CHAPTER 2: KINEMATICS - Science at Concordia Shanghai

Physics: AP® Edition Instructor Solutions Manual Chapter 2

CHAPTER 2: KINEMATICS 2.1 DISPLACEMENT

1. Find the following for path A in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.

Solution (a) 7 m

(b) 7 m

(c)

2. Find the following for path B in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.

Solution (a) 5 m

(b) 5 m

(c)

3. Find the following for path C in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.

Solution (a)

(b)

(c)

m 7 m 0 m 7 +=-=Dx

m 5 m 12 m 7 -=-=Dx

m 31 m 3 m 2 m 8 =++

m 9

m 9 m 2 m 11 +=-=Dx


4. Find the following for path D in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.

Solution (a)

(b)

(c)

2.3 TIME, VELOCITY, AND SPEED

5. (a) Calculate Earth’s average speed relative to the Sun. (b) What is its average velocity over a period of one year?

Solution (a)

(b) After one year, Earth has returned to its original position with respect to the Sun. Thus,

6. A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the average speed of the blade tip in the helicopter’s frame of reference. (b) What is its average velocity over one revolution?

Solution (a)

(b) After one revolution, the blade returns to its original position with total displacement of 0 m.

m 8 m 2 m 6 =+

m 4

m 4 m 9 m 5 -=-=Dx

m/s 10 3.0 m/s 102.99s 3600

hour 1hours 24day 1

days 365.25m) 10(1.502

2TimeTraveled Dist. =Earth of Speed Avg.

44

11

´=´=

´´´

=

=

p

ptr

.m/s 0=v

m/s 4.52rev s/100 60m) (5.0022

elapsed time traveleddistance tipblade of speed Average ====

pptr

.m/s 0=v


7. The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present?

Solution

8. Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant?

Solution

9. On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world’s nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s?

Solution

y 10 2 years 10 1.67 m 1cm 100

cm/y 3m 10 5.00

ratekm 500 77

5

´=´=´´

==Dt

yearsmillion 10 years 101years 1083.9m 1cm 100

cm/year 6m 105.90 76

5

=´=´=´´

=Dt

hours 13.083s 3600

h 1s 104.7098s 104.7098

s 58min 1

s 60min 4h 1

s 3600h 13

44 =´´=´=

+÷øö

çèæ ´+÷

øö

çèæ ´=t

m/s 34.689s 104.7098m 101.6338 m/sin speed average

km/h 124.88h 13.0828

km 1633.8elapsed time traveleddistancekm/hin speed average

4

6

=´´

=

===


10. Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon’s orbit increases by (1%)?

Solution

11. A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction south of east, what was her average velocity? (c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?

Solution (a)

(b)

(c)

, since the total displacement was 0.

12. The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?

Solution

m1084.3 6´

yearsmillion 100 years 101years 1060.9m 1cm 100

cm/year 4m 1084.3 87

6

=´=´=´´

=Dt

°0.25

km/h 40.0h 1min 60

min 18.0km 12.0speed average =´=

.E of S 25 km/h, 34.3h 1min 60

min 18.0km 10.3

°=´=DD

=txv

km/h 3.20h 7.5km 12.02speed average =´=

0=v

seconds. 0.061m/s 18m 1.1

speed average traveleddistanceelapsed time ===


13. Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light .

Solution

14. A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity (a) for each of the three intervals and (b) for the entire motion.

Solution (a) For each interval, .

(b) For the full interval, we need displacement.

(Note: this is different from the average of the 3 interval velocities, which is 2.77 m/s.)

m/s) 1000.3( 8´

km 000,3842

km 000,7672

traveleddist.dist. ME

km 767,000sec56.2m/s 1000.3= timeelapsedspeed avg.= traveleddist. 8

===-

=´´´

txv =

m/s 04.4s 5.20m 21.0

m/s 71.1s 1.75m 3.00

m/s 6.00s 2.50m 15.0

3

2

1

===

-=-

==

===

txv

txv

txv

sm/ 3.49s 9.45

m 33s 5.20s 1.75s 2.50m 21.0m 3.00m 15.0

timetot.ntdisplaceme

avg ==+++-

==v


15. The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as having a single electron in a circular orbit in diameter. (a) If the average speed of the electron in this orbit is known to be , calculate the number of revolutions per second it makes about the nucleus. (b) What is the electron’s average velocity?

Solution (a)

(b) , since there is no net displacement per revolution.

2.4 ACCELERATION

16. A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

Solution

17. Professional Application Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a) acceleration and (b) deceleration. Express each in multiples of by taking its ratio to the acceleration of gravity.

1.06´10-10 mm/s 1020.2 6´

rev/s106.61onm/revoluti1033.3

m/s 10 2.20evolutiondistance/rspeed average

1s srevolution

onm/revoluti103.33 ionm)/revolut10(1.06 revolution/n/revolutio2evolutiondistance/r

m/s 1020.2time

distance speed average

15

10

6

1010

6

´=´

´==

´=´=

==

´==

-

--p

pp dr

m/s 0=v

20f m/s 4.29s 7

m/s 0 m/s 30.0=

-=

D-

=tvva

g )m/s(9.80 2


Solution (a)

(b)

18. A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration?

Solution (a)

(b)

19. Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in and in multiples of ?

