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Given, $$ mdv/dt = mg - kv $$

Question is:

Find the velocity $v(t)$ that satisfies this initial value problem. Also, by letting $t$ approach positive infinity, determine the terminal velocity of the ball.

I first separated the equation and got $$ dv/dt = g - k/mv $$

I have no clue how to approach this question. Can someone please help?

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  • $\begingroup$ Are we to assume that the variables are $v$ (dependent) and $t$ (independent) and that $m$, $g$, and $k$ are real constants? And is the equation $m\frac{dv}{dt}=mgkv$? If so, why doesn't the $m$ immediately cancel? $\endgroup$ – Rory Daulton Mar 22 '15 at 22:59
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then terminal velocity is the constant velocity $v_\infty$ which is given by $$kv_\infty = m \, g \to v_\infty = \frac{mg} v.$$ we can now write the differential equation as $$\frac{dv }{dt} = \frac k m(v_\infty - v) \to \frac{dv}{v_\infty - v} = \frac k mdt$$ and can ve integrated to yield $$\ln(v_\infty - v) = -\frac k mt + \ln C \to v = v_\infty - Ce^{-kt/m}$$

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  • $\begingroup$ how is terminal velocity is the constant velocity V∞ = g/km ? $\endgroup$ – Steve Kim Mar 23 '15 at 1:53
  • $\begingroup$ @SteveKim, how are you going to terminal velocity? $\endgroup$ – abel Mar 23 '15 at 2:02
  • $\begingroup$ i have no idea ... or is v∞ = g/km the only formula? $\endgroup$ – Steve Kim Mar 23 '15 at 2:19
  • $\begingroup$ @SteveKim, sorry for the typos. hopefully everything is is fixed now. $\endgroup$ – abel Mar 23 '15 at 14:11
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Set $u=g-k/m·v$. Then $$ u'=-\frac km ·v'=-\frac km·u $$ which is the well known equation for exponential growth resp. decay.

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$$v'+\frac kmv=g$$ is a first order ordinary linear equation of constant coefficients.

You can use the integrand factor $e^{\frac kmt}$ to get

$$(ve^{\frac km})'=ge^{\frac kmt},$$ and $$ve^{\frac kmt}=\frac{mg}ke^{\frac kmt}+C.$$

or $$v=\frac{mg}k+Ce^{-\frac kmt}.$$

The integration constant is found from

$$v_0=\frac{mg}k+C.$$

The speed at infinity is more directly obtained by setting $v'=0$ (steady regime), so that

$$v=\frac{mg}k.$$

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  • $\begingroup$ Setting $v'=0$ happens to work for this particular example, but it is not a general approach. Finding the extrema of the velocity is not the same as finding the terminal velocity. $\endgroup$ – Demosthene Mar 23 '15 at 21:18

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