Say there is a particle moving at $50~\text{m/s}$ and modeled with a function of acceleration such that:

$$a = - 0.5v$$

(this is derived from a force as a function of velocity)

$$F = 50v = 100\cdot a$$

Then say I wanted to know what v was equal to at the time t = 4 seconds.

I integrated both sides and get:

$$30 - 0.5Δv t = 0.5\Delta x$$

is this correct? I now seem to not have enough variables to solve the equation for v at time $t = 5$. What other equation am I missing? Did I get this equation incorrectly? I can not use UAM equations because this particle is not accelerated at a constant rate, so I must derive kinematic equations from this circumstance?

Any help is appreciated. Thanks.

  • $\begingroup$ Sorry I'm not following too well, but when you say you are integrating both sides, what equation are you integrating, and with respect to what variable? Many thanks $\endgroup$
    – Involute
    Sep 22, 2015 at 6:03
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    $\begingroup$ I note someone has voted to close the question on the grounds it's homework-like, and I sympathise with that assessment. I can say only that knowing how to tackle this sort of differential equation is a valuable skill that once learned will serve you well over and over again. That's why I think it's worth an answer even though it does veer dangerously close to homework. $\endgroup$ Sep 22, 2015 at 6:31
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    $\begingroup$ @John Rennie: Hi, sir; I voted to close it; but I wanted to do it because I couldn't understand anything from what he has provided. He integrated what? I don't know. With respect to what ? I don't know. It is really a hotchpotch. But above all, it is a Homework-question which is not asking for any PHYSICS concept (at least this can be comprehended from the body). That's why I voted to close it at the plea of dearth of any query on physics concept. Nevertheless, salute to you that you could conceive something from this junk:p $\endgroup$
    – user36790
    Sep 22, 2015 at 7:03

2 Answers 2


It isn't clear from your question exactly what you are integrating and how, but this is the way to tackle problems like this. You know that:

$$ \frac{dv}{dt} = -kv $$

The way to solve equations like this one is to rearrange it by dividing both sides by $v$ and multiplying both sides by $dt$ to get:

$$ \frac{1}{v}dv = -k\,dt $$

Now we can integrate both sides to get:

$$ \ln v = -kt + C $$

where $C$ is some constant of integration. It's probably clearer if we take the exponential of both sides to get:

$$ v = e^{-kt+C} $$

Mathematicians tend to recoil in horror when we physicists casually treat $df(t)/dt$ as if it were a simple fraction, but it works in physics!

  • $\begingroup$ +1; Yes, $\frac{dy}{dx}$ can be treated as a fraction when you are dealing with approximations, errors & finding areas at a limiting scale & mathematicians do it day in & day out:) $\endgroup$
    – user36790
    Sep 22, 2015 at 7:03
  • $\begingroup$ Ah that makes a lot of sense. Thanks a lot! I am a first year physics student and I appreciate that help, I know the information I gave was probably odd and not that great. Thanks again! $\endgroup$ Sep 22, 2015 at 13:44

Hint: Begin by showing that the general solution of the linear first-order differential equation $$ a=\frac{dv}{dt}=-0.5v $$ with constant coefficients is $$ v(t)=v(0)e^{-0.5t}. $$

  • $\begingroup$ I guess the general theme here is that when your acceleration is only a function of speed (as opposed to both speed and position, or just position), you should expected a first order DE rather than the more common 2nd order DE. $\endgroup$ Sep 22, 2015 at 6:17

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