Solution

g 5.76 5.76m/s 9.8m/s 56.4

m/s 56.4s 5.00

m/s 0m/s 282

2

2

20

=Þ==

=-

=-

=

aga

tvva

g 6.2055.20m/s 9.8m/s 201

m/s 201s 1.40

m/s 282m/s 0

2

2

20

=¢Þ==¢

-=-

=-

=

aga

tv'v'a'

2m/s 1.40

s 1.43m/s 1.40

m/s 0m/s 2.002

0 =-

=-

=avvt

20 m/s 2.50s 0.800

m/s 2.00m/s 0-=

-=

-=

tvva

2m/s g )m/s(9.80 2

g 11.1m/s 9.8m/s 108

m/s 108s 60.0

m/s 0m/s 106.50

2

2

23

0

=Þ=

=-´

=-

=

aga

tvva


2.5 MOTION EQUATIONS FOR CONSTANT ACCELERATION IN ONE DIMENSION

20. An Olympic-class sprinter starts a race with an acceleration of . (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.

Solution (a)

(b) Assuming the acceleration is constant, we know . We can create a graph by plugging in a few different t-values, say t = 1, 2, 3, 4, 5:

Time (s) Position

(m) 0 0 1 2.25 2 9 3 20.25 4 36 5 56.25

21. A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is , and 1.85 ms elapses from the time the ball first

touches the mitt until it stops, what was the initial velocity of the ball?

2m/s 4.50

m/s 8.10)s 40.2)(m/s 50.4(m/s 0 20 =+=+= atvv

22 2.25(½) tatx ==

24 m/s 10 10.2 ´ s) 10 ms (1 3-=


Solution (about 87 miles per hour)

22. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of for . What is its muzzle velocity (that is, its final velocity)?

Solution

23. (a) A light-rail commuter train accelerates at a rate of . How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of . How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in ?

Solution (a)

(b)

(c)

24. While entering a freeway, a car accelerates from rest at a rate of for 12.0 s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.

m/s 38.9s) 10)(1.85m/s 102.10( m/s 0 3240 =´´--=-= -atvv

25 m/s 10 6.20´ s 10 8.10 4-´

m/s 502 s) 10)(8.10m/s 10(6.20m/s 0 4250 =´´+=+= -atvv

2m/s 1.35

2m/s 1.65

2m/s

s 16.5km 1

m 1000s 3600

h 1m/s 1.35

km/h) 0km/h (80.02

0 =´´-

=-

=avvt

s 13.5km 1

m 1000s 3600

h 1m/s 1.65

km/h 80.0km/h 02

0 =´´-

-=

-=

avvt

20 m/s 2.68km 1

m 1000s 3600

h 1s 8.30

km/h 80.0km/h 0-=´´

-=

-=

tvva

2m/s 2.40


Solution (a)

(b) Knowns:

(c) is the unknown. We can use the equation because the

only unknown it includes is , which is what we want to solve for. First we substitute the knowns into the equation and then we solve for .

.

(d) is the unknown. We need an equation that relates our knowns to the unknown we want. We can use the equation because in it all of the variables other than are known. We substitute the known values into the equation and then solve for v: .

25. At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of . (a) How far does she travel in the next 5.00 s? (b) What is her final

velocity? (c) Evaluate the result. Does it make sense?

Solution (a)

(b)

(c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at , then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.

m 0 m/s; 0 s; 12.0 ;m/s 2.40 002 ==== xvta

x 200 2

1 attvxx ++=

xx

m 173s) )(12.0m/s (2.4021s) m/s)(12.0 (0m 0

21 222

00 =++=++= attvxx

vatvv += 0

vm/s 28.8s) m/s)(12.0 (2.40m/s 00 =+=+= atvv

2m/s 2.00

m 20.0s) )(5.00m/s 2.00(21s) m/s)(5.00 (9.00m 0

21 222

00 =-++=++= attvxx

m/s 1.00s) )(5.00m/s 2.00(m/s 9.00 20 -=-+=+= atvv

2m/s 2.00


26. Professional Application Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?

Solution (a)

(b) Knowns:

(c)

This is the best equation to use because it uses our 3 knowns to determine our unknown.

(d) Yes, the answer seems reasonable. An entire heartbeat cycle takes about one second. The time for acceleration of blood out of the ventricle is only a fraction of the entire cycle.

27. In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes , calculate the distance over which the puck accelerates.

Solution

cm 80.1 cm/s; 30.0 m/s; 0 00 =-== xxvv

s 0.120m/s) 0.300(m/s) (0

m) 2(0.01802/)(

)(

0

0

avg

0 =+

=+-

=-

=vvxx

vxx

t

s 10 3.33 2-´

m 0.799s) 10m/s)(3.33 40.0m/s (8.0021)(

21 2

0avg0 =´+=+==- -tvvtvxx


28. A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?

Solution (a)

(b)

29. Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of

for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of , how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?

