JEE Main 2024 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 31 January 2024, Shift 1, Physics Q32

Question 32 of 90 in this shift, Physics question 2 of 30, Section A.

The relation between time 't' and distance 'x' is

t=\alpha x^2+\beta x

, where

\alpha

and

\beta

are constants. The relation between acceleration

(a)

and velocity

(v)

is :

(1) $a=-2\alpha v^3$ Official answer
(2) $a=-3\alpha v^2$
(3) $a=-4\alpha v^4$
(4) $a=-5\alpha v^5$

Official answer

Option 1

NTA final key.

Chapter: Motion in a Straight Line
Topic: Kinematics in 1D
Idea tested: Implicit Time-Position Equations

Same idea in other shifts

Asked 3× in all

25 Jul 2021, Shift 2 · Q4The relation between time t and distance $x$ for a moving body is given as $t = mx^2 + nx$ , where $m$ and $n$ are constants. The…MediumSingle correct
9 Jan 2020, Shift 1 · Q21The distance $x$ covered by a paritcle in one dimensional motion varies with time t as $x^{2} = at^{2} + 2bt + c$ . If the acceleration of…HardNumerical value

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.