5h along a straight road if her average velocity is 18km/h? given find formula solution t = 2.5h x v = 18km/h

Instantaneous Velocity

• The limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero

• The instantaneous velocity indicates what is happening at every point of time

This plot shows the average velocity being measured over shorter and shorter intervals. The instantaneous velocity is tangent to the curve.

Example 5: Obtain average and instantaneous velocities from a graph.

A train moves slowly along a straight track (see graph). Find: (a) the average velocity for the total trip. Find the average velocity from the origin to the final point c. Calculate the slope of the dashed blue line:

(b) The average velocity during the first 4.00 s of motion.

(c) The average velocity during the next 4.00 s of motion.

(d) The instantaneous velocity at 2.00 s. This is the same as the average velocity found in (b), because the graph is a straight line.

x v t (18 / )(2.5 ) 45x km h h km

lim

0t

xv

t

110.00.833 / ( 0.833 )

12.0

x mv m s or ms

t s

14.001.00 / ( 1.00 )

4.00

x mv m s or ms

t s

100 / ( 0 )

4.00

x mv m s or ms

t s

1.00 /v m s

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(e) The instantaneous velocity at t = 9.00 s.

The tangent line appears to intercept the x-axis at (3.0s, 0m) and graze the curve at (9.0s, 4.5m). The instantaneous velocity at t = 9.00 s equals the slope of the tangent line through these points.

Acceleration

• Changing velocity (non-uniform) means an acceleration is present

• Acceleration is the rate of change of the velocity

Units are m/s² (SI), cm/s² (cgs), and ft/s² (US Cust) Average Acceleration

• Vector quantity • When the sign of the velocity and the acceleration are the same (either positive

or negative), then the speed is increasing • When the sign of the velocity and the acceleration are in the opposite directions,

the speed is decreasing

Average acceleration: Acceleration is the rate of change of velocity.

Typical Accelerations (m/s2)

Ultracentrifuge Bullet fired from a rifle Batted baseball Airbag deployment Bungee jump High jump Gravity on Earth Emergency stop in a car Airplane during takeoff An elevator Gravity on the Moon

3 x 106 4.4 x 105 3 x 104 60 30 15 9.80 8 5 3 1.62

4.5 00.75 /

9.0 3.0

x m mv m s

t s s

f i

f i

v vva

t t t

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Acceleration is a vector, although in one-dimensional motion we only need the sign. The previous image shows positive acceleration; here is negative 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.

Graphical Interpretation of Acceleration

• Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph

• Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph

Relationship between Acceleration and Velocity

• • • • • • Uniform velocity (shown by red arrows maintaining the same size) • Acceleration equals zero

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• Velocity and acceleration are in the same direction • Acceleration is uniform (blue arrows maintain the same length) • Velocity is increasing (red arrows are getting longer) • Positive velocity and positive acceleration •

• Acceleration and velocity are in opposite directions • Acceleration is uniform (blue arrows maintain the same length) • Velocity is decreasing (red arrows are getting shorter) • Velocity is positive and acceleration is negative

Example 6: A car accelerates along a straight road from rest to 75 km/h in 5.0 s. What is the magnitude of its average acceleration? given find formula solution (Note: Convert km/h to m/s = dividing km/h by 3.6 will give you m/s) Example 7: A car is moving along a straight road. The driver puts on the brakes when the initial velocity is 15.0 m/s, and it takes 5.0 s to slow down to 5.0 m/s. What was the car’s average acceleration? given find formula solution Example 8: A baseball player runs to the outfield. His velocity as a function of time is shown in the graph. Find his instantaneous acceleration at points A, B, C. At each point, the velocity vs. time graph is a straight line segment so instantaneous acceleration will be the slope. Acceleration at A: The acceleration equals the slope of the line connecting the points (0s, 0m/s) and 2.0s, 4.0m/s)

