An automobile of mass 2 500 kg moving at 50.0 m/s is braked suddenly with a constant braking force of 8 500 N. How far does the car travel before stopping

367.64 meters

Step-by-step explanation:

Using the kinetic equation where:

Vf ^ 2 - Vi ^ 2 = 2ax

We know that:

Vf = 0 m / s

Vi = 50 m / s

The acceleration can be calculated by means of F = M.a

F = -8500 (because it is stopping, braking force) and m = 2500 kg

We have left that a = F / M = -8500/2500 = -3.4 m / s ^ 2

Now reorganizing we have that:

x = -Vi ^ 2 / 2a

replacing the values is:

x = - (50 ^ 2) / (2 * -3.4) = 367.64 meters and this would be the distance the car travels when it stops.

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