This comes from USA Harvard-MIT Mathematics Tournament (I don't know from which year)
A particle moves along the $x$-axis such that its velocity at the $x$-position is given by the formula $v(x) = 2 + \sin(x)$. What is its acceleration at $x = \pi/6$? (Answer is $2.2$)
How do I do this if I don't know velocity as a function of time?
We have for the one-dimensional motion that
$$v = \frac{d x}{d t} = 2 + \sin x$$
The acceleration is
$$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x}$$
by the chain rule. We have $\frac{d v}{d x} = \cos x$ so that
$$a = (2 + \sin x) \cos x$$
When $x = \frac{\pi}{6}$, $\sin x = \frac{1}{2}$ and $\cos x = \frac{\sqrt{3}}{2}$. Therefore, $a \left( x = \frac{\pi}{6} \right) = \boxed{\frac{5 \sqrt{3}}{4}} \approx 2.165$.