1 0 /v m sa 2 1

2 1

v va

t t

2 75 / 20.8 /v km h m s

220.8 / 0 /4.16 /

5.0 0

m s m sa m s

s s

1 15.0 /v m s

2 5.0 /v m s

5.0t s

a2 1

2 1

v va

t t

25.0 / 15.0 /2.0 /

5.0

m s m sa m s

s

4.0 / 0

2.0 0

v m sa

t s

22.0 /m s

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Acceleration at B: , because the segment is horizontal Acceleration at C: The acceleration equals the slope of the line connecting the points (3.0s, 4.0m/s) and (4.0s, 2.0m/s) Kinematic Equations

• Used in situations with uniform acceleration (THESE ARE KEY EQUATIONS)

Notes on the equations

Shows velocity as a function of acceleration and time

• Use when you don’t know and aren’t asked to find the displacement

• Gives displacement as a function of time, velocity and acceleration • Use when you don’t know and aren’t asked to find the final velocity

Gives velocity as a function of acceleration and displacement

• Use when you don’t know and aren’t asked for the time

THE BIG SIX Missing variable

2

2

2 2

1

2

1

2

2 ( )

1x ( )

2

o

o o

o o

o o

o

x vt

v v at

x x v t at

x x v t at

v v a x x

v v t

a

x

v vo

t

0v 4.0 / 4.0 /

3.0 2.0

v m s m sa

t s s

20 /m s

22.0 / 4.0 /2.0 /

4.0 3.0

v m s m sa m s

t s s

ov v at

21

2ox v t at

2 2 2ov v a x

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Problem-Solving Hints

• Read the problem • Draw a diagram

– Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations

• Label all quantities, be sure all the units are consistent – Convert if necessary

• Choose the appropriate kinematic equation • Solve for the unknowns

– You may have to solve two equations for two unknowns • Check your results

– Estimate and compare – Check units

Example 9: You are designing an airport for small planes. One kind of plane must reach a speed of 27.8 m/s before takeoff, and is able to accelerate at . (a) If the runway is 150 m long, can this plane reach the required speed for take off? given find formula solution

This runway length is not sufficient. (b) What minimum length must the runway have?

2 2 2 ( )o ov v a x x

Example 10: How long does it take a car to cross a 30.0 m wide street after the light turns green, if the car accelerates from rest at a constant ? given find formula solution

30.0x m t 22.00 /a m s

Example 11: Airbags can protect drivers in a head-on collision at a speed of 100km/h. Estimate how fast the air bag must inflate to protect the driver. (distance car crumples is 1 m) given find formula solution To be effective, the air bag would need to inflate faster than this.

22.00 /m s

2

0

0

150

2.00 /

o

o

x

v

x m

a m s

v 2 2 2 ( )o ov v a x x 2 2 20 2(2.0 / )(150 ) 600 /m s m m s

2 2600 / 24.5 /v m s m s

2 2 2

2

(27.8 / ) 0( ) 193

2 2(2.0 / )

oo

v v m sx x m

a m s

22.00 /m s

21

2x at 2x

ta

2

2 2(30.0 )5.48

2.00 /

x mt s

a m s

100 / 2.8 /ov km h m s

0v

1x m

a

t

2

2

ova

x

ov vt

a

2 22(28 / )

390 /2 2.0

ov m sa m s

x m

2

0 28 /0.07

390 /

ov v m st s

a m s

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Example 12: A race car starting from rest accelerates at a constant rate of . What is the velocity of the car after it has traveled 30.5m? given find formula solution a = 5.00m/s2 v x = 30.5 m Example 13: A jet taking off from the deck of an aircraft carrier starts from rest and is catapulted with a constant acceleration of 31 m/s2 along a straight line. The jet reaches a velocity of 62 m/s. Find the displacement of the jet. given find formula solution

a = 31 m/s2 x 2 2 2 ( )o ov v a x x

v = 62 m/s 2 2

2

ov vx

a

2

2

(62 / ) 062

2(31 / )

m sx m

m s

Example 14: For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200. Evolution went from zero to 26.8 m/s in 3.275 s. Find the magnitude of the car’s acceleration. given find formula solution vo = 0 m/s a

v = 26.8 m/s ov v at ov v

at

226.8 / 0

8.18 /3.275

m sa m s

s

t = 3.275 s

Galileo Galilei

• 1564 - 1642 • Galileo formulated the laws that govern the

motion of objects in free fall • Also looked at:

