A sports car moving at constant speed travels 164 m in 13.77 s. If it then brakes and comes to a stop in 3.6 s. What is its acceleration in terms of "g's", where 1.00 g = 9.80 m/s2?

a = - 0.3376 g's

Explanation:

The sports car has a constant speed when travelling. Covered 164 m in 13.77 s. Thus, speed = 164/13.77 m/s

It brakes and now comes to a stop in 3.6 s.

Thus final velocity = 0 m/s

Formula for acceleration is;

a = (v - u)/t

a = (0 - (164/13.77))/3.6

a = -3.308 m/s²

In terms of g's", where 1.00 g = 9.80 m/s², we have;

a = -3.308/9.8 g's

a = - 0.3376 g's