question_answer

If the coefficient of static friction between the tyres and road is 0.5, what is the shortest distance in which an automobile can be stopped when travelling at 72 km/h?

A)  50 m          

B)  60 m

C)  40.8 m        

D)  80.16 m

Correct Answer: C

Solution :

Key Idea: Frictional force acting between road and lyres retards the motion of automobile. There is a static friction between tyres and road, so frictional force cause the retardation in velocity of a automobile. Free body diagram of automobile is shown. From Newtons third law                 \[F={{f}_{e}}=\mu R=\mu g\] where m is the mass of automobile. Also,      \[F=ma\]                 \[ma=\mu mg\] \[\Rightarrow \] a = retardation                 \[=\mu g\]                 = 0.5 g Let automobile stops at a distance \[x\], then from equation of motion                 \[{{v}^{2}}={{u}^{2}}-2ax\] Given,   \[v=0,u=72\,km/h=72\times \frac{5}{18}m/s\]                 \[=20\,m/s\]                 \[g=9.8\,\,m/{{s}^{2}}\] \[\therefore \] \[{{0}^{2}}{{(20)}^{2}}-2\times 0.5\times 9.8\,\,x\] \[\Rightarrow \] \[x=\frac{20\times 20}{2\times 0.5\times 9.8}=40.8\,m\]