In a car race on straight road,… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Mechanics

2019mcqmedium

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

Official previous-year question

Held on 9 Jan 2019 · Verified 6 Jul 2026.

Options

A
$\frac{2{a}_{1}{a}_{2}}{{a}_{1}+{a}_{2}}t$
B
$\frac{{a}_{1}+{a}_{2}}{2}t$
C
$\sqrt{{a}_{1}{a}_{2}}t$
D
$\sqrt{2{a}_{1}{a}_{2}}t$

Did you get this right?

Solution

Given,

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

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

Let us assume,

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

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

Using second equation of motion, we have

${u}_{1}t+\frac{1}{2}{a}_{1}{t}_{1}^{2}={u}_{2}t+\frac{1}{2}{a}_{2}{{t}_{2}}^{2}$

$\Rightarrow 0\times t+\frac{1}{2}{a}_{1}{t}_{0}^{2}=0\times t+\frac{1}{2}{a}_{2}{({t}_{0}+t)}^{2}$

$\Rightarrow \sqrt{\frac{{a}_{1}}{{a}_{2}}}{t}_{0}={t}_{0}+t$

$\Rightarrow {t}_{0}=\frac{t}{\sqrt{\frac{{a}_{1}}{{a}_{2}}}-1}$

From first equation of motion, we have

$\begin{matrix}{v}_{1}={a}_{1}{t}_{0}& & {v}_{2}={a}_{2}({t}_{0}+t)\end{matrix}$

$\Rightarrow v={v}_{1}-{v}_{2}=({a}_{1}-{a}_{2}){t}_{0}-{a}_{2}t$

$\Rightarrow v=({a}_{1}-{a}_{2})\frac{t}{\sqrt{\frac{{a}_{1}}{{a}_{2}}}-1}-{a}_{2}t$

$\Rightarrow v=t(\frac{{a}_{1}\sqrt{{a}_{2}}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-\frac{{a}_{2}{\sqrt{a}}_{2}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-{a}_{2})$

$\Rightarrow v=t(\frac{{a}_{1}\sqrt{{a}_{2}}-{a}_{2}{\sqrt{a}}_{1}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}})$

$\Rightarrow v=\sqrt{{a}_{1}.{a}_{2}} t$

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Solution

Given,

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

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

Let us assume,

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

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

Using second equation of motion, we have

${u}_{1}t+\frac{1}{2}{a}_{1}{t}_{1}^{2}={u}_{2}t+\frac{1}{2}{a}_{2}{{t}_{2}}^{2}$

$\Rightarrow 0\times t+\frac{1}{2}{a}_{1}{t}_{0}^{2}=0\times t+\frac{1}{2}{a}_{2}{({t}_{0}+t)}^{2}$

$\Rightarrow \sqrt{\frac{{a}_{1}}{{a}_{2}}}{t}_{0}={t}_{0}+t$

$\Rightarrow {t}_{0}=\frac{t}{\sqrt{\frac{{a}_{1}}{{a}_{2}}}-1}$

From first equation of motion, we have

$\begin{matrix}{v}_{1}={a}_{1}{t}_{0}& & {v}_{2}={a}_{2}({t}_{0}+t)\end{matrix}$

$\Rightarrow v={v}_{1}-{v}_{2}=({a}_{1}-{a}_{2}){t}_{0}-{a}_{2}t$

$\Rightarrow v=({a}_{1}-{a}_{2})\frac{t}{\sqrt{\frac{{a}_{1}}{{a}_{2}}}-1}-{a}_{2}t$

$\Rightarrow v=t(\frac{{a}_{1}\sqrt{{a}_{2}}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-\frac{{a}_{2}{\sqrt{a}}_{2}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}}-{a}_{2})$

$\Rightarrow v=t(\frac{{a}_{1}\sqrt{{a}_{2}}-{a}_{2}{\sqrt{a}}_{1}}{{\sqrt{a}}_{1}-{\sqrt{a}}_{2}})$

$\Rightarrow v=\sqrt{{a}_{1}.{a}_{2}} t$

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics