JEE Main 2021PhysicsMotion In One DimensionMediumMCQ

JEE Main 2021 — Motion In One Dimension Question with Solution

JEE Main 2021 (25 Jul Shift 2)

Question

The instantaneous velocity of a particle moving in a straight line is given as v=αt+βt2, where α and β are constants. The distance travelled by the particle between 1 s and 2 s is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B32α+73β

Step-by-step explanation

V=αt+βt2

dsdt=αt+βt2

∫S1S2 ds=∫12αt+βt2dt

S2-S1=αt22+βt3312

As the particle is not changing direction So distance = displacement. Distance =α[4-1]2+β[8-1]3

=3α2+7β3

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About this question

This is a previous-year question from JEE Main 2021, covering the Motion In One Dimension chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.