A car is moving on a circular… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

2023integerhard

A car is moving on a circular path of radius $600m$ such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of $54\mathrm{km}{h}^{-1}$ is $t(1–{e}^{–\frac{\pi }{2}})s$ . The value of $t$ is ______.

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

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Solution

Given:

Initial speed $=\frac{54\times 1000}{3600}=15m{s}^{-1}$

$|{a}_{t}|=|{a}_{c}|$

$\Rightarrow v\frac{dv}{dx}=\frac{{v}^{2}}{R} \Rightarrow {\int }_{15}^{v}\frac{dv}{v}=\frac{1}{R}{\int }_{0}^{x}dx$

$\Rightarrow v=15{e}^{\frac{x}{R}}$

$\Rightarrow \frac{dx}{dt}=15{e}^{\frac{x}{R}}$

Distance covered in first quarter of revolution $=\frac{2\pi R}{4}=\frac{\pi R}{2}$ .

Therefore,

${\int }_{0}^{\frac{\pi R}{2}}{e}^{-\frac{x}{R}}dx=15{\int }_{0}^{{t}_{0}}dt$

$\Rightarrow {t}_{0}=40(1-{e}^{-\frac{\pi }{2}})$

Hence, $t=40$ .

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Solution

Given:

Initial speed $=\frac{54\times 1000}{3600}=15m{s}^{-1}$

$|{a}_{t}|=|{a}_{c}|$

$\Rightarrow v\frac{dv}{dx}=\frac{{v}^{2}}{R} \Rightarrow {\int }_{15}^{v}\frac{dv}{v}=\frac{1}{R}{\int }_{0}^{x}dx$

$\Rightarrow v=15{e}^{\frac{x}{R}}$

$\Rightarrow \frac{dx}{dt}=15{e}^{\frac{x}{R}}$

Distance covered in first quarter of revolution $=\frac{2\pi R}{4}=\frac{\pi R}{2}$ .

Therefore,

${\int }_{0}^{\frac{\pi R}{2}}{e}^{-\frac{x}{R}}dx=15{\int }_{0}^{{t}_{0}}dt$

$\Rightarrow {t}_{0}=40(1-{e}^{-\frac{\pi }{2}})$

Hence, $t=40$ .

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Solution

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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