**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.
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Last edited by Hanga; Jul 11th 2011 at 10:01 AM.
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