# hw02

January 25, 2018 | Author: Janna Marie Pantonial | Category: Speed, Velocity, Acceleration, Temporal Rates, Spacetime

#### Description

Olson

PHY 121

homework 2

Chapter 2 3. A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and heads 3.7 km due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction of the displacement of the whale? (a) 6.9 + 1.8 + 3.7 = 12.4 km (b) (6.9, 0) + (−1.8, 0) + (3.7, 0) = (8.8, 0). The magnitude of the displacement is 8.8 km . The direction is east . 5. As the earth rotates through one revolution, a person standing on the equator traces out a circular path whose radius is equal to the radius of the earth (6.38 × 106 m). What is the average speed of this person in (a) meter per second and (b) miles per hour? The distance of travel is given by 2πr = 2π(6.38 × 106 ) = 4.01 × 107 m. The time of travel is 1 day = 86400 s. (a) The speed is (4.01 × 107 )/86400 = 464 m/s . (b)

464.1 m 3600 s 3.2808 ft mi · · · = 1040 mi/h s h m 5280 ft

7. A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. The car is a distance d away. The bear is 26 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?  x = vt  x d  tourist t= = v 4.0  x = vt  x d + 26  bear t= = v 6.0

Since the time the tourist runs equals the time the bear runs, we have d d + 26 = 4.0 6.0 1

Multiplying both sides by 12 gives us 3d = 2d + 52

d = 52 m

13. A motorcycle has a constant acceleration of 2.5 m/s2 . Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 21 to 31 m/s, and (b) 51 to 61 m/s? ∆v ∆v can be written as ∆t = . ∆t a ¯ ∆v 10 (a) ∆t = = = 4.0 s a ¯ 2.5 ∆v 10 (b) ∆t = = = 4.0 s a ¯ 2.5 The equation a ¯=

19. In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of 6.0 m/s in 1.5 s. Assuming that the player accelerates uniformly, determine the distance he runs. x = 12 (v0 + v)t = 21 (0 + 6.0)(1.5) = 4.5 m . 21. A VW Beetle goes from 0 to 60.0 mi/h with an acceleration of +2.35 m/s2 . (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to 60.0 mi/h in 0.600 s. Find the acceleration (in m/s2 ) of the dragster. h 1609.344 m m 60.0 mi · · = 26.82 h 3600 s mi s (a) Using the equation v = v0 + at = at, we get t = v/a. v 26.82 = = 11.4 s a 2.35 (b) Using the equation v = v0 + at = at, we get a = v/t.

t=

a=

v 26.82 = = 44.7 m/s2 t .600

22. (a) What is the magnitude of the acceleration of a skier who, starting from rest, reaches a speed of 8.0 m/s when going down a slope for 5.0 s? (b) How far does the skier travel in this time? 2

(a) The equatin v = a ¯t can be written as a = v/t. 8.0 v = = 1.6 m/s2 t 5.0 (b) Assuming that the acceleration is constant, we have a ¯=

x=

1 1 (v0 + v)t = (0 + 8.0)(5.0) = 20 m 2 2

23. The left ventricle of the heart accelerates blood from rest to a velocity of +26 cm/s. (a) If the displacement of the blood during the acceleration is +2.0 cm, determine its acceleration (in cm/s2 ). (b) How much time does blood take to reach its final velocity? (a) Assuming constant acceleration, we may use the equation v 2 = v02 + 2ax = 2ax which may be written as a = v 2 /(2x). a=

262 v2 = = 170 cm/s2 2x 2 · 2.0

(b) Using the equation x = 21 (v0 + v)t = 21 vt, we get t = 2x/v. t=

2x 2 · 2.0 = = 0.15 s v 26

24. A cheetah is hunting. Its prey runs for 3.0 s at a constant velocity of +9.0 m/s. Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time? For the prey, we have x = vt = (9.0)(3.0) = 27 m. For the cheetah, we have the equation x = v0 t + 21 at2 = 12 at2 which may be written as a = 2x/t2 . a=

