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Hi engrymbiff; Where c is the ratio of B to A in this case 1:2, So c = 1 / 2, c is always less than 1. I am having trouble with setting the initial conditions. Okay , got it. If A starts at y=5 and B starts at ( 0 , 0 ) we set the initial conditions as y ' (5) = 0 and y(5) = 0 Solving the DE we get: Plugging in x = 0 we get y = 10 / 3. So A will catch B at ( 0 , 10 / 3 ). This can be easily checked as being correct. http://www.mathsisfun.com/graph/functio … 3333333333 Notice that A comes from the right and doesn't start at the origin, but to make it do so should be a simple translation.
Perfect, I'm also working on it right now actually. I'll update you if I manage to solve it!
Hi engrymbiff;
Thanks bobbym!! That does help a lot!
Hi engrymbiff;
Sorry, maybe I did not explain the problem well enough. "A's direction of movement is always in the nearestdistancedirection". So, A's direction will change according to B's movement. This will result in some sort of parable for A's path. NOT just pythagoras
Let the point A be (0, 0), i.e. origin, and lets say that they meet at a point C (b, k)
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