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Beach Path math problem.?
Anyone who has walked on the beach knows that walking speed is dependent upon how far away from the ocean one walks. If you walk on the wet sand you can walk much more quickly than if you walked on the dry sand. I have a question that discusses this principal.
On a Cartesian xy plane limited by: Domain: {x | 0 <= x <= 1} Range: {y | 0 <= y <= 1} you start at the point (0,0) and you would like to travel on a defined path to (1,1) in the shortest amount of time. This sounds simple just take the path y = x because it is the shortest path so it will take the shortest amount of time, but there is a catch. Your forward speed {dS/dt} is equal to (1 - 0.75y). With this constraint in mind the path y = x would not be the fastest path. What is the fastest path. I am open to questions about the problem itself if I have not been clear. Thank you.
@Stric. You suggest the shortest path is {y = 0.375x^2}. Thank you for taking time to think about this problem, but you forgot about a critical point. The path has to go from (0,0) to (1,1), but the path you suggest goes from (0,0) to (1, 0.375).
1 Answer
- Anonymous8 years ago
Intuitively, your problem sounds very much like vertical throw in vacuum problem. Horizontal component will never diminish (since there is no air resistance) while vertical component diminishes linearly with distance travelled upwards.
The answer would then the the opposite of an actual throw [of a ball or a simialr object] where one condition must be met: that distance travelled horizontally (at constant horizontal velocity) equals to distance travelled vertically, with constant acceleration, that is with proportionally increasing velocity.
Distance covered duiring constant accelerated motion (with starting velocity of 0) is 1/2 * a*t^2
Distance covered during constant velocity motion is v * t.
t (on x axis as distance travelled horizontally is directly proportional to time spent travelling) is identical for both equations, a = -0.75 so
1/2 * 0.75 * t^2 = 0.375 * t^2
y = 0.375 * x^2