Solution (a)

(b)

(c) For part (a),

For part (b),

30. A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.

Solution (a)

2m/s 6.87s 3.90

m/s 0m/s 26.8=

-=

DD

=tva

m 3.25s) .903m/s)( 8.62m/s (021)(

21

00 =+=+=- tvvxx

2m/s 0.0500

2m/s 0.550

m/s 28.0)min 1

s 60min )(8.00m/s (0.0500m/s 4.00 20 =´+=+= atvv

s 50.9m/s 0.550

m/s 28.0m/s 02

0 =-

-=

-=

avvt

m 107.68s) )(480m/s (0.050021s) m/s)(480 0.4(

21 3222

00 ´=+=+=- attvxx

m 713)m/s 0.5502(

m/s) (28.0m/s) (02 2

2220

2

0 =-

-=

-=-

avvxx

s 1069.7m/s 0m/s 65.0

m) 2(0.250)(2 3

0

0 -´=+

=+-

=vvxxt


(b)

31. A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of , how far will it travel before becoming airborne? (b) How long does this take?

Solution (a)

(b)

32. Professional Application A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in and in multiples of

. (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of ?

Solution (a)

(b) , so that

2322

0

20

2

m/s 1045.8m) 2(0.250

m/s) (0m/s) (65.0)(2

´=-

=--

=xxvva

2m/s 0.350

m 51.4)m/s .35002(

m/s) (0m/s) (6.002 2

2220

2

0 =-

=-

=-avvxx

s 17.1m/s 0.350

m/s 0m/s 6.002

0 =-

=-

=avvt

2m/s( )2m/s 80.9 =gg

g

.9.1818.9m/s 9.80m/s 90.0

m/s 90.0m)102(2.00m/s) (0.600m/s) (0

)(2

2

2

23

22

0

20

2

gaga

xxvva

=Þ==

-=´

-=

--

= -

tvvxx )(21

00 +=- s 106.67m/s) (0m/s) (0.600

m) 102(2.00)(2 33

0

0 --

´=+

´=

+-

=vvxxt


(c)

33. An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?

Solution (a)

(b)

34. In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

Solution Knowns:

We want , so we can use this equation:

. Negative acceleration

means that the pilot was decelerating at a rate of 486 m/s every second.

35. Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

gaga

xxvva

4.0808.4m/s 9.80m/s 40.0

m/s 40.0m) 102(4.50m/s) (0.600m/s) (0

)(2

2

2

23-

22

0

20

2

=Þ==

-=´

-=

--

=

222

0

20

2

m/s 4.80m) 2(0.350

m/s) (7.50m/s) (0)(2

-=-

=--

=xxvva

s109.33m/s 7.50m/s 0

m) 2(0.350)(2 2

0

0 -´=+

=+-

=vvxx

t

m/s 54 m/s; 0 m; 3 0 === vvx

a

222

02

20

2 m/s 486m) 2(3

m/s) (54m/s 02

2 -=-

=-

=Þ+=xvvaaxvv


Solution (a)

(b)

This is times the deceleration of the pilots, who were falling from thousands of meters high!

36. An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?

Solution (a) , and rearranging

(b) , and rearranging

(c) Here, we use the fact that the train will leave the station when the nose is away from the beginning of the station.

m/s 7.7m) 3.0)(m/s 9.82(22 220

2 =--==Þ+= axvaxvv

23222

02

m/s 105.1m) 2(0.02

m/s) (7.7m/s) (02

´-=-

=-

=xvva

3 ~

2m/s 0.150

200 2

1 attvxx +=-

s 9.88m/s 0.150

m) )(210m/s 0.1502(m/s) (22.0m/s 22.0

)(2

2

22

0200

=-

-+±-=

-+±-=

axxavv

t

)(2 020

2 xxavv -+=

m/s 20.6m) )(210m/s 0.1502(m/s) (22.0)(2 220

20 =-+=-+= xxavv

m 340 = m 130 + m 210

200 2

1 attvxx +=-

s 4.16m/s 0.150

m) )(340m/s 0.1502(m/s) (22.0m/s 22.0

)(2

2

22

0200

=-

-+±-=

-+±-=

axxavv

t


(d)

37. Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

Solution (a)

(b) , and rearranging

(c) because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at during the last few meters, but substantially less, and the final velocity would be less than

.

38. A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Solution (a)

(b) Let be the time it takes the rider to reach the finish line without

m/s 5.91m) )(340m/s 0.1502(m/s) (22.0

)(222

020

2

=-+=

-+=

v

xxavv

20 m/s 32.6s 4.45

m/s 0m/s 145=

-=

-=

tvva

)(2 020

2 xxavv -+=

m/s 162m) )(402m/s 2(32.6m/s) (0 22 =+=v

maxvv >

2m/s 6.32

m/s 162

2m/s 0.500

m/s 15.0s) )(7.00m/s (0.500m/s 11.5 20 =+=+= atvv

constt


accelerating:

Now let be the distance traveled during the 7 seconds of acceleration.

We know so

Let be the time it will take the rider at the constant final velocity to complete

the race: .

So the total time it will take the accelerating rider to reach the finish line is

Finally, let be the time saved. So .

(c) Let be the time it takes for rider 2 to reach the finish line.

Therefore he finishes after the winner.

