Near the surface of the moon, the distance that an object falls is a function of time. It is given by d(t) = 2.6667t^2, where t is in seconds and d(t) is in feet

Question

If an object is dropped from a certain height, find the average velocity of the object from t = 1 sec. to t = 2

Answer 1

Hi Melissa,

Average velocity is change in position over change in time.

To calculate this, we can see where the object is at t = 1 and where the object is at t = 2.

At t = 1:

d(t = 1) = 2.667 * (1)² = 2.667 ft

At t = 2:

d(t = 2) = 2.667 * (2)² = 10.667 ft

Our change in position is 10.667 - 2.667 = 8 feet

Our change in time is 2 sec - 1 sec = 1 second

Our average velocity between t = 1 and t = 2 is then 8 feet / sec.