The driver of a car travelling at velocity v | vedclass

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Question

(a) When driver applies brakes and the car covers distance $x$ before coming to rest, under the effect of retarding force $F$

then $\frac{1}{2}m{v^

Vedclass · Accepted Answer

Brake sharply

Vedclass · Answer

(a) When driver applies brakes and the car covers distance $x$ before coming to rest, under the effect of retarding force $F$

then $\frac{1}{2}m{v^2} = Fx$ 

$&rArr;$ $x = \frac{{m{v^2}}}{{2F}}$

But when he takes turn then $\frac{{m{v^2}}}{r} = F$

$&rArr;$  $r = \frac{{m{v^2}}}{F}$

It is clear that $x = r/2$

i.e. by the same retarding force the car can be stopped in a less distance if the driver apply breaks. This retarding force is actually a friction force.

The driver of a car travelling at velocity $v$ suddenly see a broad wall in front of him at a distance $d$. He should

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