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JEE Main 2021 (31 Aug, Shift 2) · Graphs of Motion

Graphs of Motion question from JEE Main 2021 (31 Aug, Shift 2)

Q24JEE Main 2021 · 31 AUG · SHIFT 2Easy

Graphs of Motion

A particle is moving with constant acceleration 'a'. Following graph shows v2 versus x(displacement) plot. The acceleration of the particle is ___________ m/s2.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

$v^2 = u^2 + 2ax$ , so slope of the $v^2$ - $x$ graph $= 2a$ .
Slope $= \frac{80 - 20}{30 - 0} = 2$
$2a = 2 \Rightarrow a = 1$
$a = 1\text{ m/s}^2$

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