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Question: Two cars 1 & 2 starting from rest are moving with speed \({{V}_{1}}\And {{V}_{2}}\dfrac{m}{s}({{V}_{...

Two cars 1 & 2 starting from rest are moving with speed V1&V2ms(V1>V2){{V}_{1}}\And {{V}_{2}}\dfrac{m}{s}({{V}_{1}}>{{V}_{2}}), car 2 is ahead of car 1 by 'S' meters when the driver of car 1 sees car 2. What minimum retardation should be given to car 1 to avoid a collision.
A.V1−V2S B.V1+V2S C.V1+V22S D.(V1−V2)22S \begin{aligned} & A.\dfrac{{{V}_{1}}-{{V}_{2}}}{S} \\\ & B.\dfrac{{{V}_{1}}+{{V}_{2}}}{S} \\\ & C.\dfrac{{{V}_{1}}+{{V}_{2}}}{2S} \\\ & D.\dfrac{{{({{V}_{1}}-{{V}_{2}})}^{2}}}{2S} \\\ \end{aligned}

Explanation

Solution

Hint: This question can be solved by relative motion and the kinematic equation of the motion. First, we have to use the relative motion concept to find the relative velocity of car 1 with respect to car 2 and then we have to apply the kinematic equation.

Step by step answer:

It is given in the question that, the car 1 is moving with speed V1{{V}_{1}} and the car 2 is moving with velocity V2{{V}_{2}}.

Now, we need to find the relative speed of the car 1 with respect to the car 2. Assuming the car 2 is at rest, the relative speed of car 1 with respect to car 2 is given as,
V1−V2{{V}_{1}}-{{V}_{2}}

At distance 'S' car 1 driver will see car 2 so the driver of car 1 will apply the brake for retardation. So, the collision will not happen. For finding the value of retardation now we have to apply the kinematic equation of motion on car 1.

There are 3 kinematic equation we have to apply equation number 2 of the kinematic equation of motion that is

v2=u2-2aS{{\text{v}}^{\text{2}}}\text{=}{{\text{u}}^{\text{2}}}\text{-2aS}

Where u\text{u}is the initial velocity of car 1 that is

V1−V2{{V}_{1}}-{{V}_{2}}
v\text{v} is the final velocity of car 1, which is zero.
aa is the acceleration/retardation that we have to find.
S\text{S}is the distance between the car 1 and car 2.

So, when we put all those values in the equation we get,

0=(V1−V2)2−2aS a=(V1−V2)22S \begin{aligned} & 0={{({{V}_{1}}-{{V}_{2}})}^{2}}-2aS \\\ & a=\dfrac{{{({{V}_{1}}-{{V}_{2}})}^{2}}}{2S} \\\ \end{aligned}

The answer is (D)

Note: Mostly all the questions related to linear motion or relative motion can be solved by using the three equations of kinematic motion. These equations are,

v=u+at\text{v=u+at}
S=ut+12at2S=ut+\dfrac{1}{2}a{{t}^{2}}
v2=u2-2aS{{\text{v}}^{\text{2}}}\text{=}{{\text{u}}^{\text{2}}}\text{-2aS}