Q85 / 90 · JEE Main 2021 (31 Aug Shift 2) · Physics

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

Correct answer: 1

Solution

The equation of the straight line,

y = m x + c

, from the above graph we get the equation,

v^{2} = 2 x + 20

, now differentiate it with respect to

x

,

\Rightarrow 2 v \frac{d v}{d x} = 2 \Rightarrow a = \frac{2}{2} = 1 m s^{- 2}

.

Hence, the correct answer is 1.

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