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Relative Acceleration

Two cars A an...

Question

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 A is at a distance d ahead of the car B, the driver of the car A applied the brake producing a uniform retardation a. There will be no collision when

A

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

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B

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

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C

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

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D

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 $= v_{1} - v_{2},$ final relative velocity=0

From $v^{2} = u^{2} - 2 a s \Rightarrow 0 = (v_{1} - v_{2})^{2} - 2 \times a \times s$

$\Rightarrow s = \frac{(v_{1} - v_{2})^{2}}{2 a}$

If the distance between two cars 's' then collision will take place.To avoid collision $d > s$

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

where d=actual initial distance between two cars.

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