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?

1 Answer
Nov 22, 2016

-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