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Math Help - Acceleration varying with speed problem

  1. #1
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    Acceleration varying with speed problem

    Hi,

    I'm attempting to solve a problem where the torque of a motor varies with RPM linearly. I've done most of the groundwork, and come up with this equation for how acceleration would vary with the speed of the vehicle

    a = -1.2503v + 1.4533

    From this and F = ma, I think I should come up with the following differential equation:

    m dv/dt = -1.2503v + 1.4533

    From there, I'm a bit stumped...
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    Re: Acceleration varying with speed problem

    Quote Originally Posted by jmonk View Post
    Hi,

    I'm attempting to solve a problem where the torque of a motor varies with RPM linearly. I've done most of the groundwork, and come up with this equation for how acceleration would vary with the speed of the vehicle

    a = -1.2503v + 1.4533

    From this and F = ma, I think I should come up with the following differential equation:

    m dv/dt = -1.2503v + 1.4533

    From there, I'm a bit stumped...
    Let's get this into a better format:
    m \frac{dv}{dt} = -1.2503v + 1.4533

    m \frac{dv}{dt} + 1.2503v - 1.4533 = 0

    \frac{dv}{dt} + \left ( \frac{1.2503}{m} \right ) v - \frac{1.4533}{m} = 0

    This is the "standard" form for a 1st order linear differential equation. There are a number of ways to proceed from here. Can you finish it?

    -Dan
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    Re: Acceleration varying with speed problem

    You are given a, the acceleration, not the force, so there should be no "m" in the equation.

    The simplest way to solve a= \frac{dv}{dt}= -1.2503v+1.4533 is to write it as \frac{dv}{1.4533- 1.2503v}= dt
    and integerate both sides. On the left you will want to make the substitution [tex]u= 1.4533- 1.2503v[tex].
    Thanks from topsquark
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    Re: Acceleration varying with speed problem

    Thanks, that has helped a lot
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    Re: Acceleration varying with speed problem

    Quote Originally Posted by HallsofIvy View Post
    You are given a, the acceleration, not the force, so there should be no "m" in the equation.
    Thanks for the catch!

    -Dan
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