After their meeting, one ship went south and the other went west. Two hours after their meeting, they were $60$ km apart. Find the speed of each ship if the speed of one of them is known to be $6$ kmph higher than that of the other.
Let the speed of ship going west $= S_w$ kmph
Let $\;$ $S_s = \left(S_w + 6\right)$ kmph $\;\;\; \cdots \; (1)$
Initially, both the ships meet at point $O$
After $2$ hours, ship going west covers a distance $= OW$
and the ship going south covers a distance $= OS$
$\therefore \;$ Speed of ship going west $= S_w = \dfrac{OW}{2}$
$\implies$ $OW = 2 \; S_w$ $\;\;\; \cdots \; (2a)$
and speed of ship going south $= S_s = \dfrac{OS}{2}$
$\implies$ $OS = 2 \; S_s$ $\;\;\; \cdots \; (2b)$
Since both the ships are traveling perpendicular to one-another from the point $O$, $\triangle OWS$ is a right angled triangle.
Given: $\;$ Distance between the ships $= WS = 60$ km
We have from right triangle $OWS$,
$WS^2 = OW^2 + OS^2$
i.e. $\;$ $60^2 = \left(2 \; S_w\right)^2 + \left(2 \; S_s\right)^2$ $\;$ [in view of equations $(2a)$ and $(2b)$]
i.e. $\;$ $3600 = 4 \; S_w^2 + 4 \; S_s^2$
i.e. $\;$ $900 = S_w^2 + S_s^2$ $\;\;\; \cdots \; (3)$
$\therefore \;$ In view of equation $(1)$ equation $(3)$ becomes
$900 = S_w^2 + \left(S_w + 6\right)^2$
i.e. $\;$ $900 = S_w^2 + S_w^2 + 12 \; S_w + 36$
i.e. $\;$ $2 \; S_w^2 + 12 \; S_w - 864 = 0$
i.e. $\;$ $S_w^2 + 6 \; S_w - 432 = 0$
i.e. $\;$ $\left(S_w + 24\right) \left(S_w - 18\right) = 0$
i.e. $\;$ $S_w = -24$ $\;$ or $\;$ $S_w = 18$
Since the spped of a ship cannot be negative
$\implies$ Speed of the ship sailing west from $O$ is $= S_w = 18$ kmph
Substituting the value of $S_w$ in equation $(1)$ gives $\;$ $S_s = 18 + 6 = 24$
$\implies$ Speed of the ship sailing south from the point $O$ is $= S_s = 24$ kmph