Laws Of Motion Question 148
Consider a car moving on a straight road with a speed of 100 m/s. The distance at which the car can be stopped is $ [{{\mu } _{k}}=0.5] $ m
Options:
A) 400m
B) 100m
C) 1000m
D) 800m (Note: It should be minimum distance in which the car can be stopped.)
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Answer:
Correct Answer: C
Solution:
[c] Retardation due to friction $ =\mu g $ we know $ v^{2}=u^{2}+2as $
$ \therefore 0={{(100)}^{2}}-2(\mu g)s $ Or $ 2\mu gs=100\times 100 $ Or $ s=\frac{100\times 100}{2\times 0.5\times 10}=100m $
BETA
