A) Brake sharply
B) Turn sharply
C) (a) and (b) both
D) None of the above
Correct Answer: A
Solution :
When driver applies brakes and the car covers distance x before coming to rest, under the effect of retarding force F then \[\frac{1}{2}m{{v}^{2}}=Fx\]Þ \[x=\frac{m{{v}^{2}}}{2F}\] But when he takes turn then \[\frac{m{{v}^{2}}}{r}=F\]Þ \[r=\frac{m{{v}^{2}}}{F}\] 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.
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