# Giancoli Physics (5th ed) Chapter 2

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## Contents

- 1 Problems
- 1.1 1. What must be your average speed in order to travel 230 km in 3.25 hours?
- 1.2 2.
- 1.3 3. If you are driving 110 km/hr along a straight road and you look to the side for 2.0 sec, how far do you travel during this period?
- 1.4 4.
- 1.5 5. You are driving home from school steadily at 65 mph for 130 miles. It begins to rain and you slow to 55 mph. You arrive home after driving 3 hours and 20 minutes. (a) How far is your hometown from school? (b) What was your average speed?
- 1.6 6.
- 1.7 7.A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 minutes. Calculate (a) the average speed and (b) the average velocity.
- 1.8 8.
- 1.9 9.Two locomotives approach each other on parallel tracks. Each has a speed of 95 km/h with respect to the ground. If they are initially 8.5 km apart, how long will it be before they reach each other? (See Fig. 2-28)
- 1.10 10.
- 1.11 11.
- 1.12 12.
- 1.13 13.A sports car accelerates from rest to 95 km/h in 6.2 sec. What is its average acceleration in m/s/s?
- 1.14 14.
- 1.15 15.A sprinter accelerates from rest to 10.0 m/s in 1.35 sec. What is her acceleration (a) in m/s/s, and (b) in km/h/h?
- 1.16 16.
- 1.17 17.
- 1.18 18.
- 1.19 19.
- 1.20 20.
- 1.21 21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?
- 1.22 22.
- 1.23 23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?
- 1.24 24.
- 1.25 25.A car traveling 45 km/h slows down at a constant .50 m/s/s just by "letting up on the gas". Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop and (c) the distance it travels during the first and fifth seconds.
- 1.26 26.
- 1.27 27. Determine the stopping distances for an automobile with an initial speed of 90 km/h and a human reaction time of 1.0s: (a) for an acceleration a= - 4.0 m/s/s; (b) for an acceleration a= - 8 m/s/s.
- 1.28 28.
- 1.29 29.
- 1.30 30.
- 1.31 31.
- 1.32 32.
- 1.33 33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?
- 1.34 34.
- 1.35 35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?
- 1.36 36.
- 1.37 37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?
- 1.38 38.
- 1.39 39.
- 1.40 40.
- 1.41 41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?
- 1.42 42.
- 1.43 43.
- 1.44 44.
- 1.45 45.
- 1.46 46.
- 1.47 47.A rock is dropped from a sea cliff and the sound of it striking the ocean is hears 3.4 seconds later. If the speed of sound is 340 m/s, how high is the cliff?
- 1.48 48.
- 1.49 49.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?
- 1.50 50.
- 1.51 51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?
- 1.52 52.
- 1.53 53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?
- 1.54 54.
- 1.55 55.
- 1.56 56.
- 1.57 57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.
- 1.58 58.
- 1.59 59.
- 1.60 60.A person jumps from a fourth-story window 15.0 m above a firefighter's safety net. The survivor stretches the net 1.0 m before coming to rest, Fig 2-35. (a) What was the average deceleration experienced by the survivor when slowed to rest by then net? (b) What would you do to make it "safer" (that is, generate a smaller deceleration): would you stiffen or loosen the net? Explain.
- 1.61 61. The acceleration due to gravity on the Moon is about one sixth what it is on Earth. If an object is thrown vertically upward on the Moon, how many times higher will it go than it would on Earth, assuming the same initial velocity?
- 1.62 62. A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 "g's" (1.00 g = 9.80 m/s/s). Assuming uniform deceleration of this value, calculate the distance over which the front end of the car must be designed to collapse if a crash being the car to rest from 100 km/h.
- 1.63 63. A race car driver must average 200.0 km/h over the course of a time trial lasting ten laps. If the first nine laps were done at 199.0 km/h, what average speed must be maintained for the last lap?
- 1.64 64.
- 1.65 65. A first stone is dropped from the roof of a building 2.00 s after that a second stone is thrown straight down with an initial speed of 30.0 m/s, and it is observed that the two stones land at the same time. (a) How long did it take the first stone to reach the ground? (b) How high is the building? (c) What are the speeds of the two stones just before they hit the ground?
- 1.66 66.
- 1.67 67.A police car at rest, passed by a speeder traveling at a constant 110 km/hr, takes off in hot pursuit. The police officer catches up to the speeder in 700 m, maintaining a constant acceleration. Qualitatively plot the position vs time graph for both cars from the police car's start to the catch up point. b) calculate how long it took the police officer to overtake the speeder, c) calculate the required police car acceleration, and d) calculate the speed of the police car at the overtaking point.?
- 1.68 68.
- 1.69 69. Pelicans tuck their wings and free fall straight down when diving for fish. Suppose a pelican starts its dive from a height of 16.0 m and cannot change its path once committed. If it takes a fish 0.20 s to perform evasive action, at what minimum height must it spot the pelican to escape? Assume the fish is at the surface of the water.
- 1.70 70.
- 1.71 71.
- 1.72 72.
- 1.73 73. Bond is standing on a bridge, 10 m above the road below, and his pursuers are getting too close for comfort. He spots a flatbed truck loaded with mattresses approaching at 30 m/s, which he measures by knowing that the telephone poles are 20 m apart in this country. The bed of the truck is 1.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he jumps down from the bridge onto the truck, making his getaway. How many poles is it?

