In a car race, car A takes time t less than car B and passes the finishing point with a velocity of 12 m/s more than the velocity with which car B passes the finishing point. Assume that the cars A and B start from rest and travel with constant accelerations of $9 m / s^{2}$ and $4 m / s^{2}$ , respectively. If $V_{A}$ and $V_{B}$ be the final velocities of cars A and B, respectively, then

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t = 2 sec

t = 3 sec

$V_{A} = 24 m / s, V_{B} = 12 m / s$

$V_{A} = 36 m / s, V_{B} = 24 m / s$

answer is A, D.

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Detailed Solution

$9 (T - t) - 4 T = 12$

$\frac{1}{2} 9 {(T - t)}^{2} = \frac{1}{2} 4 T^{2}$

$3 T - 3 t = 2 T$

$T = 3 t$

$6 T - 4 T = 12$

$T = 6$

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