Question

An automobile of mass 2500 kg moving at 49.4 m/s is braked suddenly with a constant braking force of 8,868 N. How far does the car travel before stopping

Answer 1

344.68 m

Step-by-step explanation:

The computation of the far does the car travel before stopping is

Data provided in the question

Force = F = 8,868 N

mass = m = 2,500 kg

So,

accleration = a is

`= (-F)/(m)\\\\\= (8868)/(2500)`

a = -3.54 m/s^2

The initial speed = u = 49.4 m / s

final speed = v = 0

Based on the above information

Now applying the following formula

v^ 2- u^ 2= 2aS

Therefore

`S = (v^ 2- u^ 2)/(2a)\\\\\ = (0- 49.4^ 2)/(2* -3.54)`

= 344.68 m