Question

Interpreting Date: Bob straps on his

in-lineskatesand pushes down a hill. His velocity changes from 0
m/s atthe start to 4.5 m/s exactly 15 s later: what is Bob's
averageacceleration?

Answer 1

`0.3 m/s^2` is Bob's average acceleration.

Step-by-step explanation:

Initial velocity of the bob's in-lineskates= u = 0 m/s

Acceleration of the Bob= a = ?

Final velocity of the bob's in-lineskates= v = 4.5 m/s

Duration of time in which 0 m/s velocity is changed to 4.5 m/s = t = 15 s

Using first equation of motion: v = u + at

`4.5 m/s=0m/s+a* (15 s)`

`5.0 m/s-0m/s =a* 15 s`

`4.5 m/s=a* 15 s`

`a=(4.5 m/s)/(15 s)=0.3 m/s^2`

`0.3 m/s^2` is Bob's average acceleration.