Physics · Motion in one Dimension

JEE Main 2024 — 9 April, Shift 2 — Question 36

Two cars are travelling towards each other at speed of $20 \mathrm{~m} \mathrm{~s}^{-1}$ each. When the cars are 300 m apart, both the drivers apply brakes and the cars retard at the rate of $2 \mathrm{~m} \mathrm{~s}^{-2}$ . The distance between them when they come to rest is :

Option A:
200 m
Option B:
50 m
Option C:
100 m
Correct
Option D:
25 m

Answer: C

Step-by-step solution

$\left|\overrightarrow{\mathrm{u}}_{\mathrm{BA}}\right|=40 \mathrm{~m} / \mathrm{s}$

$\left|\overrightarrow{\mathrm{a}}_{\mathrm{BA}}\right|=4 \mathrm{~m} / \mathrm{s}$ Apply

$\left(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \text { as }\right)_{\text {relative }}$

$\mathrm{O}=(40)^{2}+2(-4)(\mathrm{S})$

$\mathrm{S}=200 \mathrm{~m}$

Remaining distance $=300-200=100 \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion