Question

An automobile is moving on a horizontal road with a speed v. If the coefficient of friction between the tyres and road is u, show that the shortest distance in which the automobile can be stopped is v^2 / 2 ug.​

Answer 1

Answer:

Consider the problem

Let

Initial speed =v

and

Final speed =u

according to question

Forceacting=frictionalforce

therefore,

ma=μ

s

mg

then

a=μ

s

g

consider traveled is d.

therefore,

u

2

−v

2

=2ad

So,

0

2

−v

2

=2ad (final speed is 0)

−v

2

=2μ

s

gd

d=−

2μg

v

2

Where negative sign indicates that force acts in the opposite direction of the force.

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