ns was wondering exactly how you would version the velocity of a fallout’s object, taking into account waiting resistance. Be afflicted with in psychic I have actually only studied basic calculus, and also have no experience with differential equations.

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Considering a general solution because that mass $m$ and drag $kv^2$, wherein $k=frac12 ho AC_D$. Let"s to speak that once $t=0$, $v=0$.

Using Newton"s 2nd Law: $mg - kv^2 = ma$, hence $a = g - frackm v^2$.

From here, ns tried to incorporate with respect to $t$, which offered $v = gt-fracktmv^2$. This, when solved for v, appears to design the velocity correctly, however I have been led to think that solutions to differential equations that this type will show off $e^x$ in some way. Is my systems correct, or have actually I integrated incorrectly? just how do we go from here to a role for v in regards to t?

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asked Dec 9 "17 at 23:36

Daniel GibsonDaniel Gibson
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reply Dec 10 "17 in ~ 0:01

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You should account because that the reality that acceleration and velocity (as lot as position) might be —and in truth are— varying with time.

So the equation you obtained from Newton"s second law means$$a(t)=g- frac k m ^2.$$

Then, integration in $t$ would certainly imply

$$v(t)-v(t_0)=int_t_0^t a( au) d au=int_t_0^t g- frac k m ^2 d au.$$

Here you have the right to see your mistake: girlfriend omitted the truth that $v$ is not constant in $t$.

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Here is whereby differential equations come right into play. Staying clear of technicalities, we can see the your equation have the right to be rewritten as$$ig(g- frac k m ^2ig)^-1v"(t)=1,$$that is$$fracv"(t)ig(sqrtg-sqrt frac k m v(t)ig)ig(sqrtg+sqrt frac k m v(t)ig)=1.$$Then, if we use indefinite integration in $t$$$intfracv"(t)ig(sqrtg-sqrt frac k m v(t)ig)ig(sqrtg+sqrt frac k m v(t)ig)dt=int dt,$$which offers an equation the the form$$A lnleft(sqrtg-sqrt frac k m v(t) ight)+Blnleft(sqrtg-sqrt frac k m v(t) ight)=t+C.$$

Constants $A$ and also $B$ are the only numbers such that $$fracAsqrtg-sqrt frac k m v(t)+fracBsqrtg+sqrt frac k m v(t)=frac1ig(sqrtg-sqrt frac k m v(t)ig)ig(sqrtg+sqrt frac k m v(t)ig),$$

while consistent $C$ have the right to be any kind of real number in a general setting, yet has a unique value (in terms of $v_0$) if we impose the condition $v(t_0)=v_0$. Finding every these constants and also solving for $v(t)$ gets you the standard formula for this kind of trajectory, which will indeed incorporate exponential functions.