– Inclined planes – Relative motion – Thermometers – Pendulum

Free Fall

• All objects moving under the influence of gravity only are said to be in free fall – Free fall does not depend on the object’s original motion

• All objects falling near the earth’s surface fall with a constant acceleration • The acceleration is called the acceleration due to gravity, and indicated by g.

25.00 /m s

2 2 2ov v a x

2 2ov v a x 2 2(0) 2(5.00 / )(30.5 )v m s m

17.5 /m s

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Acceleration due to Gravity

• Symbolized by g • g = 9.80 m/s²

– When estimating, use g » 10 m/s2 • g is always directed downward

– toward the center of the earth • Ignoring air resistance and assuming g doesn’t vary with altitude over short vertical distances, free

fall is constantly accelerated motion

Falling Object 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.

Freely Falling Objects Free fall is the motion of an object subject only to the influence of gravity. The acceleration due to gravity is a constant, g.

In the absence of air resistance, all objects fall with the same acceleration, although this may be hard to tell by testing in an environment where there is air resistance.

Free Fall – an object dropped

• Initial velocity is zero • Let up be negative • Use the kinematic equations

– Generally use y instead of x since vertical

• Acceleration is g = 9.80 m/s2

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Example 15: A ball is dropped from a tower 70 m high. How far will the ball have fallen after 1.00s, 2.00s, and 3.00s? given find formula

Free Fall – an object thrown downward

• a = g = 9.80 m/s2

• Initial velocity 0 – initial velocity will be positive

Example 16: Suppose the ball in the previous example is thrown downward with an initial velocity of 3.00 m/s, instead of being dropped. (a) What then would be its position after 1.00s and 2.00s? (b) What would its speed be after 1.00s and 2.00s?

Example 17: A person steps off the end of a 5.00-m-high diving platform and drops to the water below. How long does it take for the person to teach the water below? given find formula solution

x = 5.0 m t 2 21 10 0

2 2o ox x v t at gt

2xt

g

g = 9.80 m/s2 2

2(5.00 )1.01sec

9.80 /

mt

m s

Free Fall -- object thrown upward

• Initial velocity is upward, so positive • The instantaneous velocity at the maximum height is zero

• a = g = -9.80 m/s2 everywhere in the motion

29.80 /a m s

1,2,&3t s

y 2

1 1

1

2oy v t at

2 2

1 1

10 (9.80 / )(1.00 ) 4.90

2y m s s m

2 2 2

2

1 1(9.80 / )(2.00 ) 19.6

2 2y at m s s m

2 2 2

3

1 1(9.80 / )(3.00 ) 44.1

2 2y at m s s m

2

1 1

1

2oy v t at

2 21(3.00 / )(1.00 ) (9.80 / )(1.00 ) 7.90

2m s s m s s m

2 21(3.00 / )(2.00 ) (9.80 / )(2.00 ) 25.6

2m s s m s s m

ov v at 23.00 / (9.80 / )(1.00 ) 12.8 /m s m s s m s

23.00 / (9.80 / )(2.00 ) 22.6 /m s m s s m s

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• The motion may be symmetrical

– Then tup = tdown

– Then v = -vo • The motion may not be symmetrical

– Break the motion into various parts • Generally up and down

Example 18: A person throws a ball upward into the air with an initial velocity of 15.0m/s. Calculate (a) How high it goes (b) How long the ball is in the air before it comes back to the hand? (c) How long does it take the ball to reach the maximum height? (d) What is the velocity of the ball when it returns to the throwers hand? (e) Calculate at what time the ball passes a point 8.00 m above the person’s hand.

