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A bullet shot underwater given a velocity function

  1. Jan 12, 2012 #1
    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. jcsd
  3. Jan 12, 2012 #2
    Do you know how to go from a velocity equation to a position equation?
    That is: [tex] v \rightarrow x [/tex] ?
    Hint: Use the fact that: [tex] v = \frac{dx}{dt} [/tex]
    How do you get x?
     
  4. Jan 12, 2012 #3
    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]
     
  5. Jan 12, 2012 #4
    Right idea, lets step through it.
    [tex] x(t) = \int v_o e^{-bt} dt = C - \frac{v_o}{b} e^{-bt} [/tex]
    Got that right?
    Now you need to use your initial conditions to solve for C right?
    Use
    [tex] x(0) = x_o [/tex]
    What is C? Now what is [tex] x(t) [/tex]?
     
  6. Jan 13, 2012 #5
    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?
     
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