2x 2 · 27 = = 6.0 m/s2 t2 3.02

25. A jetliner, traveling northward, is landing with a speed of 69 m/s. Once the jet touches down, it has 750 m or runway in which to reduce its speed to 6.1 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing. The known variables are v0 = 69 m/s, v = 6.1 m/s, x = 750 m. Starting with the equation v 2 = v02 + 2ax, we solve for a: a=

v 2 − v02 6.12 − 692 = = −3.1 = 3.1 m/s2 south 2x 2 · 750

27. A speed ramp at an airport is basically a large conveyor belt on which you can stand and be moved along. The belt of one ramp moves at a constant speed such that a person 3

who stands still on it leaves the ramp 64 s after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of 0.37 m/s2 , he covers the same distance as the ramp does, but in one-fourth the time. What is the speed at which the belt of the ramp is moving? Here is what is given. ramp

Clifford

t = 64 s

t = 64/4 = 16 s

a=0

a = 0.37 m/s2 v0 = 0 x = 21 at2

x = vt

Working with Clifford’s variables, we have x=

1 1 2 at = (0.37)(162 ) = 47.36 m 2 2

Next, for the ramp, we re-write the equation x = vt to find v. v=

x 47.36 = = 0.74 m/s t 64

28. A drag racer, starting from rest, speeds up for 402 m with an acceleration of +17.0 m/s2 . A parachute then opens, slowing the car down with an acceleration of −6.10 m/s2 . How fast is the racer moving 3.50 × 102 m after the parachute opens? We will divide the problem into a first part before the parachute opens and a second part after the parachute opens. first part

second part

v0 = 0 a = 17.0 m/s2

a = −6.10 m/s2

x = 402 m

x = 350 m

Using the equation v 2 = v02 + 2ax on the first part, we have v=

q p v02 + 2ax = 0 + 2(17.0)(402) = 116.9 m/s 4

The final velocity of the firt part is equal to the initial velocity of the second part. So using the same equation v 2 = v02 + 2ax applied to the second part, we get v=

q

v02 + 2ax =

p

116.92 + 2(−6.10)(350) = 96.9 m/s

29. Suppose a car is traveling at 20.0 m/s, and the driver sees a traffic light turn red. After 0.530 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 7.00 m/s2 . What is the stopping distance of the car, as measured from the point where the driver first notices the red light? There will be two parts to this problem: before and after the brakes are applied. The distance of the first part is given by x = vt = (20.0)(0.530) = 10.6 m Solving the equation v 2 = v02 + 2ax, we get the distance for the second part. v 2 − v02 0 − 20.02 x= = = 28.6 m 2a 2(−7.00) The total distance is 10.6 + 28.6 = 39.2 m . 31. A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.5 km away? For the first car, we can solve the equation x = vt to find t. t=

x 2500 = = 75.8 s v 33

For the second car, we solve the equation x = 21 at2 to find a. a=

2x 2(2500) = = 0.87 m/s2 2 t 75.862

33. Along a straight road through town, there are three speed-limit signs. They occur in the following order: 55, 35, and 25 mi/h, with the 35-mi/h sign being midway between the other two. Obeying these speed limits, the smallest possible time tA that a driver can 5

spend on this part of the road is to travel between the first and second signs at 55 mi/h and between the second and third signs at 35 mi/h. More realistically, a driver could slow down from 55 to 35 mi/h with a constant decceleration and then do a similar thing from 35 to 25 mi/h. This alternative requires a time tB . Find the ratio tB /tA . The route is divided into two equal parts of unknown distance x. Following plan A, speed is constant between signs. Therefore, we may use the equation x = vt. Solving for t gives us t = x/v. So, between the first two signs, the time is x/55 and between the next two signs, the time is x/35. The total time under plan A is tA = x/55 + x/35. For plan B, we must use the equation x = 21 (v0 +v)t. Solving this equation for t gives us t = 2x/(v0 + v). The total amount of time spent in plan B is tB = 2x/(55 + 35) + 2x/(35 + 25). The ratio of times is tB = tA

2x 90 x 55

+ +

2x 60 x 35

=

2 90 1 55

+ +

2 60 1 35

= 1.2

35. A train has a length of 92 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time t = 14 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 28 s, the car is again at the rear of the train. Find the magnitudes of (a) the car’s velocity and (b) the train’s acceleration. We will begin the problem when the car is at the back end of the train and end the problem when the car is at the front of the train. The train satisfies the equation x = 21 at2 = 12 a142 = 98a. The final velocity of the train is v = 14a. Since the car travels 92 m farther than the train, its distance of travel is x + 92. The car’s velocity is equal to the final velocity of the train: v = 14a. Using the equation “distance equals speed times time,” we have x + 92 = vt = v14 = (14a)(14) = 196a. Re-writing the equation gives us x = 196a − 92. We now have two equations: x = 98a x = 196a − 92 Combining the equations gives us 98a = 196a−92. Solving this equation gives a = .9388 = .94 m/s2 . From the equation v = 14a, we see that v = 14(.9388) = 13.14 = 13 m/s . 6

37. A penny is dropped from rest from the top of the Sears Tower in Chicago. Considering that the height of the building is 427 m and ignoring air resistance, find the speed with which the penny strikes the ground. We are given v0 = 0, x = −427 m, and a = −9.80 m/s2 . From the equation v = v02 + 2ax, we have q p v = − v02 + 2ax = − 0 + 2(−9.80)(−427) = −91.5 The speed is |v| = 91.5 m/s . 41. From her bedroom window a girl drops a water-filled balloon to the ground, 6.0 m below. If the balloon is released from rest, how long is it in the air? Use the equation x = 21 at2 to get t=

r

2x = a

r

2(−6.0) = 1.1 s −9.80

42. At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 4.6 m/s. A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball? We use the equation v = v0 + at with v = 0, v0 = 4.6 m/s, and a = −9.80 m/s2 . Solving for t we have t=

v − v0 0 − 4.6 = = 0.47 s a −9.80

43. Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground? Pellet A leaves its gun, goes to its maximum height, and then starts to fall downward. When it returns to the height from which it was fired, it is in the same situation as pellet B when it is first fired from its gun. Therefore the difference in times for the two pellets is the amount of time it took for pellet A to leave the gun, go to the maximum height, and return to the gun. 7

If we consider the time from when pellet A leaves the gun to its highest point, we have v0 = 30.0 m/s, v = 0, and a = −9.80 m/s2 . Re-writing the equation v = v0 + at shows that t=

0 − 30.0 v − v0 = = 3.061 s a −9.80

Doubling this time gives us 6.12 s . 45. A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. the time it takes for the ball to fall halfway to the ground is 1.2 s. Find the time it takes for the ball to fall from rest all the way to the ground. For half-way, we can use the equation x = 21 at2 to get 1 1 x = (−9.80)(1.22 ) 2 2 Solving this equation gives us x = −14.112 m. Going back to the equation x = 21 at2 , we now find the time for the total distance:

t=

r

2x = a

r

2(−14.112) = 1.7 s −9.80

47. A ball is thrown straight upward and rises to a maximum height of 12.0 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value? We are given v = 0, x = 12.0 m, and a = −9.80 m/s2 . Using the equation v 2 = v02 + 2ax, we can write v0 =

p p v 2 − 2ax = 0 − 2(−9.80)(12.0) = 15.34 m/s

To find where the ball is at half its original speed, re-write the equation v 2 = v02 + 2ax to get v 2 − v02 x= = 2a

¡1

2

¢2 · 15.34 − 15.342 = 9.00 m 2(−9.80)

48. Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach 8

their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is 25.0 m/s and that the second arrow is fired 1.20 s after the first. Determine the initial speed of the second arrow. Since the arrows have the same acceleration (−9.80 m/s2 ) and have the same final velocity (v = 0) at the same time, from the equation v = v0 + at, we see that the two arrows must also have the same velocity at t = 1.20 s. For arrow number one at t = 1.20 s, v = v0 + at = 25.0 − 9.80(1.20) = 13.2 m/s . Since the two arrows have the same velocities, this is also the initial velocity of arrow number two. 51. A log is floating on swiftly moving water. A stone is fropped from rest from a 75-mhigh bridge and lands on the log as it passes under the bridge. If the log moves with a constant speed of 5.0 m/s, what is the horizontal distance between the log and the bridge when the stone is released? The falling stone is described by the equation x = 21 at2 . From this we see that the time of the fall is

t=

r

2x = a

r

2 · −75 = 3.91 s −9.80

The log’s motion is described by x = vt from which we have x = vt = (5.0)(3.91) = 20 m

53. A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.50 s after the stone is dropped. How deep is the hole? Let x be the depth of the hole. For the falling rock we have x = 21 at21 . Solving for t1 gives us

t1 =

r

2x = a

r

2x 9.80

For sound we have the equation x = vt2 = 343t2 . Solving for t2 : t2 =

x 343

Since t1 + t2 = 1.50, we make substitutions and write 9

r

2x x + = 1.50 9.80 343

To help solve this equation, let x = y 2 : r

y2 2y 2 + = 1.50 9.80 343 0.4518y + 0.002915y 2 = 1.50 0.002915y 2 + 0.4518y − 1.50 = 0 Using the quadratic formula we get

y=

−0.4518 +

p

0.45182 − 4(0.002915)(−1.50) = 3.252 2(0.002915)

Recall that x = y 2 . x = 3.2522 = 10.6 m

55. A ball is dropped from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths. 2 2 We can use the p equation v p= v0 + 2ax to find the final velocity of the ball that was dropped: v = v02 + 2ax = 0 + 2(−9.8)(−24) = 21.7 m/s.

Next, let x be the position where the two balls cross as measured from the bottom of the cliff. For the falling ball: x = 24 + 12 (−9.8)t2 . For the upward ball: x = 21.7t + 21 (−9.8)t2 . Setting these equal: 1 1 24 + (−9.8)t2 = 21.7t + (−9.8)t2 2 2 24 = 21.7t t = 1.106 s 10

Then x = 24 + 12 (−9.8)t2 = 24 − 4.9 · 1.1062 = 18.0 m. So the balls cross 6.0 m below the top of the cliff. 65. The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of 39 m/s (88 mi/h). To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent. Starting with v 2 = v02 + 2ax, we get v=

q

v02 + 2ax =

p

0 + 2(9.80)(99.4) = 44.1 m/s

68. A hot-air balloon is rising upward with a constant speed of 2.50 m/s. When the balloon is 3.00 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground? For the compass we have v0 = 2.50 m/s, x = −3.00 m, and a = −9.80 m/s2 . 1 x = v0 t + at2 2 1 −3.00 = 2.50t + (−9.80)t2 2 4.90t2 − 2.50t − 3.00 = 0 Using the quadratic formula:

t=

2.50 +

p

2.502 − 4(4.90)(−3.00) = 1.08 s 2(4.90)

71. Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initial velocity of +5800 m/s, while rocket B has an initial velocity of +8600 m/s. After a time t both rockets are again side by side, the displacement of each being zero. The acceleration of rocket A is −15 m/s2 . What is the acceleration of rocket B? Rocket A x = v0 t + 21 at2 0 = 5800t + 12 (−15)t2 7.5t2 − 5800t = 0 11

t(7.5t − 5800) = 0 t = 0 or 7.5t − 5800 = 0 t = 5800/7.5 = 773 s Rocket B x = v0 t + 12 at2 0 = 8600 · 773 + 12 a · 7732 0 = 6650000 + 299000a a=

−6650000 = −22 m/s2 299000

12