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.

  • $\begingroup$ Acceleration is $dv/dt$, not $dv/ds$. $\endgroup$ – zipirovich Nov 4 '18 at 20:14
  • $\begingroup$ @zipirovich, I agree with you, so how do I proceed from here? $\endgroup$ – Jane T Nov 4 '18 at 20:35
  • 2
    $\begingroup$ Hint: Chain rule. $\endgroup$ – Robert Israel Nov 4 '18 at 20:40

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.


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