Math Help - Acceleration varying with speed problem

1. 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...

2. Re: Acceleration varying with speed problem

Originally Posted by jmonk
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

3. 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].

4. Re: Acceleration varying with speed problem

Thanks, that has helped a lot

5. Re: Acceleration varying with speed problem

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

-Dan