## Problems

### 1. What must be your average speed in order to travel 230 km in 3.25 hours?

distance = 230 km and time = 3.25 hours so average speed is 230/3.25, or

70.8 km/hr.

### 2.

- (0.60 hr - 36 minutes)

### 3. If you are driving 110 km/hr along a straight road and you look to the side for 2.0 sec, how far do you travel during this period?

The only tricky part of this problem is to remember to keep your time units the same. Divide your speed in km/hr by 3.6 and you get m/s.

110/3.6 = 30.555556 m/s

Now use 30.555556 as your velocity and 2.0 as your time and solve for displacement.

v = s/t, so s = vt = (30.555556 m/s)(2.0 s) =61 m

### 4.

- (105 km/h, 29 m/s, 95 ft/s)

### 5. You are driving home from school steadily at 65 mph for 130 miles. It begins to rain and you slow to 55 mph. You arrive home after driving 3 hours and 20 minutes. (a) How far is your hometown from school? (b) What was your average speed?

(a) First, figure out how long it took to travel the first 130 miles. v = s/t. v = 65 and s = 130. Solve for t and you get that the first 130 miles took 2 hours. Since the whole trip took 3 hours and 20 minutes, you know that 1 hour and 20 minutes was spent driving in the rain at 55 mph. Since you know time (1.3333 hrs) and speed (55 mph), you can figure out how much distance you covered in the second part of the trip. v = s/t. v = 55 and t = 1 1/3 (1 hour and 20 minutes). Solve for s and you find that you drove 73.333 miles in the rain.

You can add these distances together to find out how far you traveled altogether

73.333 + 130 = 203.333 miles =

203 miles(b) Average speed is total distance/total time. You just figured out the total distance (203.333 miles) and the total time was given (3.3333 hours). Divide these two. Average speed = 60.999 mph =

61 mph.

### 6.

- (about 300 m/s)

### 7.A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 minutes. Calculate (a) the average speed and (b) the average velocity.

The important thing to notice here is that average speed means

distance/time and average velocity meansdisplacement/time.(a) First we must figure out how much distance the jogger covered. Eight times around .25 miles = 2 miles. The book wants your answer in m/s so we must convert 2 miles to meters. 1 mile = 1.61 km, so 2 miles = 3.22 km. 3.22 km*1000 = 3220 m.

Now we need to know how much time it took. We need to take the time they gave us (12.5 minutes) and convert it to seconds. 12.5 * 60 = 750 sec.

Now we can figure out speed. distance/time = 3220 m / 750 sec =

4.29 m/s(b) Now we have to consider the

displacement, not the distance. Remember that displacement means how far (and in what direction) you are now from where you started. If someone jogs eight complete laps, then they end up right where they first started. Since displacement = 0,velocity = 0.

