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Rectilinear Motion solved questions
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CHAPTER 17Rectilinear Motion
17.1 The equation of free fall of an object (under the influence of gravity alone) is s = s0 + vat — I6t2, where sa is the
initial position and va 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 16t2 feet after t seconds.
I To say that the object is released from rest means that the initial velocity v0 = 0, so its position after t
seconds is s0 — 16t2. The difference between that position and its initial position s0 is 16f2.
17.2 How many seconds does it take the object released from rest to fall 64 feet?
I By Problem 17.1, 64 = 16f2. Hence, t2 = 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 = 16f2, so t2 = 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=16f2, t2 =25,
t = 5. The velocity v = D,s. Since s = s0 - I6t2, 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 = s0 + v0t - I6t2. Since s0 = 0 and v0 = 128, s = 128t-l6t2. 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 - I6t2 = 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 = s0 + v0t - I6t2. In this case, J0 = 480 and u 0 =-16. Thus, s = 480 - I6t - 16t2 =
16(30- t-t2) = 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
, 134 0 CHAPTER 17
I From Problem 17.8, we know that v = D,s = —16 — 32t. Since the rock is moving downward, a speed of
112 ft/s corresponds to a velocity v of -112. Setting -112=-16-32r, 32/ = 96, t = 3 seconds.
17.10 Under the same conditions as in Problem 17.8, when has the rock traveled a distance of 60 feet?
f Since the rock starts at a height of 480 feet, it has traveled 60 feet when it reaches a height of 420 feet.
Since s = 480 - 16f - 16f2, we set 420 = 480 - 16f - 16*2, obtaining 4t2 + 4f - 15 = 0, (It + 5)(2t - 3)= 0,
t=— 2.5 or f=1.5. Hence, the rock traveled 1.5 seconds.
17.11 An automobile moves along a straight highway, with its position 5 given by s = 12t3 - l&t2 + 9t- 1.5 (s in
feet, t in seconds). When is the car moving to the right, when to the left, and where and when does it change
direction?
I Since s increases as we move right, the car moves right when v = D,s>0, and moves left when
v = D,s<0. v= 36t2 -36t + 9 = 9(4f 2 -4t + 1) = 9(2t- I)2. Since i; >0 (except at t = 0.5, where
i; = 0), the car always moves to the right and never changes direction. (It slows down to an instantaneous
velocity of 0 at t = 0.5 second, but then immediately speeds up again.)
17.12 Refer to Problem 17.11. What distance has the car traveled in the one second from t = 0 to t = 1?
I From the solution to Problem 17.11, we know that the car is always moving right. Hence, the distance
traveled from t = 0 to t = 1 is obtained by taking the difference between its position at time t = 1 and its
position at time t = 0: s(l) - s(0) = 1.5 - (-1.5) = 3ft.
17.13 The position of a moving object on a line is given by the formula s = (t — l)3(t — 5). When is the object moving
to the right, when is it moving left, when does it change direction, and when is it at rest? What is the farthest to
the left of the origin that it moves?
I v = D,s = (t-l)3 + 3(t-l)2(t-5) = (t-l)2[t-l + 3(t-5)] = 4(t-l)2(t-4). Thus, u > 0 when t>
4, and v<0 when t<4 (except at t=l, when v = 0). Hence, the object is moving left when
t<4, and it is moving right when t>4. Thus, it changes direction when r = 4. It is never at rest. (To be
at rest means that s is constant for an interval of time, or, equivalently, that v = 0 for an entire interval of
time.) The object reaches its farthest position to the left when it changes direction at t = 4. When t = 4,
s = -27.
17.14 A particle moves on a straight line so that its position s (in miles) at time t (in hours) is given by
s = (4t — l)(t — I)2. When is the particle moving to the right, when to the left, and when does it change
direction? When the particle is moving to the left, what is the maximum speed that it achieves?
I v = D,s = 4(t - I) 2 + 2(t - l)(4f - 1) = 2(t - l)[2(t - 1) + 4t - 1] = 2(t - 1)(6( - 3) = 6(t - \)(2t - 1). Thus,
the key values are t = \ and ( = 0.5. When t>\, v>Q; when 0.5<«1, v<0; when f<0.5,
v>0. Thus, the particle is moving right when f<0.5 and when t>l. It is moving left when
0.5 < t< 1. So it changes direction when t = 0.5 and when t = 1. To find out what the particle's maximum
speed is when it is moving left, note that the speed is \v\. Hence, when v is negative, as it is when the particle is
moving left, the maximum speed is attained when the velocity reaches its absolute minimum. Now, D,v =
6[2(t- l) + 2 f - l ] = 6(4f-3), and D2v = 24>0. Hence, by the second derivative test, v reaches an
absolute minimum when t = 0.75 hour. When t = 0.75, i; = -0.75 mi/h. So the desired maximum speed
is 0.75 mi/h.
17.15 Under the assumptions of Problem 17.14, what is the total distance traveled by the particle from t = 0 to
f=l?
I The problem cannot be solved by simply finding the difference between the particle's positions at t = 1 and
t = 0, because it is moving in different directions during that period. We must add the distance dr traveled while
it is moving right (from t = 0 to t = 0.5) to the distance de traveled while it is moving left (from t = 0.5
to r = l ) . Now, dr = i(0.5) -s(0) = 0.25 -(-!) = 1.25. Similarly, dt = s(0.5) - s(l) = 0.25 -0 = 0.25.
Thus, the total distance is 1.5 miles.
17.16 A particle moves along the A:-axis according to the equation x = Wt - 2t2. What is the total distance covered by
the particle between t = 0 and t = 3?
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