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(B) Given: Initial velocity $u = 100\, m/s$,Final velocity $v = 0\, m/s$,Coefficient of kinetic friction $\mu_k = 0.5$,Acceleration due to gravity $g

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$1000$

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(B) Given: Initial velocity $u = 100\, m/s$,Final velocity $v = 0\, m/s$,Coefficient of kinetic friction $\mu_k = 0.5$,Acceleration due to gravity $g = 10\, m/s^2$.The retardation force acting on the car is $F = \mu_k N = \mu_k mg$.Using Newton's second law,the retardation $a = \frac{F}{m} = \mu_k g = 0.5 	imes 10 = 5\, m/s^2$.Using the kinematic equation $v^2 - u^2 = 2as$:$0^2 - (100)^2 = 2(-5)s$$-10000 = -10s$$s = \frac{10000}{10} = 1000\, m$.Therefore,the distance at which the car can be stopped is $1000\, m$.

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 ........ $m$. (Take $g = 10\, m/s^2$ and $\mu_k = 0.5$)

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