Kinematics

JEE Main level, single correct, medium

196 of 308

Question 196 of 308

In a car race on straight road, car AA takes a time tt less than car BB at the finish and passes finishing point with a speed vv more than that of car B.B. Both the cars start from rest and travel with constant acceleration a1{a}_{1} and a2{a}_{2} respectively. Then vv is equal to:

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Answer: (C) a1a2t\sqrt{{a}_{1}{a}_{2}}t

Given,
Initially both car is at rest, so u1=u2=0{u}_{1}={u}_{2}=0
Acceleration of car A&B is {a}_{1}&{a}_{2}

Let us assume,
Time of reach to destination of car AA is, t1=t0{t}_{1}={t}_{0}
Time of reach to destination of car BB is, t2=t0+t{t}_{2}={t}_{0}+t

Using second equation of motion, we have
u1t+12a1t12=u2t+12a2t22{u}_{1}t+\dfrac{1}{2}{a}_{1}{t}_{1}^{2}={u}_{2}t+\dfrac{1}{2}{a}_{2}{{t}_{2}}^{2}
⇒0×t+12a1t02=0×t+12a2(t0+t)2\Rightarrow 0\times t+\dfrac{1}{2}{a}_{1}{t}_{0}^{2}=0\times t+\dfrac{1}{2}{a}_{2}{\left({t}_{0}+t\right)}^{2}
⇒a1a2t0=t0+t\Rightarrow \sqrt{\dfrac{{a}_{1}}{{a}_{2}}}{t}_{0}={t}_{0}+t
⇒t0=ta1a2−1\Rightarrow {t}_{0}=\dfrac{t}{\sqrt{\dfrac{{a}_{1}}{{a}_{2}}}-1}

From first equation of motion, we have
v1=a1t0v2=a2(t0+t)\begin{matrix}{v}_{1}={a}_{1}{t}_{0}& & {v}_{2}={a}_{2}\left({t}_{0}+t\right)\end{matrix}
⇒v=v1−v2=(a1−a2)t0−a2t\Rightarrow v={v}_{1}-{v}_{2}=\left({a}_{1}-{a}_{2}\right){t}_{0}-{a}_{2}t
⇒v=(a1−a2)ta1a2−1−a2t\Rightarrow v=\left({a}_{1}-{a}_{2}\right)\dfrac{t}{\sqrt{\dfrac{{a}_{1}}{{a}_{2}}}-1}-{a}_{2}t
⇒v=t(a1a2a1−a2−a2a2a1−a2−a2)\Rightarrow v=t\left(\dfrac{{a}_{1}\sqrt{{a}_{2}}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-\dfrac{{a}_{2}{\sqrt{a}}_{2}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-{a}_{2}\right)
⇒v=t(a1a2−a2a1a1−a2)\Rightarrow v=t\left(\dfrac{{a}_{1}\sqrt{{a}_{2}}-{a}_{2}{\sqrt{a}}_{1}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}\right)
⇒v=a1.a2t\Rightarrow v=\sqrt{{a}_{1}.{a}_{2}}t

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