Question

Two cars $A$ and $B$ are travelling in the same direction with velocities $v_A$ and $v_B\left(v_A>v_B\right)$. When the car is at a distance $s$ behind car $B$, then the drives of the car $A$ applies the brakes producing a uniform retardation $a$, there will be no collision when

• A. $s \leq \frac{v_A-V_B}{2}$
• B. $s \leq \frac{\left(v_A-v_B\right)^2}{2 a}$
• C. $s \leq\left(v_A-v_B\right)^2$
• D. $s=\frac{\left(v_A-v_B\right)^2}{a}$

Answer 1

Answer: (B)

For no collision, the speed of car $A$ should be reduced to $v_B$ before the cars meet, i.e. final relative velocity of car $A$ with respect to car $B$ is zero, i.e.
$v_{\text {relative }}=0$
Here, initial relative velocity, $u_r=v_A-v_B$
Relative acceleration, $a_r=-a-0=-a$
Let relative displacement be $s_r$.
Thus, from third equation of motion, we get
$ v_{\text {relative }}^2=u_r^2+2 a_r s_r=\left(v_A-v_B\right)^2-2 a s_r$
$ \Rightarrow \quad s_r=\frac{\left(v_A-v_B\right)^2}{2 a}$
For no collision, $s \leq s_r$
$\text { i.e., } s \leq \frac{\left(v_A-v_B\right)^2}{2 a}$