એક કારચાલક v વેગથી જઈ રહ્યો છે અને અચાનક ત

When driver applies brakes and the car covers distance $x$ before coming to rest, under the effect of retarding force $F$

then $\frac{1}{2}m{v^2} = Fx$ 

$⇒$ $x = \frac{{m{v^2}}}{{2F}}$

But when he takes turn then $\frac{{m{v^2}}}{r} = F$

$⇒$  $r = \frac{{m{v^2}}}{F}$

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.