When the other racer reaches the finish line, the winner has been traveling at for 4.2 seconds, so the other racer finishes

behind the other racer.

39. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

Solution There are two parts to the race: an acceleration part and a constant speed part. First, we need to determine how long (both in distance and time) it takes the

motorcycle to finish accelerating. During acceleration, .

26.1ss 26.09m/s 11.5m 300

0const ====

vxt

d

s 7.00 =t

m 92.8 m 92.75 s) )(7.00m/s 0.5(0.500 s) m/s)(7.00 (11.5 21 222

0 ==+=+= attvd

t¢

s 13.8 s 13.82 m/s 15.0

m 92.75m 300 ==-

=-

=¢vdxt

Ts. 20.8 s 20.82 s 13.82 s 7 ==+=+= t t T

T* s 5.27 s 20.82s 26.09 =-=T*

2t

s 4.2 s 20.817 s 25.0 difference times; 25.0m/s 11.8m 295

20

2 =-=-===¢¢

= Ttvxt

s 4.2

m/s 15 m 63 = m/s) m/s)(15 2.4(=x

s 4mph 60

=a


At constant velocity, .

Now, we complete the calculation by determining how much time is spent on the course at max speed.

40. (a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

Solution (a) There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into (while

accelerating) and (with constant speed), where .

When the speed is constant, , so .

When accelerating, , so . Plugging into

the previous equation, we get his maximum speed:

vmax =183.58 mph = at1 ⇒ t1 =18.358 mph 4.00s60.0 mph"

#$

%

&'=12.239 s

x1 = vavgt1 =183.58 mph+ 0 mph

2"

#$

%

&'(12.239 s) = (91.79 mph) 1 h

3600 s"

#$

%

&'(12.239 s)

x1 = 0.312 mi

mi 4.688 mi 0.312 mi 5.00 2 =-=x

s 104s 91.94s 12.239

so s, 91.94h 1

s 3600h 0.02554mph 183.58mi 4.688

21total

max

22

=+=+=

====

tttvxt

1x

2x m 10021 =+ xx

s 6.69=t 12 m 100s) (6.69 xvvtx -===

s 00.3=t vvtatx s) 50.1(21

21 2

1 === 1x

( ) ( ) m/s 2.12s 50.1s 69.6

m 100s 69.6s 50.1m 100 =+

=Þ=- vvv


Therefore, his acceleration was

(b) Similar to part (a), we can plug in the different values for time and total distance:

2.7 FALLING OBJECTS

41. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be .

Solution Knowns:

To find displacement we use , and to find velocity we use

.

(a)

(b)

(c)

The ball is almost at the top.

2m/s 07.4s 3.00

m/s 2.12===

tva

( ) ( ) m/s 2.11s 50.1s 30.16

m 200s 30.16s 50.1m 200 =+

=Þ=- vvv

0 0 =y

m/s 15.0 m; 0 ;m/s 9.8 gravity todueon accelerati 002 +==-=== vyga

200 2

1 attvyy ++=

atvv += 0

m/s 10.1s) )(0.500m/s 9.8(m/s) (15.0

m 6.28s) )(0.500m/s 9.8(21s) m/s)(0.500 (15.0m 0

21

2101

22

211001

=-+=+=

=-++=

++=

atvv

attvyy

m/s 5.20s) )(1.00m/s 9.8(m/s) (15.0

m 10.1s) )(1.00m/s 9.8(21s) m/s)(1.00 (15.0m 0

21

2202

22

222002

=-+=+=

=-++=

++=

atvv

attvyy

m/s 300.0s) )(1.50m/s 9.8(m/s) (15.0

m 11.5s) )(1.50m/s 9.8(21s) m/s)(1.50 (15.0m 0

21

2303

22

233003

=-+=+=

=-++=

++=

atvv

attvyy


(d)

The ball has begun to drop.

42. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Solution Knowns:

To find displacement we use , and to find velocity we use

.

(a)

(b)

(c)

m/s 60.4s) )(2.00m/s 9.8(m/s) (15.0

m 4.01s) )(2.00m/s 9.8(21s) m/s)(2.00 (15.0m 0

21

2404

22

244004

-=-+=+=

=-++=

++=

atvv

attvyy

m 0 m/s; 14.0 ;m/s 9.8 002 =-=-= yva

200 2

1 attvyy ++=

atvv += 0

m/s 18.9s) )(0.500m/s 9.8(m/s) 14.0(

m 23.8s) )(0.500m/s 9.8(21s) m/s)(0.500 14.0(m 0

21

2101

22

211001

-=-+-=+=

-=-+-+=

++=

atvv

attvyy

m/s 8.32s) )(1.00m/s 9.8(m/s) 14.0(

m 9.18s) )(1.00m/s 9.8(21s) m/s)(1.00 14.0(m 0

21

2202

22

222002

-=-+-=+=

-=-+-+=

++=

atvv

attvyy

m/s 28.7s) )(1.50m/s 9.8(m/s) 14.0(

m 0.32s) )(1.50m/s 9.8(21s) m/s)(1.50 14.0(m 0

21

2303

22

233003

-=-+-=+=

-=-+-+=

++=

atvv

attvyy


(d)

(e)

43. A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Solution

44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.

