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-$$\frac{Cm*e^{-mt/k}}{k}$$

 

[rwisz]

I understand your confusion about the negative sign in the position function. However, this negative sign is not incorrect and is actually necessary to accurately represent the motion of the object.

Let's take a closer look at the equation for position, x(t)= -5m/k e^(-kt/m). The negative sign in front of the equation represents the direction of motion. In this case, since the object is moving along a horizontal surface, the positive direction would be to the right and the negative direction would be to the left. This means that when the object is moving to the right, the position would be positive, but when it is moving to the left, the position would be negative. The negative sign in the equation accounts for this change in direction.

Additionally, the negative sign also represents the initial position of the object. In this case, the initial position is assumed to be at x=0, and as the object moves to the right, the position becomes positive, but as it moves to the left, the position becomes negative.

To better understand this concept, imagine a car moving along a straight road. If we define the starting point as x=0, and the car moves to the right, the position would be positive. But if the car turns around and starts moving in the opposite direction, the position would become negative. This is why the negative sign is necessary in the position function.

In conclusion, the negative sign in the position function is not incorrect, but it represents the direction of motion and the initial position of the object. Without it, the equation would not accurately represent the motion of the object.