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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?

StopReadingThisUsername
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    Obviously, velocity can't linearly depend on time. Because the acceleration depends on velocity. You need to know that $a=dv/dt$, that's all. Then solve the ODE – Yuriy S Jul 27 '16 at 09:15
  • Besides, $v$ is a variable (or a function) and $V$ is a constant. You can't just set $V=v$ as you did. Which level are you studying math at? – Yuriy S Jul 27 '16 at 09:21

1 Answers1

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$v=v_0+at$ 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 $V$. $$T-0=\int_{40}^V\frac {dv}{\dfrac{-v^2-6400}{200}}$$ $$T=-200\int_{40}^V\frac {dv}{v^2+6400}$$ $$T=200\int_{V}^{40}\frac {dv}{v^2+80^2}$$

entrelac
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