# If $a=s^3$, how do I find $v$?

Let $a=s^3$. Given that $v=3$ when $s=4$, find $v$.

This question really threw me off, because the acceleration is defined in terms of the displacement, and not in terms of time like most kinematics questions are that I have met. I didn't know how to solve it, and the book's worked answer didn't make much sense to me. My question is: if you were to see a question like this, how would you approach it?

• I'm confused. How can an acceleration be equal to the cube of a displacement? The premise is not dimensionally correct. – Patrick Stevens Nov 20 '16 at 11:33
• @PatrickStevens In this case, dimensional consistency is not critical, because they are talking about the numerical values. For example, the equation relating the displacement from the ground of an object free falling vertically in a vacuum is $s = s_0 - \frac 12gt^2$, but if I set $s_0 = 10m, g = 10ms^{-2}$ and considered only the numerical values, I could simplify to $s = 10 - 5t^2$ which doesn't look dimensionally consistent either. – Deepak Nov 20 '16 at 11:50
• @Deepak Ugh, that's an abuse of notation I really don't like! The "simplification" I'd much prefer is $s = (10m) - (5ms^{-2}) t^2$, because then it doesn't matter what units we give $s$ and $t$ (as long as they're distancey and timey respectively). – Patrick Stevens Nov 20 '16 at 11:53
• @PatrickStevens Haha, space-wacey timey-wimey. :) :) Hope you got the reference. Anyway, I agree, physicists abuse notation like nobody's business (except mathematicians'!!). – Deepak Nov 20 '16 at 11:59

There is a standard "trick" involving chain rule that's used to relate $a$ and $s$ and $v$ independent of $t$. It's useful in deriving the classical kinetic energy formula, for example. It goes like this:

$$\frac{ds}{dt} = v$$

$$\frac{dv}{dt} = a$$

Dividing and applying chain rule, you get:

$$\frac{ds}{dv} = \frac va$$

Separating variables,

$$vdv = ads$$

Integrating,

$$\int_3^{v_f}vdv=\int_4^{s_f}ads$$

$$\int_3^{v_f}vdv=\int_4^{s_f}s^3ds$$

Note that I specified the lower bounds as the initial conditions to avoid the constant of integration. The upper bounds are time-variable (the $f$ subscript refers to "final"). In a physics context, they would just name them $v$ and $s$ (identical to the variable names), but this is an abuse of notation.

So you get:

$$\frac 12 v_f^2 - \frac {3^2}{2} = \frac 14 s_f^4 - \frac{4^4}{4}$$

And I'm sure you can proceed from here.

• Wow, nice answer! This is actually what the book did, but the book didn't explain it at all. Thank you! – Skeleton Bow Nov 20 '16 at 11:52
• @SkeletonBow Happy to help. :) – Deepak Nov 20 '16 at 11:54

Velocity is the antiderivative of acceleration. Then use $v=3$ and $s=4$ so as to find the value of the constant of integration.

• I tried this, but I still don't know how to get to the answer from here, unfortunately. This is something along the lines of what my book did, but it's still kind of confusing. – Skeleton Bow Nov 20 '16 at 10:57
• What part do you get stuck on? – gfppoy Nov 20 '16 at 10:58
• I can't even get past the first bit... so I know that, as you said, $\int v \,\mathrm{d}t=a$, but that is with respect to $t$. The $s$ in the question throws me off – Skeleton Bow Nov 20 '16 at 11:00
• You can think of $s$ as a function of $t$; that is, $s=s(t)$ – gfppoy Nov 20 '16 at 11:04
• Okay, then $v=\int a \,\mathrm{d}t = \int s^3 \,\mathrm{d}t$, which doesn't help, does it? – Skeleton Bow Nov 20 '16 at 11:07