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If $g$ is the acceleration due to gravity and $\beta$ is a known value, solve for the time $t$ it takes an object to reach a given velocity if the acceleration of the object is given by
$$a(t) = g - \beta v(t)$$

We know that when $a(t) = 0$, $v(t) = \frac{g}{\beta}$, and when $t=0$, $v(t)=0$


A possible solution

We can solve this first-order linear ordinary differential equation with a solution that looks somewhat like

$$v(t) = c_1 e^{-b t}+\frac{g}{\beta}$$

where $c_1 = -\frac{g}{\beta}$

We can solve for $t$ from there.


Is there a (simpler) way to solve this problem without using differential equations?

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    $\begingroup$ Nope, the very statement is a differential equation ($a = \dot v$). So any other method of solving this will just be a method of solving a differential equation. $\endgroup$ – Sebastian Riese May 15 '15 at 21:57
  • $\begingroup$ All methods you may find are all about solving the differential equation. For example you can directly use the method of separation of variable to express the time as a function of $v$. $\endgroup$ – Joelafrite May 15 '15 at 22:14
  • $\begingroup$ Yes, you direct integration. See answer below: $\endgroup$ – ja72 May 15 '15 at 22:22
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In reality you have $a(v) = g - \beta v$ and you want $v(t)$ and $x(t)$. This can be achived with direct integration, instead of coefficient matching and diff. equs.

$$ t = \int\limits_0^v \frac{1}{a(v)}\,\mathrm{d}v= \int\limits_0^v \frac{1}{g-\beta v}\,\mathrm{d}v = -\frac{1}{\beta} \ln \left(1- \frac{\beta v}{g} \right)$$

or

$$ v(t) = \frac{g}{\beta} \left(1 - \mathrm{e}^{-\beta t} \right)$$

Now

$$ x(t) = \int\limits_0^t v(t)\,\mathrm{d}t =\frac{g}{\beta^2} \left( \mathrm{e}^{-\beta t} + \beta t - 1 \right)$$

Alternatively you can use directly $$x = \int \limits_0^v \frac{v}{a(v)}\,\mathrm{d}v$$

For more details see https://physics.stackexchange.com/a/15620/392

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