Consider a car moving on a straight road with a speed of 100ms−1. The distance at which car can be stopped is [μk=0.5]
A
800m
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B
1000m
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C
100m
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D
400m
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Solution
The correct option is B1000m From third equation of Motion, we have v2=u2+2as
Given, v=0 [car is stopped]
As friction provide the retardation a=−μg,u=100ms−1 ∴(100)2=2μgs ⇒s=100×1002×0.5×10 =100×1005×2=1000m