A particle of mass m is constrained to move o | vedclass

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Question

When the particle is left of $E,$ then $F$ is in right direction, i.e positive. Also its magnitude is constant. Therefore, a $=\mathrm{d} \mathrm{v} /

Vedclass · Accepted Answer

Vedclass · Answer

When the particle is left of $E,$ then $F$ is in right direction, i.e positive. Also its magnitude is constant. Therefore, a $=\mathrm{d} \mathrm{v} / \mathrm{dt}=$ positive constant $dy/dt$ represent the slope of $v\, vs$ t graph. There at te its velocity will increase up to the point when the particle has reached $E.$ After it has reached $\mathrm{E}$, the force on particle is in left direction i.e negative. So the slope of $v\, vs$ $t\, graph$ is negative constant So the velocity keeps on decreasing and after becoming zero is starts moving in left direction (velocity negative) and velocity will keep on decreasing till it has reached point $E.$ Also when the particle returns to $E$ the displacement is zero, hence area under the graph from the point when it first reached $E$ and the point when it again reached $E$ is zero.

So option $A$ is correct.

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