Kinematics

JEE Main level, numerical answer, medium

95 of 308

Question 95 of 308

A particle is moving with constant acceleration a.a. Following graph shows v2{v}^{2} versus xx (displacement) plot. The acceleration of the particle is _________ ms−2.m{s}^{-2}.

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Answer: 1

The equation of the straight line, y=mx+cy=mx+c, from the above graph we get the equation, v2=2x+20{v}^{2}=2x+20, now differentiate it with respect to xx,
⇒2vdvdx=2⇒a=22=1ms−2\Rightarrow 2v\dfrac{dv}{dx}=2\Rightarrow a=\dfrac{2}{2}=1m{s}^{-2}.

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