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(C) Let the two cars be $A$ and $B$. The initial velocity of car $A$ is $u_A = 20 \,m \,s^{-1}$ and car $B$ is $u_B = -20 \,m \,s^{-1}$.The relative i

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$100$

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(C) Let the two cars be $A$ and $B$. The initial velocity of car $A$ is $u_A = 20 \,m \,s^{-1}$ and car $B$ is $u_B = -20 \,m \,s^{-1}$.The relative initial velocity of car $B$ with respect to car $A$ is $u_{BA} = u_B - u_A = -20 - 20 = -40 \,m \,s^{-1}$.The magnitude of relative initial velocity is $|u_{BA}| = 40 \,m \,s^{-1}$.Both cars retard at $a = 2 \,m \,s^{-2}$. Let $a_A = -2 \,m \,s^{-2}$ and $a_B = 2 \,m \,s^{-2}$.The relative acceleration of car $B$ with respect to car $A$ is $a_{BA} = a_B - a_A = 2 - (-2) = 4 \,m \,s^{-2}$.Using the equation of motion $v^2 = u^2 + 2as$ for relative motion, where final relative velocity $v_{BA} = 0$:$0^2 = (-40)^2 + 2(-4)S$$0 = 1600 - 8S$$8S = 1600 \implies S = 200 \,m$.This is the relative distance covered by the cars before coming to rest.The initial distance between the cars was $300 \,m$.Therefore, the remaining distance between them when they come to rest is $300 \,m - 200 \,m = 100 \,m$.

Two cars are travelling towards each other at a speed of $20 \,m \,s^{-1}$ each. When the cars are $300 \,m$ apart, both the drivers apply brakes and the cars retard at the rate of $2 \,m \,s^{-2}$. The distance between them when they come to rest is: (in $\,m$)

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