Question

Initially car $A$ is $10.5m$ ahead of car $B$. Both starts 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. $t=2\sqrt{5}sec$
• B. none
• C. $t=21sec$
• D. $20sec$

Answer 1

The correct option is C $t=21sec$

Let they meet at C.
For car A:
$x=v_{A}t=10t...\left(i\right)$

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

For car B:
$x+10.5=\frac{1}{2}{a}_B{t}^2...\left(ii\right)$

From (𝑖) and (𝑖𝑖)
$\Rightarrow 10t+10.5=\frac{1}{2}1t^2$

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

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

$\Rightarrow t=21s$