Page 46 - How To Solve Word Problems In Calculus

(b) When t = 3, its position x = 199. The particle’s instanta-
neous velocity is determined using the derivative.
dx 2
v = = 48t − 3t
dt
When t = 3, v = 117 in/sec. Since this is a positive number,
it represents the speed as well.
2
3
(c) When t = 20, its position x = 24(20) − 20 + 10 = 1610.
Its velocity,
dx 2
v = = 48t − 3t
dt

When t = 20, v =−240 in/sec. The negative velocity indi-
cates that the particle is moving in the negative direction.
The speed of the particle is 240 in/sec.
(d) v = 48t − 3t 2

0 = 48t − 3t 2

0 = 3t(16 − t)
t = 0 t = 16

The velocity of the particle is 0 when t = 0 and when
t = 16. When t = 0, x = 10 and when t = 16, x = 2058.
(e) The particle begins at rest at x = 10. For the first 16
seconds, the particle has a positive velocity and moves
in the positive direction. When t = 16, the particle
stops momentarily (v = 0) at x = 2058. It then moves
in the negative direction and returns to x = 1610 when
t = 20.
The acceleration of a moving object is the rate of change of

its velocity with respect to time. Thus a(t) = v (t). If s(t) repre-
sents the position of the particle at time t, then a(t) = s (t).

EXAMPLE 7
Compute the acceleration of the particle in Example 6 at times
t = 3, 5, 10, and 15.

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