Question

A sports car moving at a constant velocity travels 135m in 5.0s. If it then braks and comes to a stop in 4.3 sec, what is the magnitude of its acceleration (assume constant) in m/s^2, and in g's (g= 9.8m/s^2)? How far does the car move while stopping?

I got the magnitude of its acceleration 0.64m/s^2 but I do not know how to figure out how far the car moves whild stopping.

If you check my answr and answer the distance questions I will give five stars!

Answer 1

Vf - Vo = aT

where

Vf = final velocity = 0 (when the car stops)
Vo = initial velocity = 135/5.0 = 27 m/sec
a = acceleration
T = time interval = 4.3 sec (given)

Substituting values,

0 - 27 = a(4.3)

and solving for "a"

a = - 27/4.3

a = -6.279 m/sec^2

NOTE the negative sign attached to the acceleration. It simply means that the car was (obviously) slowing down when the brakes were applied.

<< and in g's (g = 9.80 m/s2)? >>

a = 6.279/9.8 = 0.6407(g)

now s = ut + 1/2 a t^2

s = 27 x 4.3 + 1/2 x ( -6.279) x 4.3^2

s = 58.05 m