Physics Help Forum Acceleration(t and n) of boy in skateboard in parabola

 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.

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