The curve between net force and time is shown | vedclass

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(D) From Newton's second law,$F = ma = m(dv/dt)$,which implies $dv = (F/m) dt$.Integrating both sides,the change in velocity $\Delta v = \int (F/m) dt

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(D) From Newton's second law,$F = ma = m(dv/dt)$,which implies $dv = (F/m) dt$.Integrating both sides,the change in velocity $\Delta v = \int (F/m) dt = (1/m) \int F dt$.Since the particle starts from rest,the velocity $v(t)$ at any time $t$ is proportional to the area under the force-time graph up to that time.$1$. Initially,$F$ is constant and non-zero,so the acceleration $a = F/m$ is constant. This means the velocity increases linearly with time $(v = at)$.$2$. During the interval where $F$ increases and then decreases (the triangular pulse),the acceleration $a$ also increases and decreases. Consequently,the slope of the velocity-time graph $(dv/dt = a)$ increases and then decreases,resulting in a curved (concave up) shape for the $v-t$ graph.$3$. After the pulse,$F$ returns to its initial constant value,so the acceleration becomes constant again,and the velocity-time graph becomes a straight line with the same slope as the initial part.Comparing this behavior with the given options,Graph $D$ correctly shows a linear increase,followed by a curved section where the slope increases and decreases,and finally a linear increase with the original slope.

The curve between net force and time is shown. Initially,the particle is at rest. Which of the following best represents the resulting velocity-time graph of the particle?

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