### 8.

- (10.4 m/s, +3.5 m/s)

### 9.Two locomotives approach each other on parallel tracks. Each has a speed of 95 km/h with respect to the ground. If they are initially 8.5 km apart, how long will it be before they reach each other? (See Fig. 2-28)

<Fig 2-28>

Since their velocities are the same and they are headed directly towards each other, we can logically assume that they will meet at a point that is halfway between where they started. This point is 8.5/2 = 4.25 km away from each train. So all we really need to worry about is how long it takes one train to travel 4.25 km.

v

_{av}= s/t

v = 95 km/h and s = 4.25 km, so solve for time and you will get t = .044737 hrs. .044737 * 60 =2.7 min.

### 10.

- (4.43 h, 881 km/h)

### 11.

(55 km/hr, and 0)

### 12.

- (6.73 m/s)

### 13.A sports car accelerates from rest to 95 km/h in 6.2 sec. What is its average acceleration in m/s/s?

a

_{av}= Dv/DtFirst we need to change 95 km/hr to m/s. 95/3.6 = 26.3889 m/s

Dv means change in velocity. If we started at 0 m/s and ended at 26.3889 m/s, then Dv = + 26.3889 m/s. We know time is 6.2 seconds. a = v/t. 26.3889 / 6.2 =

4.3 m/s/s.

### 14.

- (5.2 s)

### 15.A sprinter accelerates from rest to 10.0 m/s in 1.35 sec. What is her acceleration (a) in m/s/s, and (b) in km/h/h?

(a) Change in velocity is from 0 to 10, so it's 10.0 m/s. Time is 1.35 sec. a = v/t = 10/1.35 =

7.41 m/s/s(b) Now we have to convert our velocity to km/h and our time to hours.

Multiply by 3.6 to go from m/s to km/h. 10.0*3.6 = 36 km/h

Divide seconds by 3600 to get hours. 1.35/3600 = .000375 hrs.

Now we can use the acceleration formula. a = v/t. v = 36 km/h and t = .000375 hr.

36/.000375 = 96000 km/h/h =

9.60*10^4 km/h/h

### 16.

- (-6.3ms
^{-2}, .64 g)

### 17.

### 18.

this becomes this if u = 0 v = u + at v = at v = ut + ^{1}/_{2}at^{2}v = ^{1}/_{2}at^{2}v ^{2}= u^{2}+ 2asv ^{2}= 2ass = s =

### 19.

- (2.2ms
^{-2}, 110 m)

### 20.

- (-2.4 ms
^{-2})

### 21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?

- -The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.

- -A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:
- (30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s

- The initial velocity here is zero, so we can drop it from the equation.

- 900= 2(3.0 m/s/s) s

- s= 1.5x102 m

### 22.

- (4.41 ms
^{-2}, 2.61 s)

### 23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?

- -To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.

- -We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.

- a= (25 m/s) / (5 sec)
- a= - 5 m/s/s, because it is slowing down

- -Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.

- (0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s

- 0= 625 + (-10) s

- -625= (-10) s

- s= 62.5 m

### 24.

- (33 ms
^{-1})

### 25.A car traveling 45 km/h slows down at a constant .50 m/s/s just by "letting up on the gas". Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop and (c) the distance it travels during the first and fifth seconds.

-Change 45 km/h into m/s

45 km/h ((1 hr)/(3600 s)) ((1000 m)/(1 km)) = 12.5 m/s

-Remember that slowing down is a negative acceleration. So, a = - .5 m/s/s

-(a) The initial velocity (u) is 12.5 m/s, the acceleration is -.5 m/s/s, and the final velocity (v) is 0 m/s, because we want it to stop. Using the formula v2 = u2 + 2as, calculate the distance it takes to stop.

02 = (12.5)2 +2(-.5)s

s = 156.25 m or 1.6 x 102 m

-(b) To find the time it takes to stop, put the information in to the formula: v = u + at.

0 = 12.5 + (-.5)t

t = 25 s

-(c) The distance the car traveled during the first second can be found by using the formula s = ut + 1/2 at2, and substituting 1 s in for t.

s = (12.5)(1) + (.5)(-.5)(1)2

s = 12.25 m or 12 m

-During the fifth second, you have to realize that the distance traveled during this time is the distance between the fifth and sixth seconds. So to find the distance during only the fifth second, you take the distance traveled up to the sixth second, and subtract the distance traveled up to the fifth second.

s = (12.5)(6) + (.5)(-.5)(6)2 s = 66 m

s = (12.5)(5) + (.5)(-.5)(5)2 s = 56.25 m

Now take the distance up to the sixth second and subtract the distance up to the fifth.

