GeminiHeat wrote:
Two ants leave simultaneously from a spot on the floor. One of them crawls due north at 2 inches per second, and the other crawls due west at a rate one-third faster than the first ant's rate. Approximately how many feet apart will they be after 6 seconds?
A. 83
B. 1
C. 1.33
D. 1.67
E. 20
Let the starting point be O
After 6 seconds:
Distance traveled by the ant travelling north = \(2 * 6 = 12\) inches
(Let A be the end point of this ant)
Distance traveled by the ant travelling north = \(12 + 12 * 1/3 = 16\) inches
(Let B be the end point of this ant)
Observe that OAB forms a right-triangle with \(angle AOB = 90^o\)
Thus, distance between the 2 ants (from Pythagoras' theorem) = \(\sqrt{12^2 + 16^2} = 20\) inches
Thus, distance in feet = \(20/12 = 1.67\) feet