RECTILINEAR MOTION 0 135
I The velocity i> = D,x = 10-4t. Thus, v>0 when t<2.5, and u<0 when t>2.5. Hence, the
particle is moving right for t < 2.5 and it is moving left for t > 2.5. The distance d r that it covers while it is
moving right from t = 0 to < = 2.5 is *(2.5) - ;t(0) = 12.5 -0= 12.5. The distance d e that it covers while
it is moving left from t = 2.5 to t = 3 is *(2.5) - x(3) = 12.5 - 12 = 0.5. Hence, the total distance is
d, + 17.17
A rocket was shot straight up from the ground. What must its initial velocity have been if it returned to earth in
20 seconds?
I Its height s = s 0 + v 0 t — 16f
2
. In this case, s 0 = 0 and v 0 is unknown, so s = v 0 t — 16t
2 . We are told
that s = 0 when t = 20. Hence, 0 = v a (20) -16(20)
2
, u 0 = 320ft/s.
17.18
Two particles move along the x-axis. Their positions f(t) and g(t) are given by f(t) = 6t — t
2 and g(t) =
t
2 — 4t. (a) When do they have the same position? (ft) When do they have the same velocity? (c) When they
have the same position, are they moving in the same direction?
I (a) Set 6t-t
2 = t
2 -4t. Then t
2 -5t = 0, t(t-5) = 0, t = 0 or f = 5. (ft) The velocities are
f'(t) = 6-2t and g'(t) = 2t-4. Setting 6-2t = 2t-4, we have t = 2.5. (c) When they meet at
t = 0, f'(t)=f'(0) = 6 and g'(t) = g'(0)=-4. Since /'(O) and g'(0) have opposite signs, they are moving
in opposite directions when t = 0. When they meet at t = 5, f'(t)=f'(5)=-4 and g'(t) = g'(5) = 6.
Hence, when t = 5, they are moving in opposite directions.
17.19 A particle moves along the x-axis according to the equation x= j/
3 - \ cos 2t + 3.5. Find the distance traveled
between t = 0 and t = ir/2.
I The velocity v = t
2 + sin2t. For Q0, and, therefore, u>0. Hence, the particle
moves right between t = 0 and t = IT 12. So the distance traveled is x(irl2) - x(0) = (tr
3 /24 + 4) - 3 =
7T
3 /24+l.
17.20
A ball is thrown vertically into the air so that its height s after t seconds is given by
maximum height.
17.21
A particle moving along a straight line is accelerating at the constant rate of 3 m/s
2 . Find the initial velocity v a if
the displacement during the first two seconds is 10 m.
I The acceleration a = D,v=3. Hence, v = 3t+C. When f = 0, v = v 0 . So C = v 0 . Thus, v =
3t+v 0 , but v = D,s. So s = \t
2 + v 0 t + K. When t = 0, s = s 0 , the initial position, so K = s a .
Thus, s = \t
2 + v 0 t + s 0 . The displacement during the first two seconds is s(2) - s(Q) = (6 + 2v 0 + s 0 ) — s 0 =
6 + 2v 0 . Hence, 6 + 2u 0 = 10, i; 0 = 2m/s.
17.22
A particle moving on a line is at position s = t
3 — 6t
2 + 9t — 4 at time t. At which time(s) t, if any, does it
change direction?
I v = D,s = 3t
2 -I2t + 9 = 3(t
2 -4t + 3) = 3(t- l)(f-3). Since the velocity changes sign at (=1 and
t = 3, the particle changes direction at those times.
17.23
A ball is thrown vertically upward. Its height 5 (in feet) after t seconds is given by s = 40t — I6t
2 . Find
(a) when the ball hits the ground, (ft) the instantaneous velocity at t = 1, (c) the maximum height.
I v = D,s = 40 - 32t, D
2 s = -32. (a) To find out when the ball hits the ground, we set s = 40t - 16(
2 = 0.
Then t = 0 or f = 2.5. So the ball hits the ground .after 2.5 seconds, (ft) When t = l, u = 8ft/s. (c)
Set i>=40 — 32t = 0. Then f = 1.25. Since the second derivative is negative, this unique critical number
yields an absolute maximum. When f=1.25, s = 25ft.
17.24
A diver jumps off a springboard 10 feet above water with an initial upward velocity of 12ft/s. Find (a) her
maximum height, (ft) when she will hit the water, (c) her velocity when she hits the water.
Setting
we obtain
Since the second derivative is negative, we must have a relative maximum at t = 9, and, since that is the unique
critical number, it must be an absolute maximum. At t = 9, 5 = 9.
