Page 44 - How To Solve Word Problems In Calculus

This quantity is the derivative of s with respect to t:

ds
v =
dt
EXAMPLE 4
A bicycle travels along a straight road. At 1:00 it is 1 mile from
the end of the road and at 4:00 it is 16 miles from the end of
the road. Compute (a) its average velocity from 1:00 to 4:00
and (b) its instantaneous velocity at 3:00.

Solution
(a) Let s represent the bicycle’s position relative to the end of
the road.

s 16 − 1 15
v av = t = 4 − 1 = 3 = 5 mi/h

(b) We cannot solve this part of the problem since we do not
know the bicycle’s location at every point in time. There
is not enough information given to compute its instanta-
neous velocity.

In order to compute instantaneous velocity we need the bi-
cycle’s position as a function of time.

EXAMPLE 5
2
A bicycle travels along a straight road. At t o’clock it is t miles
from the end of the road. Compute (a) its average velocity
from 1:00 to 4:00 and (b) its instantaneous velocity at 3:00.

Solution
(a) Observe that as far as the average velocity is concerned, the
2
problem is identical to Example 4. If s = t , when t = 1,
s = 1 and when t = 4, s = 16.

s 16 − 1 15
v av = = = = 5 mi/h
t 4 − 1 3

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