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 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum Jul 11th 2011, 07:40 AM #1 Junior Member   Join Date: Sep 2008 Posts: 6 Acceleration(t and n) of boy on skateboard in parabola Hello all! Im having trouble with this question in my course classical Mechanics. The question: A boy rides a skateboard on the concrete surface of an empty drainage canal described by the equation y = 0.03x^2. He starts at y = 6m and the magnitude of his velocity is approximated by v = sqrt(2g(6-y)) m/S. What is his normal and tangential components of his acceleration when he reaches the bottom? My attempt: g = 9.81 (for the textbook) a = dv/dt t_hat + v^2/p n_hat t_hat: As velocity is not a component of time but rather of y I assume that I derivate it with respect to y instead of t. a = dv/dy = g/(sqrt(12g-2gy)) a(0) = g/sqrt(12g) which should be the acceleration in y=0 n_hat: v(0) = sqrt(12g) p =.. ? The problem: 1) Because this is a parabola I don't know what p is. In general for a circular motion it would be the radius. Here, as it is a parabola, i'm not completely certain what value to use. 2) I don't know if the acceleration in t_hat is correct, as I don't have any answer to this question. Could someone please calculate it and tell me if i'm way off or what I have missed. 3) I have not used the fact that the parabolas equation is y = 0.03x^2. I'm not quite sure how to incorporate this in my solution, but I guess that It has something to do with v^2/p. Thanks in advance for any help you can give. Last edited by Hanga; Jul 11th 2011 at 10:01 AM.  Tags accelerationt, boy, parabola, skateboard Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post kiffren Advanced Electricity and Magnetism 0 Jun 4th 2012 08:19 PM ginarific Periodic and Circular Motion 1 Jan 31st 2010 07:43 AM