Question

The table below shows the distance d(t) in feet that an object travels in t seconds: t (second) d(t) (feet) 2 64 4 256 6 576 8 1024 What is the average rate of change of d(t) between 2 seconds and 6 seconds, and what does it represent?

Answer 1

The distance covered at time t is listed below:
Distance at 2 sec = d(2) = 64 feet
Distance at 4 sec = d(4) = 256 feet
Distance at 6 sec = d(6) = 576 feet
Distance at 8 sec = d(8) = 1024 feet

We are to find the average rate of change between 2 seconds and 6 seconds. The average rate of change will be:

$\frac{d(6)-d(2)}{6-2} \\ \\ = \frac{576-64}{4} \\ \\=128$

Therefore, the average rate of distance between 2 seconds and 6 seconds is 128 feet per second. This represents the speed of the object. So the speed of the object between 2 and 6 seconds was 128 feet per second. 

Answer 2

Average rate of change of d(t) between 2 seconds and 6 seconds:
Rave=[d(6)-d(2)] / (6-2)
Rave=[d(6)-d(2)] / (4)

t=6→d(6)=576
t=2→d(2)=64
Replacing in the equation above:
Rave=(576-64) / (4)
Rave=(512) / (4)
Rave=128 feet/second

Answer: The average rate of change of d(t) between 2 seconds and 6 seconds is 128 feet/second and represents the average speed between 2 and 6 seconds