Particle B moves such that its velocity $v\>ms^{-1}$ is related to its
displacement $s\>$m, by the equation $v(s)=arcsin(\sqrt{s})$. Find
the acceleration of particle B when $s=0.1 $m.
My attempt:
$
\frac{dv}{ds}=\frac{1}{2\sqrt{s(1-s)}}=\frac{1}{2\sqrt{0.1(0.9)}}=\frac{5}{3}
$, but this is not the answer.
Best Answer
By the Chain Rule: $$\frac{dv}{dt}=\frac{dv}{ds}\cdot\frac{ds}{dt}=\frac{dv}{ds}\cdot v=\frac{\arcsin\left(\sqrt{s}\right)}{2s\sqrt{s(1-s)}},$$ and you can plug in now.