Q 11-02-103JEE MainJEE Main 2021 (31 Aug, Shift 2)Easy
A particle is moving with constant acceleration $a$. Following graph shows $v^2$ versus $x$ (displacement) plot. The acceleration of the particle is ______ $\text{m s}^{-2}$.
Numerical value type. Enter your answer.
Answer: 1
$v^2 = u^2 + 2ax$, so the slope of the $v^2$–$x$ graph is $2a$.
$$2a = \frac{80 - 40}{30 - 10} = 2 \Rightarrow a = 1\ \text{m s}^{-2}$$
Solution by Sreeraj P, M.Sc Physics