Differential Equation to Position

[schaefera]

Say you have know the fact that, as something moves along a horizontal surface, a retarding force F= -kv acts on it. Obviously you can use a differential equation to solve for velocity as a function of time; let's assume you get something like v(t)=5 e^(-kt/m). I know that this works out right.

But now, suppose you want to integrate velocity as a function of time to get position. You'd get something like x(t)= -5m/k e^(-kt/m). Why is there a negative sign in the position function? If the object is moving toward the positive direction, shouldn't the sign of position be positive, as it approaches some point at which is stops moving? What's going on with the negative from integration?

 

[zush]

The problem with your differential equation is that you're not including the initial velocity in your equation. If you solve with that, you'll notice that your result for position is not actually the position, but the "position difference" or how much the position varies from what would happen in the event of no retarding force.

 

[schaefera]

Do you mean I'm not including initial velocity in the v(t) or the x(t) equation? I was assuming 5 would be like the initial velocity, because once you have the form v(t)=Ce^(-kt/m), and letting t=0 you get v(0)=C=5.

But I get what you're saying about the 'position difference.'

How do you go from a differential equation like -kv=m(dv/dt) to position (or is it possible, even)?

 

[zush]

What you are saying about v(0)=C is correct but you are missing something. Whenever you integrate an equation there is always a constant which is added onto that equation. So when you solve the first differential equation, the one for velocity, the constant added on is the initial velocity. When you plug that into your second equation to solve for position you get: x(t)=(-Cm/k)*e^(-mt/k)+C*t= v(0)*t-\frac{Cm*e^{-mt/k}}{k}