I know, I know. On the can, this problem seems simple. Just take $\int_a^bf(x)\mathrm{d}x$ and write is as $\int f(x)\mathrm{d}x$. However, when I tried to do that on an Engineering Dynamics problem, it didn't work very well.
We have a motorcycle going around a path with an acceleration of $-0.001s\space \mathrm{m/s}$. The distance of the path happens to be $150\pi$. The initial speed is 25. Since we know $a\mathrm{d}s=v\mathrm{d}v$, That readily lends itself to this integral equation:
$$-\int_0^{150\pi}0.001s\mathrm{d}s=\int_{25}^vv\mathrm{d}s$$ This give the correct answer as provided in the book. However, I also attempted the following combination to get the change in $v$:
$$\int_0^s0.001s\mathrm{d}s=\int_0^vv\mathrm{d}s$$which solves to $v=\sqrt{0.001}\times s$. This renders a likely solution of a change of $14\space\mathrm{m/s}$ in $v$ -the actual change is closer to 5. This is obviously wrong.
I know why the first integral equation does work, but would someone be so kind as to tell me why the second one fails?
What is wrong with assuming $0$ rather than using, say, $v_1$ and $v_2$?
I need to understand my mistake so I don't make it in the future! Thanks!
$$v_1 - v_0 = \int_{v_0}^{v_1} dv = \int_{s_0}^{s_1} \frac{dv}{ds} ds = \int_{s_0}^{s_1} \frac{dv}{dt} \frac{dt}{ds} ds = \int_{s_0}^{s_1} \frac{a}{v} ds$$
Does that answer your question?