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I'm having real problems trying to solve the following:

"The particle is moving in the plane. The trajectory is given by $y=cx-bx^2$, where $c$ and $b$ are positive constants. The acceleration of the particle is always constant and given by $\underline{a}=-\underline{j}$. Find the speed of the particle at the origin O (0,0)."

I am aware that trajectory differentiates to velocity which differentiates to acceleration but I can't work out how to apply it in this case... I'm used to seeing trajectory given in terms of $r(t)=...$ and I can't deduce how to convert it into this form using the information given. I'm really hoping this is actually very simple and I'm missing one key idea!


migration rejected from Sep 21 '13 at 16:33

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closed as off-topic by Qmechanic Sep 21 '13 at 16:33

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In order to find speed, you should consider the vertical and horizontal speeds. The overall speed is the dot product of these vectors. So the problem then becomes, what is the y velocity at origin (0,0)? What is the x velocity at origin(0,0)? – Neil Apr 26 '12 at 9:56