JEE Main 2017PhysicsMotion In One DimensionEasyMCQ

JEE Main 2017 — Motion In One Dimension Question with Solution

JEE Main 2017 (08 Apr Online)

Question

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?

Choose an option

Show full solutionCorrect option: C
Correct answer
C

Step-by-step explanation

Let acceleration be a.
Given that a=-C
We know that, a=vdvdx=-C
vdv=-Cdx

Integrating the above equation,
v22=-Cx+k.

Thus, the velocity-distance graph is

The distance cover by the object is x= -v22C+kC.

Thus, with the increase in distance, velocity decreases.

Now, a=-C=dvdt

⇒dv=-Cdt⇒v=-Ct+u , straight line

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About this question

This is a previous-year question from JEE Main 2017, 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.