Page 47 - How To Solve Word Problems In Calculus

Solution
2
Since v(t) = 48t − 3t , a(t) = v (t) = 48 − 6t.

When t = 3, a = 30 in/sec 2 Since the units of velocity in
this example are inches per
2 second, the units of accel-
When t = 5, a = 18 in/sec
eration would be inches per
When t = 10, a =−12 in/sec 2 second per second. This is
usually written in/sec/sec or
When t = 15, a =−42 in/sec 2 in/sec . Other common units
2
2
of acceleration are ft/sec ,
2
mi/h , and meters/sec .
2
EXAMPLE 8
The height (in feet) at any time t (in seconds) of a projectile
thrown vertically is

2
h(t) =−16t + 256t

(a) What is the projectile’s average velocity for the first
5 seconds of travel?
(b) How fast is the projectile traveling 6 seconds after it is
thrown? How high is it?
(c) How fast is the projectile traveling 10 seconds after it is
thrown? How high is it?
(d) When is the maximum height reached by the projectile?
What is its maximum height?
(e) When does the projectile return to the ground and with
what velocity?
(f ) Describe the motion of the projectile.

Solution
2
(a) h(t) =−16t + 256t. When t = 0, h = 0 and when t = 5,
2
h =−16(5) + 256(5) = 880 ft.
h 880 − 0
v av = = = 176 ft/sec
t 5 − 0

(b) Since we are looking for the instantaneous velocity we
dh dh
compute the derivative . v = =−32t + 256. When
dt dt

34