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 from work- energy theorem \[\frac{1}{2}m{{v}^{2}}=Fx\] \[\Rightarrow \] \[x=\frac{m{{v}^{2}}}{F}\] But when he makes turn, then \[\frac{m{{v}^{2}}}{r}=F\] \[\Rightarrow \] \[r=\frac{m{{v}^{2}}}{F}\] It is clear that \[x=\frac{r}{2}\]i.e, by the same retarding force the car can be stopped in a less distance if the driver apply brakes. This retarding force is actually a friction force.
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