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Standard XII

Physics

Uniform and Non Uniform

Two cars 1 an...

Question

Two cars 1 and 2 start moving with speeds $V_{1}$ and $V_{2}$ $m / s$ such that $V_{1} > V_{2}$ . Car 2 is ahead of car 1 by $s$ meters when the driver of the car 1 sees car 2. What minimum retardation should be given to car 1 to avoid collision?

\frac{V_{1} - V_{2}}{s}

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\frac{(V_{1} - V_{2})^{2}}{s}

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\frac{V_{1} - V_{2}}{2 s}

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\frac{(V_{1} - V_{2})^{2}}{2 s}

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Solution

The correct option is D $\frac{(V_{1} - V_{2})^{2}}{2 s}$
Car 1 is moving with relative velocity $V_{1} - V_{2}$ w.r.t Car 2
So, using $v^{2} = u^{2} + 2 a s \to$ distance
$v =$ final relative velocity
$u =$ initial relative velocity
$0 = (V_{1} - V_{2})^{2} + 2 (- a) s$
$a = \frac{(V_{1} - V_{2})^{2}}{2 s}$

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Two cars 1 and 2 start moving with speeds V1 and V2 m/s such that V1>V2. Car 2 is ahead of car 1 by s meters when the driver of the car 1 sees car 2. What minimum retardation should be given to car 1 to avoid collision?

The correct option is D (V1−V2)22sCar 1 is moving with relative velocity V1−V2 w.r.t Car 2So, using v2=u2+2as→ distancev= final relative velocityu= initial relative velocity0=(V1−V2)2+2(−a)sa=(V1−V2)22s

Two cars 1 and 2 start moving with speeds $V_{1}$ and $V_{2}$ $m / s$ such that $V_{1} > V_{2}$ . Car 2 is ahead of car 1 by $s$ meters when the driver of the car 1 sees car 2. What minimum retardation should be given to car 1 to avoid collision?