JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 25 July 2021, Shift 2, Physics Q4

Question 4 of 90 in this shift, Physics question 4 of 30, Section A.

The 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 retardation of the motion is : (Where

v

stands for velocity)

(1) $2\ mv^3$ Official answer
(2) $2\ nv^3$
(3) $2\ mnv^3$
(4) $2\ n^2v^3$

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

31 Jan 2024, Shift 1 · Q32The relation between time 't' and distance 'x' is $t=\alpha x^2+\beta x$ , where $\alpha$ and $\beta$ are constants. The relation between…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.