Question

During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a 3 second time period. What was his average acceleration over that 3 second period?

Answer 1

Answer:

$6.67ft/s^2$

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a=$\frac{v-u}{t}{t}$

Using the formula

Average acceleration,a=$\frac{38-18}{3}ft/s^2$

Average acceleration,a=$\frac{20}{3}ft/s^2$

Average acceleration,a=$6.67ft/s^2$

Hence, the average acceleration=$6.67ft/s^2$