Find its
Dts=0
I The velocity i> = D,x = 10-4t. Thus, v>0 when t<2.5, and u<0 when t>2.5. Hence, the
particle is moving right for t < 2.5 and it is moving left for t > 2.5. The distance d r that it covers while it is
moving right from t = 0 to < = 2.5 is *(2.5) - ;t(0) = 12.5 -0= 12.5. The distance d e that it covers while
it is moving left from t = 2.5 to t = 3 is *(2.5) - x(3) = 12.5 - 12 = 0.5. Hence, the total distance is
d, + 17.17
A rocket was shot straight up from the ground. What must its initial velocity have been if it returned to earth in
20 seconds?
I Its height s = s 0 + v 0 t — 16f
2
. In this case, s 0 = 0 and v 0 is unknown, so s = v 0 t — 16t
2 . We are told
that s = 0 when t = 20. Hence, 0 = v a (20) -16(20)
2
, u 0 = 320ft/s.
17.18
Two particles move along the x-axis. Their positions f(t) and g(t) are given by f(t) = 6t — t
2 and g(t) =
t
2 — 4t. (a) When do they have the same position? (ft) When do they have the same velocity? (c) When they
have the same position, are they moving in the same direction?
I (a) Set 6t-t
2 = t
2 -4t. Then t
2 -5t = 0, t(t-5) = 0, t = 0 or f = 5. (ft) The velocities are
f'(t) = 6-2t and g'(t) = 2t-4. Setting 6-2t = 2t-4, we have t = 2.5. (c) When they meet at
t = 0, f'(t)=f'(0) = 6 and g'(t) = g'(0)=-4. Since /'(O) and g'(0) have opposite signs, they are moving
in opposite directions when t = 0. When they meet at t = 5, f'(t)=f'(5)=-4 and g'(t) = g'(5) = 6.
Hence, when t = 5, they are moving in opposite directions.
17.19 A particle moves along the x-axis according to the equation x= j/
3 - \ cos 2t + 3.5. Find the distance traveled
between t = 0 and t = ir/2.
I The velocity v = t
2 + sin2t. For Q
moves right between t = 0 and t = IT 12. So the distance traveled is x(irl2) - x(0) = (tr
3 /24 + 4) - 3 =
7T
3 /24+l.
17.20
A ball is thrown vertically into the air so that its height s after t seconds is given by
maximum height.
17.21
A particle moving along a straight line is accelerating at the constant rate of 3 m/s
2 . Find the initial velocity v a if
the displacement during the first two seconds is 10 m.
I The acceleration a = D,v=3. Hence, v = 3t+C. When f = 0, v = v 0 . So C = v 0 . Thus, v =
3t+v 0 , but v = D,s. So s = \t
2 + v 0 t + K. When t = 0, s = s 0 , the initial position, so K = s a .
Thus, s = \t
2 + v 0 t + s 0 . The displacement during the first two seconds is s(2) - s(Q) = (6 + 2v 0 + s 0 ) — s 0 =
6 + 2v 0 . Hence, 6 + 2u 0 = 10, i; 0 = 2m/s.
17.22
A particle moving on a line is at position s = t
3 — 6t
2 + 9t — 4 at time t. At which time(s) t, if any, does it
change direction?
I v = D,s = 3t
2 -I2t + 9 = 3(t
2 -4t + 3) = 3(t- l)(f-3). Since the velocity changes sign at (=1 and
t = 3, the particle changes direction at those times.
17.23
A ball is thrown vertically upward. Its height 5 (in feet) after t seconds is given by s = 40t — I6t
2 . Find
(a) when the ball hits the ground, (ft) the instantaneous velocity at t = 1, (c) the maximum height.
I v = D,s = 40 - 32t, D
2 s = -32. (a) To find out when the ball hits the ground, we set s = 40t - 16(
2 = 0.
Then t = 0 or f = 2.5. So the ball hits the ground .after 2.5 seconds, (ft) When t = l, u = 8ft/s. (c)
Set i>=40 — 32t = 0. Then f = 1.25. Since the second derivative is negative, this unique critical number
yields an absolute maximum. When f=1.25, s = 25ft.
17.24
A diver jumps off a springboard 10 feet above water with an initial upward velocity of 12ft/s. Find (a) her
maximum height, (ft) when she will hit the water, (c) her velocity when she hits the water.
Setting
we obtain
Since the second derivative is negative, we must have a relative maximum at t = 9, and, since that is the unique
critical number, it must be an absolute maximum. At t = 9, 5 = 9.
Find its
Dts=0