1.1. HOW DO WE MEASURE VELOCITY?
5
we observe that the average velocity of the object on the interval [a, a + h] is
AV [a,a+h] =
s(a + h) − s(a)
h
,
with the denominator being simply h because (a + h) − a = h. Initially, it is fine to think
of h being a small positive real number; but it is important to note that we allow h to
be a small negative number, too, as this enables us to investigate the average velocity of
the moving object on intervals prior to t = a, as well as following t = a. When h < 0,
AV [a,a+h] measures the average velocity on the interval [a + h, a].
To attempt to find the instantaneous velocity at t = a, we investigate what happens as
the value of h approaches zero. We consider this further in the following example.
Example 1.1. For a falling ball whose position function is given by s(t) = 16 − 16t 2 (where
s is measured in feet and t in seconds), find an expression for the average velocity of the
ball on a time interval of the form [0.5, 0.5 + h] where −0.5 < h < 0.5 and h 0. Use this
expression to compute the average velocity on [0.5, 0.75] and [0.4, 0.5], as well as to make
a conjecture about the instantaneous velocity at t = 0.5.
Solution. We make the assumptions that −0.5 < h < 0.5 and h 0 because h cannot be
zero (otherwise there is no interval on which to compute average velocity) and because the
function only makes sense on the time interval 0 ≤ t ≤ 1, as this is the duration of time
during which the ball is falling. Observe that we want to compute and simplify
AV [0.5,0.5+h] =
s(0.5 + h) − s(0.5)
(0.5 + h) − 0.5
.
The most unusual part of this computation is finding s(0.5 + h). To do so, we follow the
rule that defines the function s. In particular, since s(t) = 16 − 16t 2 , we see that
s(0.5 + h) = 16 − 16(0.5 + h)
2
= 16 − 16(0.25 + h + h
2 )
= 16 − 4 − 16h − 16h
2
= 12 − 16h − 16h
2 .
5
we observe that the average velocity of the object on the interval [a, a + h] is
AV [a,a+h] =
s(a + h) − s(a)
h
,
with the denominator being simply h because (a + h) − a = h. Initially, it is fine to think
of h being a small positive real number; but it is important to note that we allow h to
be a small negative number, too, as this enables us to investigate the average velocity of
the moving object on intervals prior to t = a, as well as following t = a. When h < 0,
AV [a,a+h] measures the average velocity on the interval [a + h, a].
To attempt to find the instantaneous velocity at t = a, we investigate what happens as
the value of h approaches zero. We consider this further in the following example.
Example 1.1. For a falling ball whose position function is given by s(t) = 16 − 16t 2 (where
s is measured in feet and t in seconds), find an expression for the average velocity of the
ball on a time interval of the form [0.5, 0.5 + h] where −0.5 < h < 0.5 and h 0. Use this
expression to compute the average velocity on [0.5, 0.75] and [0.4, 0.5], as well as to make
a conjecture about the instantaneous velocity at t = 0.5.
Solution. We make the assumptions that −0.5 < h < 0.5 and h 0 because h cannot be
zero (otherwise there is no interval on which to compute average velocity) and because the
function only makes sense on the time interval 0 ≤ t ≤ 1, as this is the duration of time
during which the ball is falling. Observe that we want to compute and simplify
AV [0.5,0.5+h] =
s(0.5 + h) − s(0.5)
(0.5 + h) − 0.5
.
The most unusual part of this computation is finding s(0.5 + h). To do so, we follow the
rule that defines the function s. In particular, since s(t) = 16 − 16t 2 , we see that
s(0.5 + h) = 16 − 16(0.5 + h)
2
= 16 − 16(0.25 + h + h
2 )
= 16 − 4 − 16h − 16h
2
= 12 − 16h − 16h
2 .