# Velocity and position from forces with mixed dependences

1. Sep 10, 2015

### QuantumCurt

1. The problem statement, all variables and given/known data

This is for a junior level classical mechanics course. We're currently working with Newton's Laws.

A particle of mass m moves under the influence of a force $\vec{F}(t,v)=-ktv \hat{v}$, where k is a positive constant. At time t=0, $v=v_0 \hat{x}$ at the origin, where $v_0$ is a positive constant.

a. Calculate the velocity of the particle $\vec{v}(t)$ for times t>0.

b. How much time does it take for the particle to stop?

c. What is the maximum x-position reached by the particle? Hint: you will need to look up $\int_{0}^{\infty} (something) dt$ on Wolfram Alpha, which is a tool that you should take the time to learn how to use!

2. Relevant equations

$$\vec{F}=m \frac{dv}{dt}$$

3. The attempt at a solution

For part a I figured that since F=m dv/dt, I could simply rewrite this as
$$-ktv \hat{v}=m \frac{dv}{dt}$$
And then solve for v

$$\frac{dv}{dt}=\frac{-ktv \hat{v}}{m}$$

$$\int{dv}=\int{\frac{-ktv \hat{v}}{m}}dt$$

$$\int{dv}=\frac{-kv \hat{v}}{m} \int{t}~dt~\hat{v}$$

So,

$$v=-\frac{kv~\hat{v}}{2m}t^2+v_0 \hat{x}$$

Where $v_0 \hat{x}$ is the constant of integration evaluated at time t=0.

Then for part b, I used this equation and set v=0, and arrived at

$$t=\sqrt{\frac{2v_0m}{kv}}$$

This seemed okay to me, but for part c it specifies that we're going to want to look up an integral. Obviously I can integrate velocity to obtain position, but the simplicity of the integral, along with the recommendation to use Wolfram Alpha, makes me think I'm doing something wrong. If anyone has any input, it would be greatly appreciated.

2. Sep 10, 2015

### fzero

First off, you should be clear that you have vectors on both sides of the equation, though maybe you just had problems making this clear in the LaTeX codes. Second, and more important for this problem, here $v=v(t)$, so we can't just pull it out of the integral. Instead you should have separated variables before the integration by writing

$$\frac{1}{v} \frac{dv}{dt}=\frac{-kt}{m}.$$

You can then integrate both sides of this. For part (c), it looks like you will get an incomplete Gaussian integral, which is why they suggest using wolframalpha to get a numerical result for it.

3. Sep 10, 2015

### QuantumCurt

I actually realized my mistake shortly after posting this. It was indeed what you're referring to. I managed to get it figured out though. Thank you!