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Question

(C) To avoid a collision,the relative velocity of car $A$ with respect to car $B$ must become zero before or at the moment the distance between them b

Vedclass · Accepted Answer

$S \geqslant \frac{(V_A - V_B)^2}{2a}$

Vedclass · Answer

(C) To avoid a collision,the relative velocity of car $A$ with respect to car $B$ must become zero before or at the moment the distance between them becomes zero. Let $u_r$ be the initial relative velocity: $u_r = V_A - V_B$. Let $a_r$ be the relative acceleration: $a_r = -a - 0 = -a$. Using the kinematic equation $v^2 = u^2 + 2as$,where $v$ is the final relative velocity $(0)$,$u$ is the initial relative velocity $(V_A - V_B)$,and $s$ is the relative displacement $S'$ required to stop: $0^2 = (V_A - V_B)^2 + 2(-a)S'$. Solving for $S'$,we get $S' = \frac{(V_A - V_B)^2}{2a}$. For no collision to occur,the initial distance $S$ must be greater than or equal to the stopping distance $S'$. Therefore,$S \geqslant \frac{(V_A - V_B)^2}{2a}$.

Two cars $A$ and $B$ are travelling in the same direction with velocities $V_A$ and $V_B$ $(V_A > V_B)$. When the car $A$ is at a distance $S$ behind the car $B$,the driver of the car $A$ applies the brakes producing a uniform retardation $a$. There will be no collision when:

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