JEE Main 2019 — Physics Mechanics

Given,

Initially both car is at rest, so ${u}_{1}={u}_{2}=0$u1​=u2​=0

Acceleration of car A&B is {a}_{1}&{a}_{2}

Let us assume,

Time of reach to destination of car $A$A is, ${t}_{1}={t}_{0}$t1​=t0​

Time of reach to destination of car $B$B is, ${t}_{2}={t}_{0}+t$t2​=t0​+t

Using second equation of motion, we have

$${u}_{1}t+\frac{1}{2}{a}_{1}{t}_{1}^{2}={u}_{2}t+\frac{1}{2}{a}_{2}{{t}_{2}}^{2}$$u1​t+21​a1​t12​=u2​t+21​a2​t2​2

$$\Rightarrow 0\times t+\frac{1}{2}{a}_{1}{t}_{0}^{2}=0\times t+\frac{1}{2}{a}_{2}{({t}_{0}+t)}^{2}$$⇒0×t+21​a1​t02​=0×t+21​a2​(t0​+t)2

$$\Rightarrow \sqrt{\frac{{a}_{1}}{{a}_{2}}}{t}_{0}={t}_{0}+t$$⇒a2​a1​​​t0​=t0​+t

$$\Rightarrow {t}_{0}=\frac{t}{\sqrt{\frac{{a}_{1}}{{a}_{2}}}-1}$$⇒t0​=a2​a1​​​−1t​

From first equation of motion, we have

$$\begin{matrix}{v}_{1}={a}_{1}{t}_{0}& & {v}_{2}={a}_{2}({t}_{0}+t)\end{matrix}$$v1​=a1​t0​​​v2​=a2​(t0​+t)​

$$\Rightarrow v={v}_{1}-{v}_{2}=({a}_{1}-{a}_{2}){t}_{0}-{a}_{2}t$$⇒v=v1​−v2​=(a1​−a2​)t0​−a2​t

$$\Rightarrow v=({a}_{1}-{a}_{2})\frac{t}{\sqrt{\frac{{a}_{1}}{{a}_{2}}}-1}-{a}_{2}t$$⇒v=(a1​−a2​)a2​a1​​​−1t​−a2​t

$$\Rightarrow v=t(\frac{{a}_{1}\sqrt{{a}_{2}}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-\frac{{a}_{2}{\sqrt{a}}_{2}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-{a}_{2})$$⇒v=t(a​1​−a​2​a1​a2​​​−a​1​−a​2​a2​a​2​​−a2​)

$$\Rightarrow v=t(\frac{{a}_{1}\sqrt{{a}_{2}}-{a}_{2}{\sqrt{a}}_{1}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}})$$⇒v=t(a​1​−a​2​a1​a2​​−a2​a​1​​)

$$\Rightarrow v=\sqrt{{a}_{1}.{a}_{2}} t$$⇒v=a1​.a2​​t