Solution (a) Knowns:

(b)

m/s 6.33s) )(2.00m/s 9.8(m/s) 14.0(

m 6.47s) )(2.00m/s 9.8(21s) m/s)(2.00 14.0(m 0

21

2404

22

244004

-=-+-=+=

-=-+-+=

++=

atvv

attvyy

m/s 5.83s) )(2.50m/s 9.8(m/s) 14.0(

m 6.65s) )(2.50m/s 9.8(21s) m/s)(2.50 14.0(m 0

21

2505

22

255005

-=-+-=+=

-=-+-+=

++=

atvv

attvyy

m/s 59.4m) )(1.25m/s 9.802(m/s) (0)(2

)(2)(222

02

0

022

0020

2

=--=--=

--=Þ-+=

yyavv

yyavvyyavv

m 0 s; 1.8 m/s; 1.40 ;m/s 9.80 02 ==-=-= ytva

m 18 s) )(1.8m/s 9.800.5( s) m/s)(1.8 1.40( m 0 21

22

200

=-+-+=

++-= attvyy


45. A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.


(b) At the highest point in the jump, . We can use the equation

because the only unknown it includes is , which is what we

want to solve for. First we substitute the knowns into the equation and then we solve for .

Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.

(c) If is the time for the dolphin to reach its peak height, then is the time the dolphin is out of the water.

46. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water?

Solution (a) Knowns: , so we use the

equation . Rearranging,

m 0 m/s; 13.0 ;m/s9.80 002 ==-= yva

m/s 0 =v

avv

yy2

20

2

0-

=- y

y

y− y0 =v2 − v0

2

2a⇒ y = v

2 − v02

2a+ y0 =

(0 m/s)2 − (13.0 m/s)2

2(− 9.80 m/s2 )+ 0 m = 8.62m

t t2

s 2.65 2 and s, 1.3265m/s 9.8

m/s 13.0m/s 02

00 ==

--

=-

=Þ+= tavv

tatvv

m/s 00.4,m/s 80.9 m, 0 m, 80.1 02

0 =-=== vayy

200 2

1 attvyy +=-


(b)

(c) and rearranging

Since the diver must be moving in the negative direction,

47. (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Solution

(a) Knowns:

Since we know , , , and and want to find , we can use the equation

.

, so the cliff is

8.26 m high.

(b) Knowns:

s 1.14m/s 9.80

m) )(1.80m/s 9.802(m/s) (4.00m/s 4.00

)(2

2

22

0200

=-

--±-=

--±-=

ayyavv

t

m 0.816)m/s 9.802(

m/s) (4.00m/s) (02 2

2220

2

0 =--

=-

=-avvyy

)(2 020

2 yyavv -+=

m/s 16.7/sm 51.28

m) 1.80)(m/s 9.802(m/s) (4.00)(222

220

20

±=±=

--+±=-+±= yyavv

m/s. 7.16 -=v

20 m/s 9.8 m/s; 8.00 m; 0 s; 2.35 -=+=== avyt

t y 0v a 0y

y = y0 + v0t +12at 2

m 8.26s) )(2.35m/s 9.80(21s) m/s)(2.35 8.00(m) (0 22 -=-+++=y

200 m/s 9.80 m/s; 8.00 m; 8.26 m; 0 -=-=== avyy


Now we know , , , and and want to find , so we use the equation

again. Rearranging,

48. A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

Solution Knowns:

49. You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Solution Knowns:

y 0y 0v a t

y = y0 + v0t +12at 2

s 0.717 s 2.35or s 0.717m/s 9.80

m/s 15.03m/s 8.00)m/s 9.80(

m) 0m )(8.26m/s (9.802m/s) 8.00(m/s) 8.00(

)5.0(2))(5.0(4

2

2

22

0200

=Þ-=-

±=

--+-±--

=

--±-=

tt

t

ayyavv

t

200 m/s 80.9 m/s; 0.11 m; 80.1 m; 20.2 -==== avyy

s 2.28 m/s 9.80

m) )(0.40m/s 9.802( m/s) (11.0 m/s 11.0

) (2

2

22

0200

=-

-+±-=

--±-=

axxavv

t

m 7.00 m/s; 15.0 ;m/s 9.80 02 ==-= yva


So and , so the total time between passing the branch is .

50. A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Solution (a)

(b)

51. Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution (a)

So the rock falls 11.0 m in the 1.50 s before the hiker sees the rock. When he finally sees the rock, it is 94.0 m above his head.

(b) , so the rock will take 4.63 s to fall

the full distance. Thus, the hiker will have to move out of the way before the rock strikes the hiker's location, ignoring the height of person.

