Kinematics

JEE Main level, single correct, medium

109 of 308

Question 109 of 308

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

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Answer: (C)

Let acceleration be aa.
Given that a=−Ca=-C
We know that, a=a=vdvdx=−C\dfrac{vdv}{dx}=-C
vdv=−Cdxvdv=-Cdx
Integrating the above equation,
v22=−Cx+k\dfrac{{v}^{2}}{2}=-Cx+k.
Thus, the velocity-distance graph is

The distance cover by the object is x=−v22C+kCx=-\dfrac{{v}^{2}}{2C}+\dfrac{k}{C}.
Thus, with the increase in distance, velocity decreases.
Now, a=−C=dvdta=-C=\dfrac{dv}{dt}
⇒dv=−Cdt⇒v=−Ct+u,(straight line)\Rightarrow dv=-Cdt \Rightarrow v=-Ct+u,\left(\text{straight line}\right)

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