question_answer

If the coefficient of static friction between the types 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)  40.8m             

D)  80.16m

Correct Answer: C

Solution :

 Free body diagram of automobile is shown By Newtons third law \[0.5=f=\mu R=\mu mg\] Where, m is the mass of automobile \[F=ma\] \[\mu mg=ma\] \[a=\mu g=0.5g\] Let automobile stops at a distance\[x,\]than from equation of motion \[{{v}^{2}}={{u}^{2}}-2ax\] \[v=0\,u=72\,km/h=72\times \frac{5}{18}=20\,m/s\] \[g=9.8\,m/{{s}^{2}}\] \[{{(0)}^{2}}={{(20)}^{2}}-2\times 0.5\times 9.8x\] \[x=\frac{20\times 20}{2\times 0.5\times 9.78}=40.8\,m\]