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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.