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

Answer:

344.68 m

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

$= \frac{-F}{m}\\\\\= \frac{8868}{2500}$

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

$S = \frac{v^ 2- u^ 2}{2a}\\\\\ = \frac{0- 49.4^ 2}{2\times -3.54}$

= 344.68 m