Consider a car moving along a straight horizontal road with a speed of 72kmph. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (Takeg=10ms2)
A
30m
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B
40m
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C
72m
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D
20m
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Solution
The correct option is B40m Given u=72kmph=20m/s;
For stopping the car
Final velocity v=0;
As only friction is acting a=fm=−μmgm=−μg=−0.5×10=−5ms2
Using equation of kinematics, v2=u2+2as ∴s=(v2−u2)2a=(0)−(20)22×(−5)=40m