I'm studying acceleration as a function of velocity and displacement as part of my mechanics module, but I'm not sure how to integrate this acceleration expression.
If you had a situation were $a = \frac{1}{s + 2}$ and you wanted to find $s$ when $v$ is equal to a certain value, you would have to integrate the above equation like $\int v dv = \int \frac{1}{s + 2} ds$, or as far as I understand anyway. But if you bring $s + 2$ to negative index form you get $s^{-1} + 2^{-1}$, and if you were to integrate that expression you would get powers of $0$ and it would not work. Obviously I'm doing it wrong, but that is my thought process.
$\endgroup$ 12 Answers
$\begingroup$$$\int x^{-1}=\log(x)+C$$ Also remember that $$\frac{1}{s+2} \not= s^{-1}+2^{-1}$$
Let me know if that didn't answer your question.
$\endgroup$ 7 $\begingroup$I think you should try to gain a deeper understanding of what's going on here.
First of all, what you wrote: $\displaystyle \int vdv = \int \frac{1}{s+2}ds$ is, in fact, correct (some others seem confused by this).
To see why, consider $\displaystyle I = \int a ds$. Put $\displaystyle a = \frac{dv}{dt}$, to give $\displaystyle I = \int \frac{dv}{dt} ds$. Rearrange to give: $\displaystyle I = \int \frac{ds}{dt} dv$. Finally, use $\displaystyle v = \frac{ds}{dt}$ to get $\displaystyle I = \int vdv$. Hence $\displaystyle \int a ds = \int v dv$, and this can be used in the derivation of the kinetic energy formula, for instance.
Next, you seem to have some confusion about how to compute $\displaystyle \int \frac{1}{s+2} ds$. First off, note that in general, $\displaystyle \frac{1}{a + b} \neq a^{-1} + b^{-1}$. So you can't split up the denominator that way.
Instead, try the substitution $\displaystyle x = s+2$. This gives $\displaystyle ds = dx$ and the integral becomes $\displaystyle \int\frac{1}{x} dx = \ln x + c = \ln (s+2) + c$ ($\displaystyle c$ is an arbitrary constant that you can avoid by employing definite integrals with the appropriate bounds, a method often preferred when dealing with physical problems).
$\endgroup$ 2
