Two cars 1 and 2 move with velocities $v_{1}$ and $v_{2}$ respectively, on a straight road in same direction. When the cars are separated by a distance d, the driver of car 1 applies brakes and the car moves with uniform retardation $a_{1}$ . Simultaneously, car 2 starts accelerating with $a_{2}$ . If $v_{1} {> v}_{2}$ . Find the minimum initial separation between the cars to avoid collision between them.

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Solution

Relative velocity approach
Using $v_{r e l}^{2}$ = $u_{r e l}^{2}$ + 2 $a_{r e l}$ . $s_{r e l}$ , we can obtain the same result.
Substitute: $v_{r e l}$ = $v_{12}^{'}$ = $v_{1}^{'}$ - $v_{2}^{'}$ , $u_{r e l}$ = $v_{1} - v_{2}$ ,
$a_{r e l}$ = $a_{12}$ = (- $a_{1}) - a_{2}$ = -( $a_{1} + a_{2}$ ) and $s_{r e l}$ = $s_{12}$ = -d,
to obtain ( $v_{1}^{'} - v_{2}^{'})^{2} = (v_{1} - v_{2})^{2} - 2 (a_{1} + a_{2}) d$
Since $v_{1}^{'} < v_{2}^{'}$ to avoid collision, we have
$d = \frac{(v_{1} - v_{2})^{2} - (v_{1}^{'} - v_{2}^{'})^{2}}{2 (a_{1} + a_{2})}$
Maximum distance d for collision is possible when $v_{1}^{'} = v_{2}^{'}$ ,
then, $d_{m a x} = \frac{(v_{1} - v_{2})^{2}}{2 (a_{1} + a_{2})}$

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Two cars 1 and 2 move with velocities $v_{1} a n d v_{2}$ respectively on a straight road in the same direction. When the cars are separated by a distance d, the driver of the car 1 applies brakes and the car moves with uniform retardation $a_{1}$ . Simultaneously, the car 2 starts accelerating with $a_{2}$ . If $v_{1} > v_{2}$ , find the minimum initial separation between the cars to avoid collision between them.