2. A particle moves along the $x$-axis in such a way that its velocity at position $x$ is given by the formula $v(x)=2+\sin x$. What is its acceleration at $x=\frac{\pi}{6}$ ?
Solution:
$5 \sqrt{3} / 4$

Acceleration is given by $a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{d v}{d x} \cdot v=\cos x \cdot(2+\sin x)=5 \sqrt{3} / 4$.