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Challenging! Find maximum deceleration and speed (not max)

  1. Dec 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Given $$\frac{d^2s}{dt^2}=-Ae^{s/B}\biggl(\frac{ds}{dt}\biggr)^2$$ Show that maximum deceleration (where [itex] A,B [/itex] are constants, [itex]v=v_0[/itex], and [itex]s=-\infty[/itex]) is [itex]\frac{v_0^2}{2eB}[/itex].

    Use the substitution [itex]v=\frac{ds}{dt}[/itex].

    2. Relevant equations
    See above.


    3. The attempt at a solution
    Using the substitution, I get [itex]\frac{d^2s}{dt^2}=-Ae^{s/B}(v)^2[/itex]. Furthermore, using [itex]v=v_0[/itex] and [itex]s=-\infty[/itex] and using the original function, I get [itex]v=e^{-ABe^{s/B}}v_0[/itex]. Combining these two equations, I get
    $$\frac{d^2s}{dt^2}=-Ae^{(s/B)-(2ABe^{s/B})}v_0^2$$.
    With [itex]s=-\infty[/itex], I end up with:
    $$\frac{-Av_0^2}{e^{2AB}}$$, which isn't what's given in the problem. Not sure what I'm doing wrong.
     
    Last edited: Dec 18, 2013
  2. jcsd
  3. Dec 18, 2013 #2

    SteamKing

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    Your Latex came out garbled. Try hitting the Preview button before you post to make sure everything formats properly.
     
  4. Dec 18, 2013 #3
    Ok, I've fixed it now. Do you have any suggestions for my physics problem?
     
  5. Dec 19, 2013 #4
    Why do you think acceleration is maximum at ##s=-\infty##? I don't see how you can find a maximum only by inspection. You will have to differentiate the acceleration function and set the derivative to zero.
     
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