Explanation

(d) From equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = ut + \frac{1}{2}at^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0 + \frac{1}{2} \times 10t^{2} = 5t^{2}\ \ \ \ \ \ \ \ldots(i)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x + 3 = 0 + \frac{1}{2} \times 10(t + 0.5)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x + 3 = 5\left( t^{2} + \frac{1}{4} + t \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Subtract Eq. (i) from Eq. (ii)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 = 5\left( \frac{1}{4} + t \right) = \frac{5}{4} + 5t$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 - \frac{5}{4} = 5t$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{7}{4} = 5t\ \Rightarrow t = \frac{7}{20}\ s$
Now, $\ \ \ \ \ \ \ \ \ \ \ \ v = u + at$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = 0 + 10 \times \frac{7}{20} = 3.5\ {ms}^{- 1}$
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