JEE Main 2019PhysicsMotion In One DimensionMediumMCQ

JEE Main 2019 — Motion In One Dimension Question with Solution

JEE Main 2019 (09 Apr Shift 2)

Question

The position of a particle as a function of time t, is given by xt=at+bt2-ct3 where a, b and c are constants. When the particles zero acceleration, then its velocity will be:

Choose an option

Show full solutionCorrect option: A
Correct answer
Aa+b23c

Step-by-step explanation

x(t)=at+bt2-ct3
velocity v(t)=ddt(x)=ddtat+bt2-ct3
=a+2bt-3ct2
Acceleration =ddtv(t)=2b-6tc
acceleration=0⇒2b-6tc=0
t=b3c
∴ velocity when t=b3c,
vt=b3c=a+2bb3c-3b3c2
=a+2b23c-b23c
=a+b23c

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About this question

This is a previous-year question from JEE Main 2019, 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.