Initially car A is 10.5 m ahead of car B. Both start moving at time t = 0 in the same direction along a straight line. The velocity time graph of two cars is shown in figure. The time when the car B will catch the car A, will be

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t=21sec

t= $2 \sqrt{5}$ sec

t=20sec

None of these

answer is A.

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Detailed Solution

$x_{A} = x_{B} 10.5 + 10 t = \frac{1}{2} a t^{2} \Rightarrow a = \tan 45^{0} = 1 m / s^{2} t^{2} - 20 t - 21 = 0 \Rightarrow t = \frac{20 \pm \sqrt{400 + 84}}{2} \Rightarrow t = 21 s e c .$

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