CHAPTER 17
Rectilinear Motion
17.1
The equation of free fall of an object (under the influence of gravity alone) is s = s 0 + v a t — I6t
2 , where s a is the
initial position and v a is the initial velocity at time t - 0. (We assume that the j-axis is directed upward away
from the earth, along the vertical line on which the object moves, with s =0 at the earth's surface, s is
measured in feet and t in seconds.) Show that, if an object is released from rest at any given height, it will have
dropped 16t
2 feet after t seconds.
I To say that the object is released from rest means that the initial velocity v 0 = 0, so its position after t
seconds is s 0 — 16t
2 . The difference between that position and its initial position s 0 is 16f
2
.
17.2
How many seconds does it take the object released from rest to fall 64 feet?
I By Problem 17.1, 64 = 16f
2
. Hence, t
2 = 4, and, since t is positive, t = 2.
17.3
A rock is dropped down a well that is 256 feet deep. When will it hit the bottom of the well?
I If t is the time until it hits the bottom, 256 = 16f
2
, so t
2 = 16, t = 4.
17.4
Assuming that one story of a building is 10 feet, with what speed, in miles per hour, does an object dropped from
the top of a 40-story building hit the ground?
f Let t be the time until the object hits the ground. Since the building is 400 feet tall, 400=16f
2 , t
2 =25,
t = 5. The velocity v = D,s. Since s = s 0 - I6t
2 , v = -32f. When t = 5, i; = -160. Thus, the speed
\v\ is 160ft/s. To change to mi/h, we calculate as follows:
In particular, when x = 160 ft/s, the speed is about 108.8 mi/h.
17.5
A rocket is shot straight up into the air with an initial velocity of 128 ft/s. How far has it traveled in 1 second?
I The height s = s 0 + v 0 t - I6t
2 . Since s 0 = 0 and v 0 = 128, s = 128t-l6t
2 . When t=l, s = 112ft.
17.6
In Problem 17.5, when does the rocket reach its maximum height?
I At the maximum value of 5, v = D,s = 0, but v = 128 - 3>2t. Setting v = 0, we obtain t = 4 seconds.
17.7
In Problem 17.5, when does the rocket strike the ground again and what is its velocity when it hits the ground?
I Setting s = 0, 128t - I6t
2 = 0, 16t(8 - t) = 0, / = 0 or t = 8. So the rocket strikes the ground again
after 8 seconds. When / = 8, the velocity v = 128- 32t= 128-256= -128 ft/s. The velocity is negative
(because the rocket is moving downward) and of the same magnitude as the initial velocity (see Problem 17.28).
17.8
A rock is thrown straight down from a height of 480 feet with an initial velocity of 16 ft/s. How long does it take
to hit the ground and with what speed does it hit the ground?
I The height s = s 0 + v 0 t - I6t
2 . In this case, J 0 = 480 and u 0 =-16. Thus, s = 480 - I6t - 16t
2 =
16(30- t-t
2 ) = 16(6+ t)(5-t). Setting 5 = 0, we obtain t=-6 or t = 5. Hence, the rock hits the
ground after 5 seconds. The velocity v = D,s = -16 - 32f. When t = 5, v = -16 - 160 = -176, so the
rock hits the ground with a speed of 176 ft/s. (The minus sign in the velocity indicates that the rock is moving
downward.)
17.9
Under the same conditions as in Problem 17.8, how long does it take before the rock is moving at a speed of
112 ft/s?
133
Rectilinear Motion
17.1
The equation of free fall of an object (under the influence of gravity alone) is s = s 0 + v a t — I6t
2 , where s a is the
initial position and v a is the initial velocity at time t - 0. (We assume that the j-axis is directed upward away
from the earth, along the vertical line on which the object moves, with s =0 at the earth's surface, s is
measured in feet and t in seconds.) Show that, if an object is released from rest at any given height, it will have
dropped 16t
2 feet after t seconds.
I To say that the object is released from rest means that the initial velocity v 0 = 0, so its position after t
seconds is s 0 — 16t
2 . The difference between that position and its initial position s 0 is 16f
2
.
17.2
How many seconds does it take the object released from rest to fall 64 feet?
I By Problem 17.1, 64 = 16f
2
. Hence, t
2 = 4, and, since t is positive, t = 2.
17.3
A rock is dropped down a well that is 256 feet deep. When will it hit the bottom of the well?
I If t is the time until it hits the bottom, 256 = 16f
2
, so t
2 = 16, t = 4.
17.4
Assuming that one story of a building is 10 feet, with what speed, in miles per hour, does an object dropped from
the top of a 40-story building hit the ground?
f Let t be the time until the object hits the ground. Since the building is 400 feet tall, 400=16f
2 , t
2 =25,
t = 5. The velocity v = D,s. Since s = s 0 - I6t
2 , v = -32f. When t = 5, i; = -160. Thus, the speed
\v\ is 160ft/s. To change to mi/h, we calculate as follows:
In particular, when x = 160 ft/s, the speed is about 108.8 mi/h.
17.5
A rocket is shot straight up into the air with an initial velocity of 128 ft/s. How far has it traveled in 1 second?
I The height s = s 0 + v 0 t - I6t
2 . Since s 0 = 0 and v 0 = 128, s = 128t-l6t
2 . When t=l, s = 112ft.
17.6
In Problem 17.5, when does the rocket reach its maximum height?
I At the maximum value of 5, v = D,s = 0, but v = 128 - 3>2t. Setting v = 0, we obtain t = 4 seconds.
17.7
In Problem 17.5, when does the rocket strike the ground again and what is its velocity when it hits the ground?
I Setting s = 0, 128t - I6t
2 = 0, 16t(8 - t) = 0, / = 0 or t = 8. So the rocket strikes the ground again
after 8 seconds. When / = 8, the velocity v = 128- 32t= 128-256= -128 ft/s. The velocity is negative
(because the rocket is moving downward) and of the same magnitude as the initial velocity (see Problem 17.28).
17.8
A rock is thrown straight down from a height of 480 feet with an initial velocity of 16 ft/s. How long does it take
to hit the ground and with what speed does it hit the ground?
I The height s = s 0 + v 0 t - I6t
2 . In this case, J 0 = 480 and u 0 =-16. Thus, s = 480 - I6t - 16t
2 =
16(30- t-t
2 ) = 16(6+ t)(5-t). Setting 5 = 0, we obtain t=-6 or t = 5. Hence, the rock hits the
ground after 5 seconds. The velocity v = D,s = -16 - 32f. When t = 5, v = -16 - 160 = -176, so the
rock hits the ground with a speed of 176 ft/s. (The minus sign in the velocity indicates that the rock is moving
downward.)
17.9
Under the same conditions as in Problem 17.8, how long does it take before the rock is moving at a speed of
112 ft/s?
133