22

22

200

20

m/s 9.80m/s 9.37m/s 15.0

m/s 9.80m) )(7.00m/s 9.802(m/s) (15.0m/s 15.0

221

-±-

=-

-+±-=

+±-=

Þ+=

aayvv

t

attvy

s 0.58 1 =t s 2.49 2 =t s 91.1

m/s 7.00m) )(2.50m/s 9.802(022 220

20

2 =--=-=Þ+= ayvvayvv

s 0.714m/s 7.00

m) 2(2.5022 0

0avg ==

+=Þ÷

øö

çèæ +

==vvyttvvtvy

m 11.0s) )(1.50m/s 9.80(21

21 222

0 ==+= attvy

s 4.63m/s 9.80

m) 2(105221

22 ===Þ=

aytaty

s 13.3s 50.1s 63.4 =-


52. An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Solution (a)

(b)

(c) First find the total time to fall:

Next, we find the distance traveled up to the last 1 second of flight:

, so the distance traveled in the last

second will be the difference: .

53. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.


, downward.

(b) Let = the time for the sound to travel to the tourist.

Let = the total time before the tourist can react.

Let = the time it takes the rock to reach the bottom.

m 90.4s) )(1.00m/s 9.80(21

21 222

0 ==+= attvx

m/s 38.3m) )(75.0m/s 2(9.802 220

2 ==Þ+= vaxvv

s 3.91m/s 9.80

m) 2(75.0221

22 ===Þ=

aytaty

m 41.5s) )(2.91m/s (9.8021

21 222 === aty

m 33.5 m 41.5 m 75.0 =-=x

.m/s 9.80 m, 250 m, 0 m/s, 0 200 -=-=== ayyv

m/s 70.0m) 250)(m/s 9.802(m/s) (0)(2 220

20 -=--+±=-+±= yyavv

st

s 0.7463m/s 335m 250cliff ofheight

===s

s vt

t

s 1.046 s 0.300 s 0.7463 imereaction t s =+=+= tt

bt


Now subtract

54. A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball’s initial velocity?

Solution The time it takes the ball to reach the top of the window, at a height of 9.5 m, will be the amount of time it takes to reach the bottom of the window, plus the time to traverse the window, 1.30 s. So let's denote the time to the top of the window as

, and the time to the bottom of the window . We then have .

We will call the height to the bottom of the window , and the height to the top of

the window .

Then, we have

We also have an equation for ,

So now we have 3 equations in 3 unknowns ( , , and ). Solving, we get a

value of

55. Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

Solution (a)

7.143s.m/s 9.8

m/s 0m/s 7.002

0b =

---

=-

=gvv

t

s 6.10 s 1.046s 7.143 b =-=- tt

TOPt Bt s 1.30+= BTOP tt

1y

2y

2TOP

2TOP0

200,22 ))(m/s 80.9(

21)( + 0 = m 50.9

21 ttvattvyy -+®++=

1y2

B2

B02

B2

B01 ))(m/s (4.90 = m 50.7))(m/s 4.90 = ttvt(tvy -®-

Bt BOTt 0v

.m/s 14.5 0 =v

.m 19.6s) )(2.0000m/s 9.800.5(21 222

00 -=-=+=- attvyy


(b) Let be the depth of the well, at the bottom of the well. Let = the speed of sound and = the time for the sound to travel from the

bottom to the top of the well. Let = the time for the rock to reach the bottom.

(i)

(ii)

(iii)

Adding equations (ii) and (iii):

Using the quadratic formula:

56. A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms

. (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Solution (a)

(b)

h hyy -== m, 00

2v 2t

1t

s 2.0000 = = + 21 ttt

212

1 ath =-

) s (2.0000 = = 1222 tvtvh -

664mm/s) (332.00)m/s 4.90(0m/s) (232.00m/s) s)(332.00 (2.0000)m/s 9.800.5(

)0000.2(210

121

21

21

2

1221

+--=

-+-=

-+=

tttt

tsvat

m 18.5s) )(1.944m/s 9.800.5(21

s. 1.944m/s 9.80

m) )(664m/s 4.904(m/s) 332(m/s 332

2221

2

22

1

=--=-=

=-

---±=

ath

t

s) 1000.8( 5-´

m/s 5.42m) 1.50)(m/s 9.802(m/s) (0)(2

)(222

020

020

2

-=--+±=-+±=

Þ-+=

yyavv

yyavv

m/s 33.5m) )(1.45m/s 9.802(m/s) (0)(2

)(222

02

0

020

2

+=--±=-+±=

Þ-+=

yyavv

yyavv


(c)

(d) The period of compression occurs when the ball goes from to

. From part (c), . So,

57. A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Solution (a) Knowns: (at top of ascent)

.

Maximum height = .

(b)

, which is

a height of 262 m.

(c)

58. A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms

(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

255

0

0

m/s 101.34s 108.00

m/s) 5.422(m/s 5.331´+=

´--

=-

=

Þ+=

-tvva

atvv

m/s 5.42 = 0 -vm/s 0 = v 25 m/s 10 × 1.344 = a

m 101.09)m/s 102(1.344

m/s) (0m/s) 5.422(2

425

22220

0-´=

´--

=-

=-avvyy

m/s 0 = ;m/s 9.80 = m/s; 10.0 = 20 vav -

m 5.10)m/s 9.802(

m/s) 0.(10m/s 02 2

220

2

=--

=-

=avvy

m 305 = m 5.1 + m 300

m 38.4s) )(4.00m/s 9.80(21s) m/s)(4.00 (10.0

21 222

0 -=-+=+= attvy

m/s 29.2s) )(4.00m/s 9.80(m/s 10.0 20 -=-+=+= atvv

.8.91s m/s 9.8

m/s 77m/s 10m/s 9.80

m) )(300m/s 9.802(m/s) (10.0m/s 10.0)(221)(

21+

2

2

220

200

200

200

=-

+-=

---±-

=--±-

=

Þ++-Þ+=

t

ayyavv

t

attvyyattvyy

s). 1050.3( 3-´


Solution (a)

(b)

(c)

(d) The period of compression occurs when the ball goes from to

. From part (c), .

