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(A) Given acceleration $v'(t) = a - kt^2$. Since the particle starts from rest, $v(0) = 0$.Integrating with respect to $t$, we get $v(t) = \int (a - k

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$\frac{2}{3} \sqrt{\frac{a^3}{k}}$

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(A) Given acceleration $v'(t) = a - kt^2$. Since the particle starts from rest, $v(0) = 0$.Integrating with respect to $t$, we get $v(t) = \int (a - kt^2) dt = at - \frac{k}{3}t^3 + C$.Using $v(0) = 0$, we find $C = 0$, so $v(t) = at - \frac{k}{3}t^3$.To find the maximum velocity, we set the acceleration to zero: $a - kt^2 = 0$, which gives $t^2 = \frac{a}{k}$, or $t = \sqrt{\frac{a}{k}}$ (since $t > 0$).Substituting this value of $t$ into the velocity equation:$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \left(\sqrt{\frac{a}{k}}\right)^3$$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \cdot \frac{a}{k} \sqrt{\frac{a}{k}}$$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{a}{3} \sqrt{\frac{a}{k}} = \frac{2a}{3} \sqrt{\frac{a}{k}} = \frac{2}{3} \sqrt{\frac{a^3}{k}}$.

$A$ particle moving in a straight line starts from rest and the acceleration at any time $t$ is $a - kt^2$, where $a$ and $k$ are positive constants. The maximum velocity attained by the particle is

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