How to solve motion question? A particle is moving at $x=3\cos\left(2t\right)$. Find the expression for velocity in terms of $x$.
I'm not sure where to start.
 A: All right, here's a slightly cleaner version of previous answers.  We have
$$x=3\cos 2t,\frac{dx}{dt}=-6\sin 2t$$
Again, we use the trig identity
$$\sin^2u+\cos^2u=1$$
Substituting $u=2t$ we have
$$\sin^2 2t+\cos^2 2t=1$$
Now we multiply this equation by 9.
$$9\sin^22t+9\cos^22t=9$$
$$9\sin^22t+x^2=9$$
$$9\sin^22t=9-x^2$$
$$3\sin2t=\pm\sqrt{9-x^2}$$
$$\frac{dx}{dt}=-6\sin 2t=\pm2\sqrt{9-x^2}$$
The plus or minus depends on the value of $t$.
A: Since the velocity is the derivative of the displacement we are looking for $\frac{dx}{dt}$ in terms of x
$$x=3\cos(2t)$$
$$\frac{dx}{dt}=-6\sin(2t)$$
Solving for $t$ in the first equation shows that
$$t=\frac{1}{2}\arccos\left(\frac{x}{3}\right)$$
Using the following trig identity 
$$\sin^2x+\cos^2x=1$$
and solving for $\sin x$ gives
$$\sin x=\pm\sqrt{1-\cos^2x}$$
Therefore, plugging in for $t$ in the second equation gives
$$\frac{dx}{dt}=-6\sin\left(\arccos\left(\frac{x}{3}\right)\right)$$
$$\frac{dx}{dt}=\pm6\sqrt{1-\cos^2\left(\arccos\left(\frac{x}{3}\right)\right)}$$
$$\frac{dx}{dt}=\pm6\sqrt{1-\frac{x^2}{9}}$$
$$\fbox{$\frac{dx}{dt}=\pm2\sqrt{9-x^2}$}$$
A: You have:
$$\dot x(t)=-6\sin(2t)$$
From the equation:
$$x(t)=3\cos(2t)$$
you have:
$$t=\frac{1}{2}\arccos\left(\frac{x(t)}{3}\right)$$
So:
$$\dot x(t)=-6\sin\left(\arccos\frac{x(t)}{3}\right)$$
