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:
[tex]\vec r_{B/A} = \vec r_{B} - \vec r_{A}[/tex]
Where [tex]\vec r_{A}[/tex] and [tex]\vec r_{B}[/tex] 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:
[tex]\vec r_{A} = \left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i[/tex]
[tex]\vec r_{B} = \left(170\,km\right)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j[/tex]
Then,
[tex]\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[/tex]
[tex]\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[/tex]
The rate of change of the distance between the ship is constructed by deriving the previous expression:
[tex]\vec v_{B/A} = -\left(40\,\frac{km}{h} \right)\cdot i + \left(15\,\frac{km}{h} \right)\cdot j[/tex]
Its magnitude is determined by means of the Pythagorean Theorem:
[tex]\|\vec v_{B/A}\| = \sqrt{\left(-40\,\frac{km}{h} \right)^{2}+\left(15\,\frac{km}{h} \right)^{2}}[/tex]
[tex]\|\vec r_{B/A}\| \approx 42.720\,\frac{km}{h}[/tex]
The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.