So,

2.8 GRAPHICAL ANALYSIS OF ONE-DIMENSIONAL MOTION

59. (a) By taking the slope of the curve in Figure 2.72, verify that the velocity of the jet car is 115 m/s at . (b) By taking the slope of the curve at any point in Figure

2.73, verify that the jet car’s acceleration is .

Solution

t (s) x (m) 0 200 10 600 20 1500 30 2900

m/s 5.42m) 1.50)(m/s 9.802(m/s) (0)(2

)(222

020

020

2

-=--+±=-+±=

Þ-+=

yyavv

yyavv

m/s 64.4m) )(1.10m/s 9.802(m/s) (0)(2

)(222

02

0

020

2

+=--±=-+±=

Þ-+=

yyavv

yyavv

2233

0

0

m/s 8802m/s 1088.2s 1050.3

m/s) 5.422(m/s 4.643+=´=

´--

=-

=

Þ+=

-tvva

atvv

m/s 5.422 = 0 -vm/s 0 = v 23 m/s 10 × 2.88 = a

m 0.00511m 1011.5)m/s 102(2.88

m/s) (0m/s) 5.422(2

323

22220

0 =´=´

--=

-=- -

avv

yy

s20=t2m/s 5.0

01000200030004000

0 10 20 30 40posi

tion

(met

ers)

time (seconds)

position vs. time


(a)

t (s) v (m/s) 0 15 5 40 10 65 15 90 20 115 25 140 30 165

(b)

60. Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at t = 10.0 s is 0.208 m/s. Assume all values are known to 3 significant figures.

Solution

t (s) x (km) 0 3

rise (2900 600) m 115 m/srun (30 10) s

v -= = »

-

050

100150200

0 10 20 30 40

velo

city

(met

ers p

er

seco

nd)

time (seconds)

velocity vs. time

2rise (140 65) m 5.0 m/srun (25 10) s

a -= = »

-

0

10

20

30

0 20 40 60 80

posi

tion

(met

ers)

time (seconds)

position vs. time


10 4.8 20 7 30 9.3 40 11.7 50 14.2 60 16.7 70 19.2

61. Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

Solution

62. By taking the slope of the curve in Figure 2.75, verify that the acceleration is at .

(7 3) km .200 m/s 0.208 m/s(20 0) s

v -= = »

-

(11.7 7) m .235 0.238 m/s(40 20) s

v -= = »

-

2m/s3.2 s10=t


Solution

t (s) v (m/s) 0 165 10 207 20 228 30 239 40 246 50 249 60 250 70 250

63. Construct the displacement graph for the subway shuttle train as shown in Figure 2.30(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

Solution

0

100

200

300

0 20 40 60 80

velo

city

(met

ers p

er

seco

nd)

time (seconds)

velocity vs. time

22 m/s 3.2m/s 3.15s 0)(20m/s 165)(228

»=-

-=a

4.654.7

4.754.8

4.85

0 10 20 30

posi

tion

(km

)

time (seconds)

time vs. position


64. (a) Take the slope of the curve in Figure 2.76 to find the jogger’s velocity at . (b) Repeat at 7.5 s. These values must be consistent with the graph in

Figure 2.77.

Solution

t (s) x (m) 0 0 2.5 10 5 17.5 7.5 10 10 2.5 12.5 2.5 15 5 20 25

time (s) v (m/s) 0 4 2.5 4

s5.2=t

0

10

20

30

0 5 10 15 20 25posi

tion

(met

ers)

time (seconds)

position vs. time

-5

0

5

0 5 10 15 20 25

velo

city

(met

ers p

er

seco

nd)

time (seconds)

velocity vs. time


5 0 7.5 -3.6 10 -0.6 12.5 .7 15 3 20 4.4

(a)

(b)

65. A graph of is shown for a world-class track sprinter in a 100-m race. (See Figure 2.79). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at ? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

Solution

(a) For the first 4 s, .

(b) At 5 s, .

(c) .

(d) For the first 4 s, we know a and v, so .

For the second interval, distance remaining and

m/s 5.3s 0)-5(

m 0)7.51(=

-=v

m/s 3s 5)(10

m )5.17(2.5-=

--

=v

( )tv

s5=t

m/s 6 = avgv

m/s 12 = v

2m/s 3s 4

m/s 12==a

m 24 = x

m 76 = m 24 m 100 = -


, so So .

