Question

Initially car $A$ is $10.5m$ 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

• A. t = 21 s
• B. $t=2\sqrt{5}s$
• C. 20 s
• D. none

Answer 1

Answer: (A)

Let they meet at $C$.



For car $A:x=v_{A}t$

$\Rightarrow x=10t$

Acceleration of car $B:a_B=\tan 45^{\circ }=1m/s^2$

for car $B:x+10.5=\frac{1}{2}{a}_B{t}^2$

$\Rightarrow 10t+10.5=\frac{1}{2}1t^2$

$\Rightarrow t^2-20t-21=0$

$\Rightarrow (t-21)(t+1)=0$

$\Rightarrow t=21s$