Question

At noon, ship A is 170 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM

Answer 1

Answer:

The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.

Step-by-step explanation:

Vectorially speaking, let assume that ship A is located at the origin and the relative distance of ship B with regard to ship A at noon is:

$\vec r_{B/A} = \vec r_{B} - \vec r_{A}$

Where $\vec r_{A}$ and $\vec r_{B}$ are the distances of ships A and B with respect to origin.

By supposing that both ships are travelling at constant speed. The equations of absolute position are described below:

$\vec r_{A} = \left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i$

$\vec r_{B} = \left(170\,km\right)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j$

Then,

$\vec r_{B/A} = (170\,km)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j-\left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i$

$\vec r_{B/A} = \left[170\,km-\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j$

The rate of change of the distance between the ship is constructed by deriving the previous expression:

$\vec v_{B/A} = -\left(40\,\frac{km}{h} \right)\cdot i + \left(15\,\frac{km}{h} \right)\cdot j$

Its magnitude is determined by means of the Pythagorean Theorem:

$\|\vec v_{B/A}\| = \sqrt{\left(-40\,\frac{km}{h} \right)^{2}+\left(15\,\frac{km}{h} \right)^{2}}$

$\|\vec r_{B/A}\| \approx 42.720\,\frac{km}{h}$

The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.