Question

The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the distance traveled at time t seconds and t equals the time in seconds. What does the average rate of change of d(t) from t = 2 to t = 5 represent?

Answer 1

Average rate of change of a function d(t) between t1 and t2 can be calculated by

average rate of change

= (d(t2)-d(t1)) /(t2-t1)

Geometrically, it represents the secant between points (t1, d(t1)), and (t2, d(t2)).

Physically, it represents the total displacement between t1 and t2, divided by the elapsed time, t2-t1.

For the given problem,

t1=2

t2=5

t2-t1=5-2=3 seconds

d(t1)=d(2)=16(2^2) = 64 ft

d(t2)=d(5)=16(5^2) = 400 ft

d(t2)-d(t1) = 400-64 = 336 ft

Average rate of change (of displacement)

= (d(t2)-d(t1)) / (t2-t1)

= (336/3)

= 112 ft / s.

Answer 2

Assuming that by "16t2" you mean $16t^2$:

First of all, let's compute the position at the two required times:

$d(2)=16\cdot 2^2 = 16*4 = 64$

$d(5)=16\cdot 5^2 = 16*25 = 400$

So, the ball traveled a total distance of $\Delta d = 400-64 = 336$, during the time $\Delta t = 5-2 = 3$.

The average rate of change is the ratio between these two:

$\frac{\Delta d}{\Delta t} = \frac{336}{3} = 112$