Question

A car going at a speed of 7 m/s can be stopped by applying brakes in a shortest distance of 10 m. Show that the total frictional force opposing the motion when brakes are applied is 1/4th of the weight of the car(g =9.8 m/s²

Answer 1

Answer:

For the motor car;

u = 7 m/s; v = 0; s = 10 m.

Using, v2 = u2 + 2as, we get

       a=v2-u2/2s=o-49/20=-2.45m/sec2

If we consider, g = 9.8 m/s2, then we get

a=-g/4

Thus, the resistance to the motion, that is resistive force is

-F=-ma=mg/4

This is one fourth the weight of the car.

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Answer 2

Answer:

this is the answer

Explanation:

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