Kinematics

JEE Main level, single correct, medium

96 of 308

Question 96 of 308

The velocity-displacement graph of a particle is shown in the figure.

The acceleration-displacement graph of the same particle is represented by :

Show answer and solution

Answer: (C)

v=−(v0x0)x+v0v=-\left(\dfrac{{v}_{0}}{{x}_{0}}\right)x+{v}_{0}
a=vdvdxa=\dfrac{vdv}{dx}
a=[−(v0x0)x+v0][−v0x0]a=\left[-\left(\dfrac{{v}_{0}}{{x}_{0}}\right)x+{v}_{0}\right]\left[-\dfrac{{v}_{0}}{{x}_{0}}\right]
a=(v0x0)2X−v02x0a={\left(\dfrac{{v}_{0}}{{x}_{0}}\right)}^{2}X-\dfrac{{v}_{0}^{2}}{{x}_{0}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Projectile motion simulationDrag the sliders and watch the numbers in this question change.Revise Kinematics & Projectile Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Kinematics (Graphs of motion in one dimension), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.