Two cars $A$ and $B$ are travelling in the same direction with velocities $v_{1}$ and $v_{2}$ $(v_{1} > v_{2})$ . 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 < \frac{(v_{1} - v_{2})^{2}}{2 a}

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

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

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

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Solution

The correct option is C $d > \frac{(v_{1} - v_{2})^{2}}{2 a}$
initial relative velocity of car A w.r.t. car B
$u_{r e l} = v_{1} - v_{2}$
Final relative velocity of car A w.r.t. car B
$v_{r e l} = 0$
as there should be no collision, both cars will finally move with same velocity
$a_{r e l} = - a$
$\Rightarrow S_{r e l} = \frac{v_{r e l}^{2} - u_{r e l}^{2}}{2 a} = \frac{(v_{1} - v_{2})^{2}}{2 a}$
To avoid collision $d \geq S_{r e l} \Rightarrow d \geq \frac{(v_{1} - v_{2})^{2}}{2 a}$

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