A body is moving through a liquid: $$a = -\frac{v^2}{200}-32\ m/s^2$$ $$v_0 = 40\ m/s$$ Find an expression for the time taken, $T$, for the body to slow down to $V\ m/s$
This is from my math class, where no prior physics knowledge is required. However, I do not know a way to do this without using the physics.
I used the SUVAT equation $v = u + at$:
$$V = 40 + (\frac{v^2}{200}-32)\cdot T$$
Isolating $T$,
$$T = -200\Big(\frac{v-40}{v^2+80^2}\Big)$$
This is close to the answer given, but not quite:
$$T = 200\int_{V}^{40}\frac{1}{v^2+80^2}dv$$
My question is, how can I get the above result using just mathematics?
$v=v_0+at$ is only applicable if $a$ is a constant, which is not the case. You must use the definition of acceleration for this case: $$\frac {dv}{dt}=a$$ $$dt=\frac{dv}a$$ $$\int_0^T dt=\int_{40}^V\frac {dv}{\dfrac{-v^2}{200}-32}$$ Time $0$ (initial time) corresponds to the initial velocity of $40$ while $T$ is the moment when the velocity reaches $V$. $$T-0=\int_{40}^V\frac {dv}{\dfrac{-v^2-6400}{200}}$$ $$T=-200\int_{40}^V\frac {dv}{v^2+6400}$$ $$T=200\int_{V}^{40}\frac {dv}{v^2+80^2}$$