Physics · Motion in Plane

JEE Main 2024 — 4 April, Shift 2 — Question 51

A bus moving along a straight highway with speed of $72 \mathrm{~km} / \mathrm{h}$ is brought to halt within 4 s after applying the brakes. The distance travelled by the bus during this time (Assume the retardation is uniform) is $\qquad$ m .

Answer: 40

Numerical answer — enter this value.

Step-by-step solution

Initial velocity

$=u=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\Rightarrow 0=20+\mathrm{a} \times 4$

$\mathrm{a}=-5 \mathrm{~m} / \mathrm{s}^{2}$

$\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}$

$\Rightarrow 0^{2}-20^{2}=2(-5) . \mathrm{s}$

$\mathrm{s}=40 \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Physics
Chapter: Motion in Plane
Topic: Relative Motion in One Dimension