66 m - 56.25 m = 9.75 m or 10 m

### 26.

- (-3.9x10
^{2}ms^{-2}, 40 g)

### 27. Determine the stopping distances for an automobile with an initial speed of 90 km/h and a human reaction time of 1.0s: (a) for an acceleration a= - 4.0 m/s/s; (b) for an acceleration a= - 8 m/s/s.

-Change 90 km/h into m/s.

- 90 km/h ((1 hr)/(3600 s)) ((1000 m)/(1 km)) = 25 m/s

- -(a) First, find the distance it takes to stop with an acceleration of - 4.0 m/s/s. Use the formula: v2 = u2 + 2as

- 02 = 252 + 2(-4.0)s

- s = 78.125 m

- -But remember that it takes one second for human reaction, so before the driver starts slowing down, the car is traveling at 25 m/s for one second. The car travels a distance of 25 m.

- -Add 25 m to the 78.125 m, and you get 103 m.

- -(b) Do the same as with a, but for an acceleration of -8.0 m/s/s.

- 02 = 252 + 2(-8.0)s

- s = 39.06 m

- -Add 39.06 m to the 25 m the car travels during the one second of human reaction to get 64 m.

### 28.

- ()

### 29.

- (24.0 s, 240 km/h)

### 30.

- (stop, dude)

### 31.

- (3.1 s)

### 32.

- (28.6 g)

### 33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?

- -First you have to change the 90 km/h to m/s

- 90 km/h (1 hr / 3600 s)(1000 m / 1 km)
- = 25 m/s

- 90 km/h (1 hr / 3600 s)(1000 m / 1 km)

- -You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.

- -Using the formula v = u + at, we can find the time needed.

- 25 m/s = 0 + (9.8)t

- t = 2.6 s

### 34.

- (60.0 m)

### 35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?

- -We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.

- -(a) To find the time, we can use the formula: s = ut + 1/2 at2

- 380 m = 0(t) + 1/2 (9.8)t2

- t2 = 77.6161

- t = 8.81 s

- -(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as

- v2 = (0)2 + 2(9.8)(380)

- 86.3 m/s

### 36.

- (32 m, 5.1 s)

### 37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?

- -If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.

- -We can use the formula: s = ut + 1/2 at2. Now put in the information we know.

- 2.7 = 0(t) + 1/2(9.8)t2

- 2.7 = 4.9t2

- t = .74 s

- -But since this is only on the way down, the time must be doubled to include the time it was going up.

- .74 x 2 = 1.48 which is about 1.5 s

### 38.

- (13 m)

### 39.

- (graphs)

### 40.

- (4.85 m/s, 0.990 s)

### 41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

- -The package and the helicopter are both 105 m above the earth, but are both going up at 5.5 m/s. First we must find out the amount of time the package spends going up, before it begins to go down to the earth. The final velocity (v) for when the package is going up is zero, because at this moment, the package isn't going up anymore, but is falling toward the ground. We can use the formula: v = u + at

- 0 = 5.5 + (-9.8)t

- -5.5 = (-9.8)t

- t = .56 s

- -Now we know how long the package spends going up. But how high does the package go above the 105 m? Remember that the initial velocity (u) is zero. For this we can use the formula: v2 = u2 + 2as.

- 02 = (5.5)2 +2(-9.8)s

- s = 1.54 m

- -Add this to the 105 m, and we get 106 .54 m.

- -How long does it take for the package to drop from this height, 106.54 m? We can use the formula: s = ut + 1/2 at2

- 106.54 = (0)t + 1/2(9.8)t2

- t = 4.66 s

- -Remember that the package was going up first and took .56 s to do that. Add 4.66 s and .56 s to get 5.22 s.

Challenge from Murray - this can be solved without dividing it into two parts - use ut+1/2at^{2} and the quadratic, or use v^{2} = u^{2} + 2as and then v = u + at

### 42.

- ()

### 43.

- ()

### 44.

- (+/- 12.8 ms
^{-1}, .735 s and 3.35 s)

### 45.

- (0.038 s)

### 46.

- (1.8 m)

### 47.A rock is dropped from a sea cliff and the sound of it striking the ocean is hears 3.4 seconds later. If the speed of sound is 340 m/s, how high is the cliff?

- -We can use the formula s = ut + 1/2 at2, but the initial velocity (u) is zero, so we are left with, s = 1/2 at2. Replacing h with s, and solving for time gives us tf (the time to fall) = (2h)/g)^1/2

- -The time it takes for the sound to be heard up at the top of the cliff can be found by the formula v = s/t or t = s/v, which is ts = h /(340 m/s).

- -The total time is 3.4 s, so ts + tf = 3.4 s

- -Substitute for ts and tf. This gives us:

- ((h /340) + (2h /g)^1/2 = 3.4)

- -This is the same as:

- 1/340) ((h)^1/2)2 + (2/g)^1/2 (h)^1/2 - 3.4 = 0

- -This is quadratic now, where x = h^1/2, so solve it using the quadratic formula (in your book on page 1042).

- h = 52 m

### 48.

- ()

### 49.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?

- -(a) If it is thrown upward, we must find the time it takes to stop going upward, and reach a final velocity (v) of zero. Using the formula v = u + at, find the time in the air.

- 0 = 12 + (-9.8)t

- t = 1.23 s

- -Now find the height it went during that 1.22 s using the formula v2 = u2 + 2as.

- 02 = (12)2 + 2(-9.8)s

- s = 7.35 m

- -Add 7.35 m to the height of the cliff, 75 m, to get 82.35 m. Now find the time it takes to fall this new height using the formula s = ut + 1/2 at2

- 82.35 m = (0)t + 1/2(-9.8)t2

- t = 4.1 s

- -Add 4.1 s to the time the stone is going upward, which is 1.23 s, and you get 5.33 s.

- -(b) The initial velocity (u) is 0 m/s and the height is 82.35 m. Using the formula v2 = u2 + 2as, find the final velocity.

- v2 = (0)2 + 2(-9.8)(82.35)

- v = - 40.2 m/s (negative because it is going down)

- -(c) The total distance it traveled would be the distance it traveled going up, 7.35 m, and the total distance it went to the bottom of the cliff, 82.35 m. So adding these together, we get, 89.7 m.

### 50.

- ()

### 51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?

- -(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s

- -(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.

- -(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s

- -(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s

- -(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s

### 52.

- ()

### 53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?

- -(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.

- -(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)

- -(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)

- -(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.

### 54.

- ()

### 55.

- ()

### 56.

- ()

### 57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.

- -See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.

- The velocity graph is going to look like this:
- interval Position Velocity
- 0-20 sec constant slope constant V (horizontal line)
- 20-30 sec curve concave upwards positive acceleration - upward sloping straight line
- 30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero
- 37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)
- 46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero

### 58.

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### 59.

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### 60.A person jumps from a fourth-story window 15.0 m above a firefighter's safety net. The survivor stretches the net 1.0 m before coming to rest, Fig 2-35. (a) What was the average deceleration experienced by the survivor when slowed to rest by then net? (b) What would you do to make it "safer" (that is, generate a smaller deceleration): would you stiffen or loosen the net? Explain.

- <Fig 2-35>

- (a) First we need to find the final velocity when the person was falling and the net caught them. Use SUVAT.

- s = - 15 m. u = 0 m/s a = - 9.8 m/s/s (gravity) v = ???

- v2 = u2 + 2as v² = 0² + 2(-9.8)(.15)

- Do the math and you should get that v = ±Ö294. Common sense says that since the person is falling, their velocity should be negative, so v = -17.14643

- Now that we have the velocity, we can solve for the acceleration of the person when they're in the net. SUVAT again.

- s = 1.0 m. u = -17.14643 v = 0 m/s (the net slows them to rest) a = ???

- v2 = u2 + 2as 0² = (.17.14643)² + 2*1.0*a

- Do math and you should get that a = -147 m/s/s, but since they asked for deceleration (meaning negative acceleration) your final answer should be 147 m/s/s or 150 m/s/s with sig figs.

