The velocity of a particle is $v=v_0+gt+Ft^2$. Its position is $x=0$ at $t=0$; then its displacement after time (t=1) is :

Option 3: $v_0+\frac{g}{2}+\frac{F}{3}$

JEE Main 2021 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 17 March 2021, Shift 2, Physics Q11

Question 11 of 90 in this shift, Physics question 11 of 30, Section A.

The velocity of a particle is

v=v_0+gt+Ft^2

. Its position is

x=0

t=0

; then its displacement after time (t=1) is :

(1) $v_0+\frac{g}{2}+F$
(2) $v_0+2g+3F$
(3) $v_0+\frac{g}{2}+\frac{F}{3}$ Official answer
(4) $v_0+g+F$

Official answer

Option 3

NTA final key.

Chapter: Motion in a Straight Line
Topic: Kinematics in 1D
Idea tested: Calculus in Kinematics

Same idea in other shifts

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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.