66. Figure 2.80 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.

Solution

time (s) x (m) 0 0 2 2 3 -3 5 -3 6 -2

(a)

time (s) v (m/s) 0 1 2 1

m/s 12 = v .s 3.6m/s 12m 76

2 ==t s 10 = s 6.3 + s 4 = + = 21total ttt

-5

0

5

0 2 4 6 8

posi

tion

(met

ers)

time (seconds)

time vs. position

-6

-4

-2

0

2

0 2 4 6 8

velo

city

(met

ers p

er

seco

nd)

time (seconds)

time vs. velocity


2 -5 3 -5 3 0 5 0 5 1 6 1 6 0

(b)

Time (s) a (m/s2) 0 0 1 0 2 -6 3 6 4 0 5 1 6 -1 7 0

TEST PREP FOR AP® COURSES

1.

Which of the following statements comparing position, distance, and displacement is correct?

a. An object may record a distance of zero while recording a non-zero displacement.

b. An object may record a non-zero distance while recording a displacement of zero.


c. An object may record a non-zero distance while maintaining a position of zero.

d. An object may record a non-zero displacement while maintaining a position of zero.

Solution (b)

2. A student is trying to determine the acceleration of a feather as she drops it to the ground. If the student is looking to achieve a positive velocity and positive acceleration, what is the most sensible way to set up her coordinate system?

a. Her hand should be a coordinate of zero and the upward direction should be considered positive.

b. Her hand should be a coordinate of zero and the downward direction should be considered positive.

c. The floor should be a coordinate of zero and the upward direction should be considered positive.

d. The floor should be a coordinate of zero and the downward direction should be considered positive.

Solution (b)

3. A group of students has two carts, A and B, with wheels that turn with negligible friction. The two carts travel along a straight horizontal track and eventually collide. Before the collision, cart A travels to the right and cart B is initially at rest. After the collision, the carts stick together.

a. Describe an experimental procedure to determine the velocities of the carts before and after the collision, including all the additional equipment you would need. You may include a labeled diagram of your setup to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure.

b. There will be sources of error in the measurements taken in the experiment both before and after the collision. Which velocity will be more greatly affected by this error: the velocity prior to the collision or the velocity after the collision? Or will both sets of data be affected equally? Justify your answer.


Solution a. Use tape to mark off two distances on the track — one for cart A before

the collision and one for the combined carts after the collision. Push cart A to give it an initial speed. Use a stopwatch to measure the time it takes for the cart(s) to cross the marked distances. The speeds are the distances divided by the times.

b. If the measurement errors are of the same magnitude, they will have a greater effect after the collision. The speed of the combined carts will be less than the initial speed of cart A. As a result, these errors will be a greater percentage of the actual velocity value after the collision occurs. (Note: Other arguments could properly be made for ‘more error before the collision’ and error that ‘equally affects both sets of measurement.’)

4.

A cart is constrained to move along a straight line. A varying net force along the direction of motion is exerted on the cart. The cart’s velocity v as a function of time t is shown in the graph above. The five labeled points divide the graph into four sections.

Which of the following correctly ranks the magnitude of the average acceleration of the cart during the four sections of the graph?

a. aCD > aAB > aBC > aDE b. aBC > aAB > aCD > aDE c. aAB > aBC > aDE > aCD d. aCD > aAB > aDE > aBC

Solution (d)


5. Push a book across a table and observe it slow to a stop.

Draw graphs showing the book’s position vs. time and velocity vs. time if the direction of its motion is considered positive.

Draw graphs showing the book’s position vs. time and velocity vs. time if the direction of its motion is considered negative.

Solution The position vs. time graph should be represented with a positively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a positive y-intercept and a negative slope. Because the final velocity of the book is zero, the line should finish on the x-axis.

The position vs. time graph should be represented with a negatively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a negative y-intercept and a positive slope. Because the final velocity of the book is zero, the line should finish on the x-axis.


6. A group of students is attempting to determine the average acceleration of a marble released from the top of a long ramp. Below is a set of data representing the marble’s position with respect to time.

[Table 2_05_01]

Position (cm) Time (s)

0.0 0.0

0.3 0.5

1.25 1.0

2.8 1.5

5.0 2.0

7.75 2.5

11.3 3.0

Use the data table above to construct a graph determining the acceleration of the marble. Select a set of data points from the table and plot those points on the graph. Fill in the blank column in the table for any quantities you graph other than the given data. Label the axes and indicate the scale for each. Draw a best-fit line or curve through your data points.

Using the best-fit line, determine the value of the marble’s acceleration.

Solution Students should graph the position against the square of the time. (Alternately, they could graph the time against the square root of the position.) The scale on each axis should be reasonable. A best-fit line should pass between nearly all points.

The slope of the line drawn on the position vs. time squared graph should be

approximately 1.25 m/s2. Given the equation , the slope of the line

is equivalent to . As a result, the acceleration is approximately 2.5 m/s2.

212o ox x v t at= + +

12a


7. Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 5 m/s. When it is 100 m closer to the surface of the asteroid, the craft reports a velocity of 8 m/s. According to their data, what is the approximate gravitational acceleration on this asteroid?

a. m/s2 b. 0.03 m/s2 c. 0.20 m/s2 d. 0.65 m/s2 e. 33 m/s2

Solution (c)

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CHAPTER 2: KINEMATICS - Science at Concordia Shanghai

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