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

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

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\,(km)/(h) \right)\cdot t\right]\cdot i`

`\vec r_(B) = \left(170\,km\right)\cdot i +\left[\left(15\,(km)/(h) \right)\cdot t\right]\cdot j`

Then,

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

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

Its magnitude is determined by means of the Pythagorean Theorem:

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

`\|\vec r_(B/A)\| \approx 42.720\,(km)/(h)`

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