The acceleration-time graph of a body is show | vedclass

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(D) From the given $a-t$ graph,it is clear that the acceleration is increasing at a constant rate.$	herefore \frac{da}{dt} = k$ (constant) $\implies

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(D) From the given $a-t$ graph,it is clear that the acceleration is increasing at a constant rate.$	herefore \frac{da}{dt} = k$ (constant) $\implies a = kt$ (by integration).Since $a = \frac{dv}{dt}$,we have $\frac{dv}{dt} = kt \implies dv = kt \, dt$.Integrating both sides: $\int dv = \int kt \, dt \implies v = \frac{1}{2}kt^2$.This shows that $v$ is dependent on time parabolically,where the graph is a parabola opening upwards.After the acceleration reaches its maximum value,it suddenly becomes zero. When acceleration is zero,the velocity of the body becomes constant.Therefore,the velocity-time graph will show a parabolic increase followed by a horizontal line representing constant velocity. This corresponds to graph $(d)$.

The acceleration-time graph of a body is shown below. The most probable velocity-time graph of the body is:

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