# How to calculate speed given trajectory and acceleration?

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!

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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

## migrated from math.stackexchange.comSep 13 '12 at 9:58

If the time-dependent parametization of the trajectory is $x=x(t)$ and $y=y(t)$ then the constraints are given by $$y''(t)=-1,\quad x''(t)=0\quad (*)$$ and $$y(t)=cx(t)-bx^2(t)\quad(**).$$ Differentiating $(**)$ two times and inserting $(*)$ into the result you get $$-1=-2b(x'(t))^2\quad(***).$$
This means that $x(t)=tD+\textrm{costant},$ where $D=1/\sqrt{2b}.$ Chosing the initial instant so that $x(0)=0,$ we get even $$x(t)=tD,\ y(t)=cDt-bD^2t^2.$$ Finally $(x'(0),y'(0))=(D,cD)$