I've got a problem with finding t values for when a velocity vector is horizontal.
There is a wheel, radius $R$ that rolls to the right along a straight line at speed $v$. The path $\vec c$ of a point on the rim is given by $$\vec c(t) = \Bigl( vt - R\sin\Bigl(\frac{vt}{R}\Bigl), R-R\cos\Bigl(\frac{vt}{R}\Bigl)\Bigl) $$
In this problem, it wants me to find the t value/s when the velocity vector of this particular point is horizontal, and the speed at this point.
So far, I've computed: $$\vec v(t) = \frac{\mathrm{d}\vec c}{\mathrm dt} = \Bigl(v - v\cos\Bigl(\frac{vt}{R}\Bigl),v\sin\Bigl(\frac{vt}{R}\Bigl)\Bigl)$$
$$\frac{\mathrm d\vec v}{\mathrm dt} = \Bigl(\frac{v^2}{R}\sin\Bigl(\frac{vt}{R}\Bigl), \frac{v^2}{R}\cos\Bigl(\frac{vt}{R}\Bigl)\Bigl)$$
I then set $$\Bigl(\frac{v^2}{R}\sin\Bigl(\frac{vt}{R}\Bigl), \frac{v^2}{R}\cos\Bigl(\frac{vt}{R}\Bigl)\Bigl) = (0,0) $$ to find when the velocity vector is horizontal i.e when the gradient of the velocity is zero.
And then I find that $$\sin\Bigl(\frac{vt}{R}\Bigl) = 0$$ and $$\cos\Bigl(\frac{vt}{R}\Bigl) = 0$$
but from here I got that $$ \frac{vt}{R} = \pi n$$ and $$ \frac{vt}{R} = \frac{\pi n}{2}$$ giving $$t = \frac{n\pi R}{v}$$ and $$t = \frac{n\pi R}{2v}$$ but from here I'm unsure if I've done it correctly because the answer only has one set of t values - $$t = \frac{n\pi R}{v}$$
Any help would be appreciated, cheers :)