Two ships left a sea port simultaneously in two mutually perpendicular directions. Half an hour later, the shortest distance between them was $15$ km and another $15$ minutes later, one ship was $4.5$ km farther from the port than the other. Find the speed of each ship.
$P$: Port
$S_1, \; S_2$: Two ships traveling in mutually perpendicular directions from port $P$
$PS_1, \; PS_2$: Position of the two ships after time $t = 0.5$ hours ($30$ minutes)
$PS_1 = x$ km, $\;$ $PS_2 = y$ km
$\therefore \;$ Speed of ship $S_1$ $= s_1 = \dfrac{x}{0.5} = 2x$ kmph
and speed of ship $S_2$ $= s_2 = \dfrac{y}{0.5} = 2y$ kmph
$S_1 S_2$: Shortest distance between ships $S_1$ and $S_2$ $= 15$ km
i.e. $\;$ $x^2 + y^2 = 15^2$ $\;\;\; \cdots \; (1)$ $\;\;$ $\left(\because \; S_1 \perp S_2\right)$
When time $= t_1 = 30 + 15 = 45$ min $= \dfrac{3}{4}$ hour
distance of ship $S_1$ from port $P$ $= d_1 = 2x \times \dfrac{3}{4} = \dfrac{3x}{2}$ km
and distance of ship $S_2$ from port $P$ $= d_2 = 2y \times \dfrac{3}{4} = \dfrac{3y}{2}$ km
Given: $\;$ $d_1 \; \text{km} = \left(d_2 + 4.5\right) \; \text{km}$
i.e. $\;$ $\dfrac{3x}{2} = \dfrac{3y}{2} + 4.5$
i.e. $\;$ $3x = 3y + 9$
i.e. $\;$ $x = y + 3$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes
$\left(y + 3\right)^2 + y^2 = 15^2$
i.e. $\;$ $y^2 + 6y + 9 + y^2 = 225$
i.e. $\;$ $2 y^2 + 6y - 216 = 0$
i.e. $\;$ $y^2 + 3y - 108 = 0$
i.e. $\;$ $\left(y - 9\right) \left(y + 12\right) = 0$
i.e. $\;$ $y = 9$ $\;$ or $\;$ $y = -12$
$\therefore \;$ We have from equation $(2)$,
when $\;$ $y = 9$, $\;$ $x = 9 + 3 = 12$
when $\;$ $y = -12$, $\;$ $x = -12 + 3 = -9$
$\because \;$ Distance covered by the ship cannot be negative,
$\therefore \;$ $x = -9$ and $y = -12$ are not valid solutions.
$\therefore \;$ Speed of ship $S_1 = s_1 = 2x = 2 \times 12 = 24$ kmph
and speed of ship $S_2 = s_2 = 2y = 2 \times 9 = 18$ kmph