JEE Main 2023PhysicsLaws of MotionEasyMCQ

JEE Main 2023 — Laws of Motion Question with Solution

JEE Main 2023 (15 Apr Shift 1)

Question

The position vector of a particle related to time t is given by r→=10ti^+15t2j^+7k^m. The direction of net force experienced by the particle is :

Choose an option

Show full solutionCorrect option: C
Correct answer
CPositive y-axis

Step-by-step explanation

Given, the position vector of the particle is

r→=10ti^+15t2j^+7k^

The velocity of the particle can be calculated as follows:

v→=dr→dt=ddt10ti^+15t2j^+7k^= 10i^+30tj^

And, the acceleration of the particle is given by

a→=dv→dt=ddt10i^+30tj^=30j^

Hence, the force on the particle can be written as

F→=ma→=30mj^

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About this question

This is a previous-year question from JEE Main 2023, covering the Laws of Motion 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.