JEE Main 2024 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 1 February 2024, Shift 2, Physics Q51

Question 51 of 90 in this shift, Physics question 21 of 30, Section B.

A particle initially at rest starts moving from reference point

x = 0

along

x

-axis, with velocity

v

that varies as

v = 4\sqrt{x}

m/s. The acceleration of the particle is ________

\mathrm{ms^{-2}}

Official answer

NTA final key.

Chapter: Motion in a Straight Line
Topic: Kinematics in 1D
Idea tested: Position-Dependent Acceleration (v dv/dx)

Same idea in other shifts

Asked 5× in all

2 Apr 2026, Shift 1 · Q28The velocity of a particle is given as $\vec v=-x\hat i+2y\hat j-z\hat k$ m/s. The magnitude of acceleration at point $(1, 2, 4)$ is…MediumSingle correct
25 Jul 2022, Shift 2 · Q51A particle is moving in a straight line such that its velocity is increasing at 5 ms $^{-1}$ per meter. The acceleration of the particle is…EasyNumerical value
27 Aug 2021, Shift 1 · Q23If the velocity of a body related to displacement $x$ is given by $v=\sqrt{5000+24x}$ m/s, then the acceleration of the body is __________…EasyNumerical value
24 Feb 2021, Shift 2 · Q2A particle is projected with velocity $v_0$ along $x$ -axis. A damping force is acting on the particle which is proportional to the square…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.