Question

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 m/s^2 and in g's g = 9.80 m/s^2

Answer 1

The car is traveling at consta$v= \frac{space}{time}= 120/5 =24 m/s$nt velocity. 
It then stops in 4 seconds moving with uniform negative acceleration. Thus
$0=v+a*t$
where a is the acceleration. Therefore the acceleration is
$a=- \frac{v}{t}=-24/4=-6 (m/s^2)$
Given in g ($g=9.80 (m/s^2)$)  the acceleration is
$a=-6/9.81 =-0.612*g$