Algebra - Word Problems: Derivation of Equations

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.

In the figure

$P$: 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

Algebra - Word Problems: Derivation of Equations

Three cars leave $A$ for $B$ in equal time intervals. They reach $B$ simultaneously and then leave for point $C$ which is $120$ km from $B$. The first car arrives there an hour after the second car, and the third car, having reached $C$, immediately reverses the direction and $40$ km from $C$ meets the first car. Find the speed of the first car.

Let the first car be $C_1$, second car be $C_2$ and the third car be $C_3$.

Let $C_1$ cover distance $AB$ in time $t_1$ hours.

Let $C_2$ start after $C_1$ at point $A$ after $t$ hours

and $C_3$ start after $C_2$ at point $A$ after another $t$ hours.

Since cars $C_1, \; C_2, \; C_3$ reach point $B$ simultaneously

$\implies$ car $C_2$ covers distance $AB$ in $\left(t_1 - t\right)$ hours

and car $C_3$ covers distance $AB$ in $\left(t_1 - 2t\right)$ hours

$\therefore \;$ Speed of first car $C_1$ is $= s_1 = \dfrac{AB}{t_1}$ kmph

Speed of second car $C_2$ is $= s_2 = \dfrac{AB}{t_1 - t}$ kmph

Speed of third car $C_3$ is $s_3 = \dfrac{AB}{t_1 - 2t}$ kmph

Given: $\;$ Distance $BC = 120$ km

Time taken by $C_1$ (with speed $s_1$) to cover distance $BC$

$= T_1 = \dfrac{BC}{s_1} = \dfrac{120}{AB / t_1} = \dfrac{120 \times t_1}{AB}$ hours

Time taken by $C_2$ (with speed $s_2$) to cover distance $BC$

$= T_2 = \dfrac{BC}{s_2} = \dfrac{120}{AB / \left(t_1 - t\right)} = \dfrac{120 \times \left(t_1 - t\right)}{AB}$ hours

Given: $\;$ $T_1 = \left(T_2 + 1\right)$ hours

i.e. $\;$ $\dfrac{120 \times t_1}{AB} = \dfrac{120 \times \left(t_1 - t\right)}{AB} + 1$

i.e. $\;$ $120 t_1 = 120 t_1 - 120 t + AB$

i.e. $\;$ $AB = 120 t$ $\implies$ $t = \dfrac{AB}{120}$ $\;\;\; \cdots \; (1)$

Let cars $C_1$ and $C_3$ meet at a point $P$.

Then, distance $CP = 40$ km and distance $BP = 80$ km

Since $C_1$ and $C_3$ meet at point $P$

$\implies$ Time taken by $C_1$ to cover distance $BP$

$\hspace{2cm}$ $=$ Time taken by $C_3$ to cover distance $BC + CP$

i.e. $\;$ $\dfrac{BP}{s_1} = \dfrac{BC + CP}{s_3}$

Substituting the values of $BP$, $BC$, $CP$, $s_1$ and $s_3$ gives

$\dfrac{80}{AB / t_1} = \dfrac{120 + 40}{AB / \left(t_1 - 2t\right)}$

i.e. $\;$ $\dfrac{80 t_1}{AB} = \dfrac{160 \left(t_1 - 2t\right)}{AB}$

i.e. $\;$ $t_1 = 2\left(t_1 - 2t\right)$

i.e. $\;$ $t_1 = 2 t_1 - 4t$

i.e. $\;$ $t_1 = 4t = 4 \times \dfrac{AB}{120}$ $\;\;\;$ [by equation $(1)$]

i.e. $\;$ $t_1 = \dfrac{AB}{30}$ hours

$\therefore \;$ Speed of first car $C_1$ $= s_1 = \dfrac{AB}{t_1} = \dfrac{AB}{AB / 30} = 30$ kmph