# A bullet shot underwater given a velocity function

1. Jan 12, 2012

### getty102

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

A bullet shot underwater will experience "viscous drag" and follow this velocity equation.
vx(t)=voxe-bt

What is the maximum position of the bullet (if you were to wait a long time, what will its position be)?

Given: vox=247m/s
b=0.53 s-1
x0=47.9m

2. Relevant equations

x(t)=d(vx)/dt, vx=d(ax)/dt

3. The attempt at a solution

I'm not sure where to start. The only thing I came up with is that when the v=0, that would be the max position although that isn't giving me the correct answer.

Last edited: Jan 12, 2012
2. Jan 12, 2012

### Winzer

Do you know how to go from a velocity equation to a position equation?
That is: $$v \rightarrow x$$ ?
Hint: Use the fact that: $$v = \frac{dx}{dt}$$
How do you get x?

3. Jan 12, 2012

### getty102

In order to go from v→x, you need to integrate

x=∫vxdt

integrating voxe-bt gave me x(t)= x0+[(-vox/b)e-bt]

4. Jan 12, 2012

### Winzer

Right idea, lets step through it.
$$x(t) = \int v_o e^{-bt} dt = C - \frac{v_o}{b} e^{-bt}$$
Got that right?
Now you need to use your initial conditions to solve for C right?
Use
$$x(0) = x_o$$
What is C? Now what is $$x(t)$$?

5. Jan 13, 2012

### getty102

Thank you, that helped alot!

I have another question relating to these types of problems

When you solve for C in an indefinite integral, will that always give you a max or min of the function?