Question

A roller coaster is moving at 25 m/s at the bottom of a hill. Three seconds later it reaches the top of the hill moving at 10 m/s. What was the acceleration of the coaster?

Answer 1

`-5m//s^2`

Explanation:

`a_(avg)=(Deltav)/(Deltat)=(v_f-v_i)/(t_f-t_i)`

Where:-

`v_f` is the final velocity
`v_i` is the initial velocity
`t_f` is the final time
`t_i` is the initial time

The final velocity of the roller coaster is `10m//s` (when it reached the top"`t=3s`"), and the initial velocity is `25m//s` (when it started moving at the bottom "`t=0s`")

So
`v_f=10m//s`
`v_i=25m//s`
`t_f=3s`
`t_i=0s`

`a_(avg)=(10-25)/(3-0)=-15/3=-5m//s^2`