Explanation

(d) Given, $x = 6t^{2} - t^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dx}{dt} = 12t - 3t^{2}\ \ \ \ \ \ \ \ \ \ldots.(i)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dx}{dt} = 0 \Rightarrow t = 4\ s$
Now, again differentiating Eq. (i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{d^{2}x}{dt^{2}} = 12 - 6t = 12 - 6(4) = - 12$
Since, $\frac{d^{2}x}{dt^{2}}$ is negative, hence $t = 4\ s$ gives the maximum value for $x - t$ curve.
Moreover, acceleration $a = \frac{d^{2}x}{xt^{2}},\ at\ t = 0,\frac{d^{2}x}{dt^{2}} = 12\ {ms}^{- 2}$