Question

Two cars $A$ and $B$ are travelling in the same direction with velocities $v_1$ and $v_2(v_1>v_2)$ respectively. When the car $B$ is at a distance $d$ ahead of the car $A$, the driver of the car $A$ applied the brake producing a uniform retardation $a$. There will be no collision if

• A. $d\le \frac{{(v_1-v_2)}^2}{2a}$
• B. $d\le \frac{v_1^2-v_2^2}{2a}$
• C. $d\ge \frac{{(v_1-v_2)}^2}{2a}$
• D. $d\ge \frac{v_1^2-v_2^2}{2a}$

Answer 1

The correct option is C $d\ge \frac{{(v_1-v_2)}^2}{2a}$

Taking the direction of motion as positive direction,
Initial relative velocity of car $A$ w.r.t. $B$ is $u_{rel}=u_A-u_B=v_1-v_2$
To avoid the collision, both cars will come to hault and hence there velocity becomes zero.
Final relative velocity of car $A$ w.r.t. $B$ is $v_{rel}=v_A-v_B=0$
Relative acceleration of car $A$ w.r.t. to car $B$ is
$a_{rel}=-a-0=-a$
To avoid collision
$\begin{array}{}d\ge S_{rel}& \text{(1)}\end{array}$
Where $S_{rel}$ is relative diaplacement of car $A$ w.r.t. car $B$
$S_{rel}=\frac{u_{rel}^2}{2a}$ ($\therefore v_{rel}=0$)
$S_{rel}=\frac{u_{rel}^2}{2a}=\frac{{(v_1-v_2)}^2}{2a}$
Hence from equation $\left(1\right)$ we get,
$d\ge \frac{{(v_1-v_2)}^2}{2a}$