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

d \leq \frac{{(v_{1} - v_{2})}_{1}^{2}}{2 a}

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d \leq \frac{v_{1}^{2} - v_{2}^{2}}{2 a}

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d \geq \frac{{(v_{1} - v_{2})}_{1}^{2}}{2 a}

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d \geq \frac{v_{1}^{2} - v_{2}^{2}}{2 a}

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Solution

The correct option is C $d \geq \frac{{(v_{1} - v_{2})}_{1}^{2}}{2 a}$
Taking the direction of motion as positive direction,
Initial relative velocity of car $A$ w.r.t. $B$ is $u_{r e l} = u_{A} - u_{B} = v_{1} - v_{2}$
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 $v_{r e l} = v_{A} - v_{B} = 0$
Relative acceleration of car $A$ w.r.t. to car $B$ is
$a_{r e l} = - a - 0 = - a$
To avoid collision
$\begin{matrix} d \geq S_{r e l} \\ (1) \end{matrix}$
Where $S_{r e l}$ is relative diaplacement of car $A$ w.r.t. car $B$
$S_{r e l} = \frac{u_{r e l}^{2}}{2 a}$ ( $∴ v_{r e l} = 0$ )
$S_{r e l} = \frac{u_{r e l}^{2}}{2 a} = \frac{{(v_{1} - v_{2})}_{1}^{2}}{2 a}$
Hence from equation $(1)$ we get,
$d \geq \frac{{(v_{1} - v_{2})}_{1}^{2}}{2 a}$

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