A) 400m
B) 100m
C) 1000m
D) 800m (Note: It should be minimum distance in which the car can be stopped.)
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 \[s=\frac{100\times 100}{2\times 0.5\times 10}=100m\]
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