2 2 2

2

0 (15.0 / )11.5

2 2( 9.80 / )

ov v m sy m

a m s

2 2 21 0 (15.0 / ) ( 9.80 / )

2oy v t at m s t m s t

2(15.0 / 4.90 / ) 0m s m s t t

2

15.0 /0 & 3.06

4.90 /

m st t s

m s

ov v at

2

15.0 /1.53

9.80 /

ov m st s

a m s

ov v at

215.0 / (9.80 / )(3.06 ) 15.0 /m s m s s m s

21

2o oy y v t at

2 218.00 0 (15.0 / ) (9.80 / )

2m m s t m s t

2 4

2

b b ac

a

2 2(4.90 / ) (15.0 / ) (8.00 ) 0m s t m s t m

2 2

2

15.0 / (15.0 / ) 4(4.90 / )(8.00 )

2(4.90 / )

m s m s m s mt

m s

0.69 2.37t s and t s

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Non-symmetrical Free Fall

• Need to divide the motion into segments • Possibilities include •

– Upward and downward portions – The symmetrical portion back to the release point and

then the non-symmetrical portion

Example 19. A lava bomb is launched straight up from a volcano. Using a stopwatch it is determined that it took 4.80 s to rise up and land on the ground. Acceleration was 9.80 m/s2 downward. What was the initial speed?

0 and ox a g

2 21 1(v )

2 2o o o ox x v t at v t gt gt t

1 1( ) 0 0 0

2 2o ox v gt t t v gt

1 10 or

2 2o ov gt v gt

21 1(9.80 / )(4.80 ) 23.52 /

2 2ov gt m s s m s

Example 20. A car and a truck are heading directly toward one another on a straight and narrow street, but they avoid a head-on collision by simultaneously applying their brakes at t = 0. What is the separation between the car and the truck when they have come to rest, given that at t = 0 the car is at x = 15m and the truck is at x = -35m. Picture the Problem: The velocity-versus-time plots of the car and the truck are shown at right. The car begins with a positive position and a negative velocity, so it must be represented by the lower line. The truck begins with a negative position and a positive velocity, so it is represented by the upper line.

Strategy: The distances traveled by the car and the truck are equal to the areas under their velocity-versus-time plots. We can determine the distances traveled from the plots and use the known initial positions to find the final positions and the final separation.

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Solution: 1. Find the final position of

the truck:

1truck 0,truck truck 2

35 m 2.5 0 s 10 m/s 22.5 mx x x

2. Find the final position of the car: 1car 0,car car 2

15 m 3.5 0 s 15 m/s 11.25 mx x x

3. Now find the separation: car truck 11.25 m 22.5 m 11.3 mx x

Example 20. A hot-air balloon lifted off the ground and is rising at a rate of 2.0m/s. One of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 13 m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?

Picture the Problem: The trajectories of the balloon and

camera are shown at right. The balloon rises at a steady rate

while the camera’s speed is continually slowing down under

the influence of gravity. The camera is caught when the two

trajectories meet.

Strategy: The equation of motion for position as a function of

time and velocity (equation 2-10) can be used to describe the

balloon, while the equation for position as a function of time

and acceleration (equation 2-11) can be used to describe the

camera’s motion. Set these two equations equal to each other

to find the time at which the camera is caught. Then find the

height of the balloon at the instant the camera is caught.

Solution: 1. Write equation 2-10 for the balloon: ,0b b bx x v t

2. Write equation 2-11 for the camera: 21

,0 20c cx v t gt

3. Set b cx x and solve for t:

21,0 ,0 2

21,0 ,0 2

0

b b c

b c b

x v t v t gt

x v v t gt

4. Multiply by −1 and insert the numbers: 2 21

2

2

0 2.5 m 13 2.0 m/s 9.81 m/s

0 2.5 11 4.9

t t

t t

5. Apply the quadratic formula and solve for t. The larger

root corresponds to the time when the camera would pass

the balloon a second time, on its way down back to the

ground.

