fitzpratik wrote:

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?

A. \(\sqrt{(4x)^2 + ( 100 + 2y)^2}\)

B. x + y

C. \(\sqrt{x^2 + y^2}\)

D. \(\sqrt{(4x)^2 + (2y)^2}\)

E. \(\sqrt{(4x)^2 + ( 100 - 2y)^2}\)

Source: NOVA GMAT

To find the distance between the two ships at 5 p.m., construct two legs of what will be a right triangle. Ship A travels away from port. Ship B travels toward port.

Compute each ship's distance covered by 5 p.m. without reference to the other.

Ship A starts at 1 p.m. and travels for t = 4 hours at rate \(x\) miles per hour.

Ship A's distance = r*t = 4x (miles). Ship A heads due west

from port. If port is D = 0, Ship A's distance is 0 + 4x =

4x Ship B starts at 3 p.m. and travels for 2 hours at rate \(y\) miles per hour.

Ship B is 100 miles due south from the port. It travels due north

to the port. It covers part of that 100 miles. For 2 hours at \(y\) mph, rt= D, it covers 2y miles.

That partial distance must be subtracted from the given distance of 100 miles. Port is D = 0. Ship B started at 100 miles away from 0. It has covered 2y miles of the 100 miles.

Ship B's 5 p.m. location, in miles, therefore will be

(100 - 2y)Ship A's 5 p.m. location, in miles:

4xThe distance each has traveled is the leg of a right triangle. The distance between them is the hypotenuse.

Using Pythagorean theorem first, \(D\) is distance between them at 5 p.m.:

\((4x)^2 + (100 - 2y)^2 = D^2\)

\(D = \sqrt{(4x)^2 + (100 - 2y)^2}\)

Answer E