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...

Re: Acceleration varying with speed problem

Quote:

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:

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

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

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

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

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 $\displaystyle a= \frac{dv}{dt}= -1.2503v+1.4533$ is to write it as $\displaystyle \frac{dv}{1.4533- 1.2503v}= dt$

and integerate both sides. On the left you will want to make the substitution [tex]u= 1.4533- 1.2503v[tex].

Re: Acceleration varying with speed problem

Thanks, that has helped a lot

Re: Acceleration varying with speed problem

Quote:

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