The distance between the ships changes at 7 P.M. by approximately 21.4 kilometers per hour.
How to find the rate of change of the distance between two ships
Let suppose that both ships travel at constant speed. After a quick reading of the statement, we find that distance between the ships (\(r\)), in kilometers, as a function of time (\(t\)), in hours, is defined by Pythagorean theorem:
\(r = \sqrt{(r_{o}-\dot x \cdot t)^{2}+(\dot y\cdot t)^{2}}\) (1)
Where:
- \(r_{o}\) - Initial distance between the two ships, in kilometers.
- \(\dot x\) - Speed of ship A, in kilometers per hour.
- \(\dot y\) - Speed of ship B, in kilometers per hour.
Now we proceed to find an expression for the rate of change of the distance between the ships (\(\dot r\)), in kilometers per hour:
\(\dot r = \frac{[(r_{o}-\dot x\cdot t)\cdot (-\dot x)+(\dot y\cdot t)\cdot \dot y]}{ \sqrt{(r_{o}-\dot x\cdot t)^{2}+(\dot y\cdot t)^{2}}}\) (2)
If we know that \(r_{o} = 150\,km\), \(\dot x = 35\,\frac{km}{h}\), \(\dot y = 25\,\frac{km}{h}\) and \(t = 4\,h\), then the rate of change between the ships is:
\(\dot r = \frac{\left\{\left[150\,km-\left(35\,\frac{km}{h}\right)\cdot (4\,h)\right]\cdot \left(-35\,\frac{km}{h} \right)+\left[\left(25\,\frac{km}{h}\right)\cdot (4\,h)\right]\cdot \left(25\,\frac{km}{h} \right)\right\}}{\sqrt{\left[150\,km-\left(35\,\frac{km}{h}\right)\cdot (4\,h)\right]^{2}+\left[\left(25\,\frac{km}{h}\right)\cdot (4\,h)\right]^{2}}}\)
\(\dot r \approx 21.393\,\frac{km}{h}\)
The distance between the ships changes at 7 P.M. by approximately 21.4 kilometers per hour. \(\blacksquare\)
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