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

`6.67ft/s^2`

Step-by-step 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=
`(v-u)/(t){t}`

Using the formula

Average acceleration,a=
`(38-18)/(3)ft/s^2`

Average acceleration,a=
`(20)/(3)ft/s^2`

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

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