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Question

Consider a car moving along a straight horizontal road with a speed of 72 kmph. 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 (Take g=10ms2)

A
30 m
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B
40 m
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C
72 m
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D
20 m
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Solution

The correct option is B 40 m
Given
u=72 kmph=20 m/s;
For stopping the car
Final velocity v=0;

As only friction is acting
a=fm=μmgm=μ g=0.5×10=5 ms2

Using equation of kinematics,
v2=u2+2as
s=(v2u2)2a=(0)(20)22×(5)=40 m

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