A sports car moving at constant speed travels 105 m in 4.7 s. If it then brakes and comes to a stop in 4.0 s,?

Question

A sports car moving at constant speed travels 105 m in 4.7 s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration? Express the answer in terms of g's, where g = 9.80 m/s2. (Take the positive direction to be the direction of travel

Answer 1

The initial speed was (105 m) / (4.7 s) = 22.34 m/s. The acceleration is (-22.34 m/s) / (4 s) = -5.59 m/s^2. Dividing this by 9.8 m/s^2 yields -0.57 g.

Answer 2

The speed of the car is v = 105/4.7 m/s.

The acceleration is given by a = (0 - v)/t

or a = -105/4.7*4 m/s2

for g = 9.8 m/s2, we have a = -0.57 g

Answer 3

Given
(u) initial velocity = S/t= 105/4.7= 22.34 m/s
(v) final velocity = 0 m/s
(t) Time = 4 s [ do not consider the previous time, only time taken to come to a stop on application of brakes ]

v=u+at
a= (v-u)/t=(0 - 22.34)/4 = -22.34/4 = - 5.585 m/s2 [ metre per second squared]

Therefore,
acceleration = -5.585 or retardation= 5.585

Also a= -5.585 = - 0.596 (9.8) = - 0.596 g = -0.6g m/s2 (approx.)

Answer 4

- ((105 m/4.7 s)/4.0 s)(1 g-sec^2/9.80 m) = - 0.5699 g