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Question: At $1$ PM, Ship $A$ leaves port heading due west at $x$ miles per hour. Two hours later, Ship $B$ is $100$ miles due south of the same port and heading due north at $y$ miles per hour. At $5$ PM, how far apart are the ships?

Solution: I do not understand the figure they've drawn in the solution. I am confused especially about the perpendicular line that Ship $B$ is traveling and find the wording difficult to grasp. The answer is $s=\sqrt{(4x^{2})+\left ( 100-2y \right )^{2}}$.

So can anyone help me understand what's going on here? Thanks. enter image description here

OGC
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1 Answers1

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Ship A traveled four hours (between 1 PM and 5 PM), and thus its distance from port is $\;4x\;$ .

Ship B traveled only two hours (from 3PM to 5PM) due north when it was $\;100\;$ kms. from port, so it covered a distance of $\;2y\;$ in these two hours and then its distance from port is now $\;100-2y\;$ .

Thus, you get the right-angled upper triangle with the two red points denoting the ships, and their distance now is easily calculable using Pythagoras Theorem.

DonAntonio
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