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Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54

Section 2.1 • Rates of Change and the Derivative 171

DEFINITION Average Velocity
The signed distance s from the origin at time t of an object moving on a line is
given by the position function s = f (t). If at time t 0 the object is at s 0 = f (t 0 ) and at
time t 1 the object is at s 1 = f (t 1 ), then the change in time is 1t = t 1 − t 0 and the
0 t 0
t  t t 0 change in position is 1s = s 1 − s 0 = f (t 1 ) − f (t 0 ). The average rate of change of
1
0
1
t 1 O s  f (t ) f (t ) position with respect to time over the interval [t 0 , t 1 ] is
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f (t ) f (t ) s
0
1
1s f (t 1 ) − f (t 0 )
= t 1 6= t 0
1t t 1 − t 0
1s
Figure 5 The average velocity is .
1t and is called the average velocity of the object over the interval [t 0 , t 1 ]. See Figure 5.
EXAMPLE 4 Finding Average Velocity
The Mike O’Callaghan–Pat Tillman Memorial Bridge spanning the Colorado River
opened on October 16, 2010. Having a length of 1900 ft, it is the longest arch bridge in
the Western Hemisphere, and its roadway is 890 ft above the Colorado River.
2
If a rock falls from the roadway, the function s = f (t) = 16t gives the distance s, in
feet, that the rock falls after t seconds for 0 ≤ t ≤ 7.458. Here 7.458 s is the approximate
Scott Prokop/Shutterstock time it takes the rock to fall 890 ft into the river. The average velocity of the rock during
its fall is
890 − 0
f (7.458) − f (0)
1s
≈ 119.335 ft/s
=
=
1t
7.458 − 0
R
NOW WORK AP Practice Problem 7. 7.458
NOTE Here the motion occurs along a The average velocity of the rock in Example 4 approximates the average velocity over
vertical line with the positive direction the interval [0, 7.458]. But the average velocity does not tell us about the velocity at any
downward.
particular instant of time. That is, it gives no information about the rock’s instantaneous
velocity.
We can investigate the instantaneous velocity of the rock, say, at t = 3 s, by
computing average velocities for short intervals of time beginning at t = 3. First we
compute the average velocity for the interval beginning at t = 3 and ending at t = 3.5.
The corresponding distances the rock has fallen are
2
2
f (3) = 16 · 3 = 144 ft and f (3.5) = 16 · 3.5 = 196 ft
Then 1t = 3.5 − 3.0 = 0.5, and during this 0.5-s interval,

1s f (3.5) − f (3) 196 − 144
Average velocity = = = = 104 ft/s
1t 3.5 − 3 0.5
Table 1 shows average velocities of the rock for progressively smaller intervals of
time.

TABLE 1

2
∆s f(t) − f(t 0 ) 16t − 144
Time Interval Start t 0 = 3 End t ∆t = =
∆t t − t 0 t − 3
2
1s f (3.1) − f (3) 16 · 3.1 − 144
[3, 3.1] 3 3.1 0.1 = = = 97.6
1t 3.1 − 3 0.1
2
1s f (3.01) − f (3) 16 · 3.01 − 144
[3, 3.01] 3 3.01 0.01 = = = 96.16
1t 3.01 − 3 0.01
2
1s f (3.0001) − f (3) 16 · 3.0001 − 144
[3, 3.0001] 3 3.0001 0.0001 = = = 96.0016
1t 3.0001 − 3 0.0001
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