22 11 11 4 4.9 2.54

2 9.8

0.26 or 2.0 s

b b act

a

t

6. Find the height of the balloon at that time: ,0 2.5 m 2.0 m/s 0.26 s 3.0 mb b bx x v t

Insight: If the passenger misses the camera the first time, she has another shot at it after 2.0 s (from the time it is

thrown) when the camera is on its way back toward the ground.

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CHAPTER 2

MOTION IN ONE DIMENSION

CONCEPTS

1. An object’s velocity can change directions when its acceleration is constant. An example of this is when an object, like a ball, is thrown straight up. The velocity is positive going up and negative going down. The acceleration will remain constant the entire time. 2. An object can have increasing speed while its acceleration is decreasing. An example would be an object released from rest in the presence of air friction. 3. An object is moving with constant acceleration in equal times its velocity changes by equal amounts. 4. When a ball is thrown straight up, its velocity at the top is zero. 5. When a ball is thrown straight up, its acceleration at the top is g (9.80 m/s2) 6. A ball, moves up along a smooth hill of ice will have the same acceleration, both up the hill and down the hill. 7. A skydiver jumps from an airplane. When she reaches terminal velocity, her acceleration is essentially zero. 8. Objects A and B both start from rest. They both accelerate at the same rate. However, object A accelerates for twice the time as object B. During the times that the objects are being accelerated, comparing the distance traveled by object A to that of object B, object A travels four times as far. 9. Objects A and B both start from rest. They both accelerate at the same rate. However, object A accelerates for twice the time as object B. Compared to the final speed of object A to that of object B, object B is twice as fast. 10. A ball is dropped from the top of a building. A second ball is thrown straight down from the same building. They are released at the same time. Neglecting air resistance the two balls accelerate at the same rate. 11. Ball A is dropped from the top of a building. One second later ball B is dropped. As time progresses the difference in their speeds remains constant. 12. Ball A is dropped from the top of a building. One second later, ball B is dropped from the same building. As time progresses, the distance between them increases. 13. Galileo postulated that at a given location on the Earth and in the absence of air resistance, all objects will fall with the same constant acceleration.

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14. Two balls are thrown straight up. The first ball is thrown with twice the initial speed of the second. Ignore air resistance. The first ball will rise four times as far. 15. Two objects are thrown from the top of a tall cliff. One is thrown up, and the other is thrown down, both with the same initial speed. Ignore air resistance, when the balls hit the ground they are traveling the same speed.

16. A horizontal line on a distance (position) versus time graph indicates that the object is at rest. 17. A moving object must undergo a change of position. 18. A horizontal line on a velocity versus time graph indicates that the object is traveling at a constant velocity, which means zero acceleration. 19. An object with zero acceleration may be in motion.

20. A horizontal line on an acceleration versus time graph indicates that the object is at a constant acceleration. 21. An object’s velocity can change direction when its acceleration is constant. Example, when a rock is thrown straight up. 22. The distance time graph on the right compares distance as a function of time an object in straight-line motion. According to the graph, the object most likely has an increasing velocity. 23. The distance time graph on the right represents the motion of an object sliding down a frictionless inclined plane. 24. If an object is traveling east with a decreasing speed, the direction of the object’s acceleration is west. 25. Galileo postulated that all objects will fall with the same constant acceleration in the absence of air or other resistance. 26. The slope of a position time graph gives instantaneous velocity. 27. The slope of a velocity versus time graph gives instantaneous acceleration. 28. Distance is to displacement as speed is to velocity. 29. If you run around a track and return to the same starting point, your average velocity is zero. (Velocity uses displacement rather than distance. The displacement in this case is zero.) 30. Acceleration is a vector quantity that represents the time-rate of change in velocity.

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31. When an object is released from rest and falls in the absence of friction its acceleration is constant. 32. Instantaneous speed is never negative. 33. Motion in the negative x direction is represented on an x vs. t plot by a downward sloping curve. 34. Oil drips at 0.5 second intervals from a truck that has an oil leak. The pattern shown represents the spacing of oil drops as the car accelerates uniformly from rest.