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 - 16f
2
, we set 420 = 480 - 16f - 16*
2
, obtaining 4t
2 + 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 = 12t
3 - l&t
2 + 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= 36t
2 -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) + 2f-l] = 6(4f-3), and D
2 v = 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 d r traveled while
it is moving right (from t = 0 to t = 0.5) to the distance d e traveled while it is moving left (from t = 0.5
to r=l). Now, d r = i(0.5) -s(0) = 0.25 -(-!) = 1.25. Similarly, d t = 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 - 2t
2 . What is the total distance covered by
the particle between t = 0 and t = 3?
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 - 16f
2
, we set 420 = 480 - 16f - 16*
2
, obtaining 4t
2 + 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 = 12t
3 - l&t
2 + 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= 36t
2 -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) + 2f-l] = 6(4f-3), and D
2 v = 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 d r traveled while
it is moving right (from t = 0 to t = 0.5) to the distance d e traveled while it is moving left (from t = 0.5
to r=l). Now, d r = i(0.5) -s(0) = 0.25 -(-!) = 1.25. Similarly, d t = 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 - 2t
2 . What is the total distance covered by
the particle between t = 0 and t = 3?