- (b) To make the deceleration less, you would want to loosen the net so that there would be a greater displacement, allowing the person a greater distance over which to slow down.

### 61. The acceleration due to gravity on the Moon is about one sixth what it is on Earth. If an object is thrown vertically upward on the Moon, how many times higher will it go than it would on Earth, assuming the same initial velocity?

- Use any formula with acceleration and displacement in it and see what happens.

- v2 = u2 + 2as Pretend that initial velocity is 10 m/s and final velocity is 0 m/s (the object reaches its peak). First work out this equation using acceleration due to gravity on Earth (a = -9.8 m/s/s). Displacement ends up being 5.102 m.

- Now do the same formula with the same initial and final velocities, but use acceleration on the moon (-9.8/6 = -1.633333). You will find that the displacement on the moon is 30.612, which coincidentally is exactly six times greater than the displacement on Earth.

### 62. A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 "g's" (1.00 g = 9.80 m/s/s). Assuming uniform deceleration of this value, calculate the distance over which the front end of the car must be designed to collapse if a crash being the car to rest from 100 km/h.

- The first thing to do is to get rid of km/h because we need m/s. Divide 100 km/h by 3.6 and you get 27.77778 m/s. We also need to know what exactly 30 "g's" is. 30*9.8 = 294 m/s/s. Now we can use SUVAT to solve for displacement.

- v2 = u2 + 2as v = 0 (car crashes to rest) u = 27.77778 m/s a = 294 s = ???

- Do the math and you will find that displacement needs to be 1.132 meters, of 1.3 m. with sig figs.

### 63. A race car driver must average 200.0 km/h over the course of a time trial lasting ten laps. If the first nine laps were done at 199.0 km/h, what average speed must be maintained for the last lap?

- We only need to use v = s/t for this problem, but change it to t = s/v. The total time is (total distance)/(average velocity). We know that the total distance is 10 laps, and we will use x to represent how many kilometers are in a lap. We also know that the driver must maintain an average of 200 km/h over the ten laps.

- Total time = (10x)/200

- We can also apply this same formula to the first nine laps that he has already completed. (Average speed for the first nine laps is 199 km/h)

- Time so far = (9x)/199

- So, logically, the time he has left is total time - time so far.......(10x)/200 - (9x)/199. This gets a little ugly, but find a common denominator and subtract these fractions and you should end up with time = (19x)/3980.

- Now, we know how much time the driver has left and we know how far he has to go (one lap, or x km.), so we can use the original equation v = s/t to figure out what his velocity must be.

- velocity = distance/time = x/(19x/3980). *Remember that we're not solving for x - they end up cancelling out

- Simplify and you get that velocity = 209.5 m/s

### 64.

- (9.8 m, 39 m)

### 65. A first stone is dropped from the roof of a building 2.00 s after that a second stone is thrown straight down with an initial speed of 30.0 m/s, and it is observed that the two stones land at the same time. (a) How long did it take the first stone to reach the ground? (b) How high is the building? (c) What are the speeds of the two stones just before they hit the ground?

- a) The stone thrown later overtakes the dropped stone just as it reaches the ground. The formula to use is
- s = ut + 1/2at2 - but therein lies the rub - we have two rocks - one dropped and one thrown down 2.00 seconds later. We have two equations of motion:
- s = 1/2at2 - for the dropped rock (u = 0)
- s = ut + 1/2at2 - for the thrown down rock - but it is a bit more complicated than that. for the thrown rock it really is
- s = u(t-2.00) + 1/2a(t-2.00)2 - because it is thrown 2.00 seconds later. Plugging in for the initial velocity and acceleration (making down positive for clarity) the displacement from the top of the cliff for the thrown rock is given by:
- s = 30.0(t-2.00) + 1/2(9.80)(t-2.00)2
- And the displacement for the dropped rock is simply:
- s = 1/2(9.80)t2 -

