# Kinematics with acceleration as a function of velocity

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.

• 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 Sep 22, 2015 at 6:03
• 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. Sep 22, 2015 at 6:31
• @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
– user36790
Sep 22, 2015 at 7:03

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!

• +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:)
– user36790
Sep 22, 2015 at 7:03
• 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! 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 $$v(t)=v(0)e^{-0.5t}.$$

• 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. Sep 22, 2015 at 6:17