Finding the formula for acceleration from $v=2s^3+5s$, where $s$ is the displacement at time $t$

Question

This is the question:

I first found $\frac{dv}{ds}=6s^2+5$, then I tried to find $\frac{ds}{dt}$ by messing about a little with implicit differentiation, but I had no luck and I therefore couldn't apply the chain rule (i.e. $a=\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}$) to find acceleration. The back of my book tells me the answer is $(6s^2+5)(2s^3+5s)$, but I fail to see how this is true as it would imply that $\frac{ds}{dt}=v$, which I can't exactly understand. Can anyone tell me what I am doing wrong?

Answer 1

$v$ and $s$ are functions of time $t$. Then \begin{align} a&=\frac{dv}{dt}\\ &=\frac{d}{dt}(2s^3+5s)\\ &=6s^2 \frac{ds}{dt} + 5\frac{ds}{dt}\\ &=(6s^2+5)\frac{ds}{dt}\\ &=(6s^2+5)v\\ &=(6s^2+5)(2s^3+5s). \end{align}

Answer 2

Hint $dv/ds=\frac{dv}{dt}.1/\frac{ds}{dt}=\frac{a}{v}$ thus accn=$\frac{a}{v}.v$ thats all

Answer 3

By chain rule: \begin{align*} \frac{dv}{dt} &=\frac{dv}{ds} \times \frac{ds}{dt}\\ a &= v\frac{dv}{ds} \\ &= (2s^{3}+5s)(2s^{3}+5s)' \\ &= (2s^{3}+5s)(6s^{2}+5) \end{align*}

Answer 4

By the chain rule, we have $$a = \frac {dv} {dt} = \frac {dv} {ds} \frac {ds} {dt} = (6s^2 + 5) v = (6s^2 + 5) (2s^3 + 5s),$$ so you're right.