Explanation

(a)
After time $t$ the distance covered by $X$ is $4t$ and $Y$ is $3t.$
Let shortest distance between $X\ and\ Y$ is $A.$ Then , by cosine law
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${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A}^{2} = (4t)^{2} + (3t)^{2} - (4t)(3t)2\cos{120{^\circ}}$
$\Rightarrow \ \ \ \ A^{2} = 16t^{2} + 9t^{2} - 24t^{2}\left( - \frac{1}{2} \right) = 37t^{2}\ \ldots.(i)$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ A = \sqrt{37t}$
If $t = 1h,then\ A = \sqrt{37}km$
Now, differentiating Eq.(i)w. r. t. $t,$we get
$2AA^{''} = 37(2t)$
After $t = 1h,\ we\ get$
2 $\sqrt{37\ }A^{''} = 2(37)$
$\Rightarrow \ \ \ A^{''} = \sqrt{37}km/h$