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(D) The equation of motion is $v^2 = u^2 + 2ax$. For a particle starting from rest $(u=0)$, this becomes $v^2 = 2ax$.Comparing this with the equation

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$0.5\,m/s^2$

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(D) The equation of motion is $v^2 = u^2 + 2ax$. For a particle starting from rest $(u=0)$, this becomes $v^2 = 2ax$.Comparing this with the equation of a straight line $y = mx$ (where $y = v^2$ and $x = x$), the slope $m$ is given by $m = \frac{dv^2}{dx} = 2a$.From the graph, the slope is $m = \frac{	ext{rise}}{	ext{run}} = \frac{10\,m^2/s^2}{10\,m} = 1\,m/s^2$.Equating the two expressions for the slope: $2a = 1\,m/s^2$.Therefore, the acceleration $a = 0.5\,m/s^2$. Since the slope is constant, the acceleration is constant for all values of $x$.

Find the acceleration of the particle at $x = 5\,m$ using the given $v^2-x$ graph, where $v$ is velocity and $x$ is displacement.

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