A sports car moving at constant velocity travels 120 m in 5.0 s. If it then brakes and comes to a stop in 4.0 s, what is the magnitude of its acceleration (assumed constant) in $\textrm{m/s}^2$, and in g's $(\textrm{g} = 9.80 \textrm{ m/s}^2)$?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The acceleration of this sports car is gonna be the change in velocity divided by time. So the final velocity is zero and the initial velocity is 120 meters divided by 5 seconds that's 24 meters per second. So we take the difference there, zero minus 24 meters per second divided by the time it spends braking— 4 seconds— and we get negative 6.0 meters per second squared and the magnitude of that acceleration is 6.0 meters per second squared. And then expressing that in units of g's there's 1 g for every 9.8 meters per second squared and so these meters per second squared's, you can think of it as a single package unit cancel and we are left with 0.61 g's is the acceleration.