Displacement and ants?

Question

an ant walks 14cmN, 8cmS then 27cmE and 9cmW. what is the displacement of the ant relative to the starting point?
Please

Answer 1

On an (x,y) graph, the ant's position is:

(0,0) start
(0,14)
(0,6)
(27,6)
(18, 6) end
where, +y = north, +x = east

At the end the ant is 18 units along the positive x-axis, and 6 units on the positive y-axis. This forms a right triangle with (0,0) at the apex. You solve the displacement (hypoteneuse) of the triangle whose known sides are 18 and 6.
hypoteneuse = sqrt(a^2 + b^2)
= sqrt(18^2 + 6^2)
= sqrt(324 + 36)
= sqrt(360)
= 18.97 which is the displacement of the ant

Hope this helps!

Answer 2

The other answers are probably what the person who asked the question is looking for, but there actually isn't enough information given to answer the question. If they're giving you directions in terms of north, east, south, and west, then one would assume you're on the surface of earth, which is a sphere- so the other answers are just approximations of the actual answer, and the actual answer depends on how far the starting point of the ant is from a pole. If it's close to the equator, then the answers given by the other posters are extremely close to the right answer. But suppose the ant started 15 cm south of the north pole. Then it would walk 14 cm north, then, 8 cm south, and would be 9 cm south of the north pole. Then by walking 27 cm east, it would walk in a circle around the pole, almost coming back to the same point it left from before going east. Then it would go about 1/3 of the way back around the pole in an arc retracing the path when it goes 9 cm west. The end result will be totally different from the results given by the other posters, which is very close to the result you get if you start anywhere arbitrarily far away from the pole.

Answer 3

14cmN-8cmS=6cmN; 27cmE-9cmW=18cmE
the distance is 3sqrt10 and the angle is arcsin (1/3)