Finding velocity vs. time, when acceleration is dependent on velocity

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To find the velocity function v(t) when acceleration a is dependent on velocity, start with the relationship a = dv/dt. Rearranging gives dt = dv/a(v), allowing integration to express time as a function of velocity. For example, with a(v) = 5v and initial velocity v0 = 10 m/s, the integration leads to t = ln(v)/5 + C. To incorporate the initial condition, substitute t=0 and v=v0 to solve for the constant C, resulting in v(t) = v0e^(5t). This approach is particularly relevant for scenarios involving velocity-dependent acceleration, such as the drag equation.
KingNothing
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Hello all. I have been struggling with this for a bit. I will present it as generally as I can.

Say I have an object with an initial velocity v0. It is only acted upon by an acceleration function, a(v), which is a function of v. How can I find a function for v(t)? Does this require a differential equation?
 
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By definition,

a = \frac{dv}{dt}

Since you know a(v), you can rearrange and integrate:

dt = \frac{dv}{a(v)}

This will give you t as a function of v (if you can do the integral), which you can in principle invert to get v(t).
 
Mute said:
By definition,

a = \frac{dv}{dt}

Since you know a(v), you can rearrange and integrate:

dt = \frac{dv}{a(v)}

This will give you t as a function of v (if you can do the integral), which you can in principle invert to get v(t).

Okay. Thanks for the start. Say for example I have a(v) = 5v, and initial velocity v0 = 10 m/s.

dt = \frac{dv}{a(v)}
t = \int \! \frac{1}{a(v)} \, \mathrm{d}v = \! \frac{ln(v)}{5}
v = e^{5t}

This looks great so far; how do I incorporate an initial value?
 
KingNothing said:
This looks great so far; how do I incorporate an initial value?

You need a constant of integration when doing the integrals.
 
pwsnafu said:
You need a constant of integration when doing the integrals.

t = \int \! \frac{1}{a(v)} \, \mathrm{d}v = \! \frac{ln(v)}{5} + C
v = e^{5t-5C}

I'm still not totally sure how to incorporate it. It doesn't really make dimensional sense to just plug it in for C.
 
KingNothing said:
t = \int \! \frac{1}{a(v)} \, \mathrm{d}v = \! \frac{ln(v)}{5} + C
v = e^{5t-5C}

I'm still not totally sure how to incorporate it. It doesn't really make dimensional sense to just plug it in for C.

You know that at t=0, v=v_0. So, plug these into your formula for v and then solve for C. Or, in this case you just find that v_0 = e^{-5C}, so you can just replace the entire e^{-5C} with v_0 to get v(t) = v_0e^{5t}.
 
Thank you guys so much! I get it now. Just an FYI, I was asking because I'm trying to use the drag equation, which when applied to an object of constant mass, is one such example of a velocity-dependent acceleration!
 
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