A bullet shot underwater given a velocity function

[getty102]

Homework Statement

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

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

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

Homework Equations

x(t)=d(v_x)/dt, v_x=d(a_x)/dt

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.

 

[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?

 

[getty102]

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

x=∫v_xdt

integrating v_oxe^-bt gave me x(t)= x₀+[(-v_ox/b)e^-bt]

 

[Winzer]

Right idea, let's 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)?

 

[getty102]

Thank you, that helped a lot!

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?