Question

A submarine left Diego Garcia five hours before a cruise ship. The ships traveled in opposite directions. The cruise ship traveled at 19 mph for seven hours. After this time the ships were 241 mi. apart. Find the submarine’s speed.

Answer 1

Answer:

The speed of the submarine is 15.429 miles per hour.

Step-by-step explanation:

Let suppose that both ships travel at constant velocities. As we know that both travel in opposite directions, it is supposed that cruise ship moves in +x direction, whereas submarine in -x direction. Kinematic equations for each sheep are described below:

Ship

$x_{Sh} = x_{o} + v_{Sh}\cdot t$

Submarine

$x_{Su} = x_{o}+v_{Su}\cdot t$

Where:

$x_{o}$ - Position of Diego Garcia island, measured in miles.

$x_{Sh}$, $x_{Su}$ - Current positions of ship and submarine, measured in miles.

$v_{Sh}$, $v_{Su}$ - Velocities of ship and submarine, measured in miles per hour.

$t$ - TIme, measured in hours.

If we know that $x_{Sh} - x_{Su} = 241\,mi$, $v_{Sh} = 19\,\frac{mi}{h}$ and $t = 7\,h$, then:

$x_{Sh} - x_{Su} = (v_{Sh}-v_{Su})\cdot t$

We clear now the velocity of submarine:

$\frac{x_{Sh}-x_{Su}}{t} = v_{Sh}-v_{Su}$

$v_{Su} = v_{Sh}-\frac{x_{Sh}-x_{Su}}{t}$

$v_{Su} = 19\,\frac{mi}{h} -\frac{241\,mi}{7\,h}$

$v_{Su} = -15.429\,\frac{mi}{h}$

Speed of the submarine is the magnitude of its velocity, which is 15.429 miles per hour.