Explanation

(c) At $D$ then the distance $CD$ is least. Therefore, $OC = (10 - 3t)$m
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$\therefore\ CD^{2} = OC^{2} + OD^{2}$
$= (10 - 3t)^{2} + (4t)^{2}$
$= 25t^{2} - 60t + 100$
or $CD = \left\lbrack (5t - 6)^{2} + 64 \right\rbrack^{1/2}$
$CD$ is least if $(5t - 6)^{2} = 0$ or $5t - 6 = 0$
or $5t = 6$ or $6 = t$/ 5 = 1.2s
So, $AC = 3 \times 1.2 = 3.6$m: $CD = (64)^{1/2}$=8m.