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**

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