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(B) The given acceleration-time $(a-t)$ graph shows that acceleration increases linearly with time $(a \propto t)$ for a certain interval and then bec

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(B) The given acceleration-time $(a-t)$ graph shows that acceleration increases linearly with time $(a \propto t)$ for a certain interval and then becomes zero (or drops to zero).$1$. For the interval where acceleration increases linearly: $a = kt$ (where $k$ is a constant).Since $a = \frac{dv}{dt}$,we have $\frac{dv}{dt} = kt$.Integrating both sides with respect to time: $\int dv = \int kt \, dt$.This gives $v = \frac{1}{2} kt^2 + C$. Since the graph starts from the origin,$C = 0$,so $v = \frac{1}{2} kt^2$.This represents a parabolic curve opening upwards.$2$. When the acceleration becomes zero,the velocity becomes constant (as $\frac{dv}{dt} = 0$).Therefore,the velocity-time graph should show a parabolic increase followed by a horizontal line representing constant velocity. This matches the graph in option $B$.

The acceleration-time graph for a particle is given. What will be the corresponding velocity-time graph?

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