- The ground happens to be where these two displacements are equal so:
- 30.0(t-2.00) + 1/2(9.80)(t-2.00)2 = 1/2(9.80)t2 (distribute)
- 30.0t-60.0 +4.9(t-2.00)2 = 4.9t2 (expand the square)
- 30.0t-60.0 +4.9(t2-4t +4.00) = 4.9t2 (distribute)
- 30.0t-60.0 +4.9t2-19.6t +19.6 = 4.9t2 (Combine like terms and put all factors of t on one side)
- 30.0t -19.6t = 60.0-19.6
- 10.4t = 40.4

- t = 40.4/10.4 = 3.8846 s = 3.88 s
- b) For the height of the building let's use the displacement of the dropped stone and plug in the time of 3.8846 s
- s = 1/2(9.80)(3.8846)2 = 73.9 m

- c) For the respective speeds I am going to use v = u + at where the acceleration of both is that of gravity:
- for the dropped stone, u = 0:
- v = u + at = 0 + 9.8(3.8846) = 38.1 m/s
- for the thrown stone, u = 30.0 m/s and t = 3.8846 - 2.00 = 1.8846 s:
- v = u + at = 30.0 + 9.80(1.8846) = 48.5 m/s

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### 67.A police car at rest, passed by a speeder traveling at a constant 110 km/hr, takes off in hot pursuit. The police officer catches up to the speeder in 700 m, maintaining a constant acceleration. Qualitatively plot the position vs time graph for both cars from the police car's start to the catch up point. b) calculate how long it took the police officer to overtake the speeder, c) calculate the required police car acceleration, and d) calculate the speed of the police car at the overtaking point.?

- First off, 110 km/hr = 30.555 m/s

- a) for the graph - here they are meeting at 700 m The speeder has a straight line plot as they are going at a constant velocity, and the police officer has a curved plot as they are accelerating uniformly:

- b) The speeder's position is given by:
- s = ut + 1/2at2 = ut (a = 0) = (30.555)t and since they meet at 700 m, the time is
- 700 = 30.555(t), t = 22.909 s = 22.9 s

- c) The police acceleration
- s = ut + 1/2at2 = 1/2at2 (u = 0)
- 700 = 1/2a(22.909)2, a = 2.6675 m/s/s = 2.67 m/s/s

- d) the police officer's speed is given by v = u + at:
- v = 0 + (2.6675)(22.90909) = 61.111 m/s = 61.1 m/s* 3.6 = 220 km/hr

### 68.

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### 69. Pelicans tuck their wings and free fall straight down when diving for fish. Suppose a pelican starts its dive from a height of 16.0 m and cannot change its path once committed. If it takes a fish 0.20 s to perform evasive action, at what minimum height must it spot the pelican to escape? Assume the fish is at the surface of the water.

- Before solving this wonderful problem, consider this poem by Ogden Nash:

- A wonderful bird is the Pelican
- His beak holds more than his belly can

- I know I can figure out the time it takes the pelican to reach the water: (s = -16.0m, u = 0, a = -9.80 m/s/s) using:
- s = ut + 1/2at2
- -16.0 = 0 + 1/2(-9.80)t2
- t = 1.807 s
- I also know that the fish needs to see the pelican .20 seconds before this or 1.807 s - .20 s = 1.607 s after the pelican has started it's descent. The pelican has gone down:
- s = ut + 1/2at2 = 0 + 1/2(-9.80)(1.607)2 = -12.65 m and is therefore 16 - 12.65 = 3.34 m above the water.

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### 72.

- (29.0 m)

### 73. Bond is standing on a bridge, 10 m above the road below, and his pursuers are getting too close for comfort. He spots a flatbed truck loaded with mattresses approaching at 30 m/s, which he measures by knowing that the telephone poles are 20 m apart in this country. The bed of the truck is 1.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he jumps down from the bridge onto the truck, making his getaway. How many poles is it?

- Bond will fall 10 - 1.5 or 8.5 meters, and I can find the time it takes to hit the truck using (s = 8.5 m (down is +), a = 9.8, u = 0)
- s = ut + 1/2at2, 8.5 = 0 + 1/2(9.80)t2, t = 1.317 s
- Since the truck is going 30.0 m/s, I can use s = ut + 1/2at2 (a = 0) = ut = 30.0(1.317) = 39.51 m which is almost two telephone poles away.