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Deriving position from a car's velocity equation.

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A driver sees the red light, and instead of braking the car, changes the gear to neutral. Suppose that velocity obeys this equation:
    [itex]v = v_0 * e^{-bt}[/itex]
    Where [itex]v_o = 20 m/s[/itex] and [itex]b = 0.150 s^{-1}[/itex]

    a)Acceleration when time equals 0;(Answer:[itex]-3.0 m/s^2[/itex])
    b)Displacement when [itex]t=5.0 s[/itex];(Answer:[itex]70.6m[/itex])
    c)Velocity when displacement is [itex]50m[/itex].(Answer: [itex]12.5m/s[/itex])

    2. Relevant equations
    It's possible to solve using only the one supplied by the problem.


    3. The attempt at a solution
    I can solve letter a, but I fail at letter b and c, so I am going to post only the attempt at those:

    b) Displacement is the integral of velocity in t, so, by integrating the velocity I get:
    [itex]x(t)=\frac{v_o}{-b}*e^{-bt}[/itex]

    and by substituting [itex]b = -0.15[/itex] and [itex]Vo = 20[/itex], I get: [itex]-62.98m[/itex]!

    c) By using the displacement equation found in letter b), I equal the displacement to [itex]50m[/itex], and try to solve for t. I find [itex]t = 6.53s[/itex].

    Then I substitute [itex]t=6.53s[/itex] in the velocity equation, and find that [itex]v = 7.50m/s[/itex]!
     
  2. jcsd
  3. Dec 9, 2012 #2

    gneill

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    Staff: Mentor

    You have a definite integral here, so there are two end points to consider. The time goes from 0s to 5s, so...
    That time value looks a bit high. Can you show your work?
     
  4. Dec 9, 2012 #3
    About letter b), God that is brilliant! Thanks a lot! I feel stupid, the initial position isn't 0, as I was assuming.

    About letter c)
    [itex]50 = \frac{20}{-0.15}*e^{-0.15t}[/itex]

    [itex]-0.375 = e^{-0.15t}[/itex]

    [itex]\ln{(-0.375)} = ln{(e)}*-0.15t[/itex]
    At this point I had a doubt, I don't know how to solve Ln(-0.375), so I sort of pretended it was Ln(0.375), I probably shouldn't do this, but what's the alternative?moving on:

    [itex]\frac{-0.981}{-.15}=t[/itex]

    [itex]t=6.5s[/itex]

    Thanks in advance!
     
  5. Dec 9, 2012 #4

    gneill

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    Staff: Mentor

    Okay, regarding (c), the same hint as for part (b) applies. For the displacement you are performing an integration of the velocity function from 0 to some (as yet) undetermined time t. Two end points. So the equation that you will set equal to 50m in order to solve for the time will look a bit different. This should clear up your negative ln() issue, too.
     
  6. Dec 9, 2012 #5
    Ohh! I GOT IT! I had a conceptual problem. I always thought that:
    [itex]x(t) = \int v(t) dt[/itex]
    and that the constant generated in the indefinite integral would be Xo. However, this is not quite right, in fact:
    [itex]x(t) =x_o + \int_{0}^{t} v(t) dt[/itex]

    This never generated any problems because most equations I studied didn't use time in an exponential, but now I finally understood my mistake.

    Using the correct assumption now, I found t = 3.13s and velocity 12.5 m/s (both correct)

    Thanks a lot gneill!
     
    Last edited: Dec 9, 2012
  7. Dec 9, 2012 #6

    gneill

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    Staff: Mentor

    You're thinking along the right lines, but you haven't quite nailed it :smile:

    When you integrate a velocity function with respect to time to yield a position function with respect to time, the constant associated with the integration (for an indefinite integration!) turns out to be the factor that's due to the initial velocity multiplied by time (vot) plus any initial offset. By performing a definite integral though, you don't have to worry about the initial velocity because it's taken into account by the initial limit of the integral; only the initial offset is needs to be accounted for. In this problem, the initial offset is zero, so it doesn't appear.

    Anyways, well done on obtaining the correct results. Good luck in your studies.
     
  8. Dec 9, 2012 #7
    Aha! I tried making sense of what you just said, and decided to try and do it by an indefinite integral to see if I understood it now, look at what I came up with:
    [tex]
    x(t)=\int v_0e^{-bt} dt \\
    x(t) = \frac{V_0}{-b}e^{-bt} + C\\
    x(t)-x(0) = (\frac{V_0}{-b}e^{-bt} + C) - (\frac{V_0}{-b}e^{0} + C))\\
    \bigtriangleup X = (\frac{-V_0}{b}e^{-bt} + \frac{V_0}{b})
    [/tex]

    Which, in fact gave me the same results as before, is this correct?

    Thanks again!
     
  9. Dec 9, 2012 #8

    gneill

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    Staff: Mentor

    Yes, that achieves the same result as evaluating the definite integral.
     
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