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  4. 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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Q. 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

Motion in a Straight Line

Solution:

Let they meet at $C$.
image
For car $A: x = v_A t$
$\Rightarrow x = 10 \,t$
Acceleration of car $B : a_B = \tan \,45^{\circ} = 1 \,m/s^2$
for car $B : x + 10.5 = \frac{1}{2} a_B t^2$
$\Rightarrow 10t + 10.5 = \frac{1}{2} 1t^2$
$\Rightarrow t^2 - 20\,t - 21 = 0$
$\Rightarrow (t - 21) (t + 1) = 0$
$\Rightarrow t = 21\,s$