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Non uniform acceleration

  1. Jul 30, 2015 #1
    Capture.PNG 1. The problem statement, all variables and given/known data
    A particle is moving along a straight line such that when it is at the origin it has a velocity of 5m/s. If it begins to decelerate at a rate of a=(-2.5v^(1/2)) m/s^2, where v is in m/s, determine the distance it travels before it stops

    2. Relevant equations


    3. The attempt at a solution
    I integrated the above expression twice with respect to time, to get an expression of s=(2(v^(5/3) t^2))/3+C1+C2. Where C1 ns C2 are constants. I've ended up with more variables than I have the know how to handle. I don't know what to do with the initial velocity. I thought of integrating with respect to v but it didn't make sense. Please help.
     
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  3. Jul 30, 2015 #2

    SteamKing

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    You should show your work.

    How did you wind up with two constants of integration without one being multiplied by something else?

    You know that when x = 0, v = 5 m/s. Did you apply these initial conditions to determine what the constants of integration must be?
     
  4. Jul 30, 2015 #3

    RUber

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    Your integral is incorrect.
    It stops when v = 0, right?
    Graphically, I get an answer of less than 3m.

    I would attack it like this:
    since acceleration is the derivative of velocity and ##a = -2.5 v^{1/2}##
    assume that velocity is of the form ##v(t) = c(t+a)^2##
    Then ##v' = 2c(t+a) = -2.5 \sqrt{ c (t+a)^2 } ##.
    Using initial information, you should be able to find a suitable solution for c and a, then solve your velocity equation for the time when v = 0.
    Last, integrate velocity from 0 to t_final to get total change in position.
     
  5. Jul 30, 2015 #4
    Thanks guys. will try it and see what I end up with.
     
  6. Jul 30, 2015 #5
    What's wrong with solving the differential equation

    $$\frac{dv}{dt}=-2.5v^{\frac{1}{2}}$$

    for v as a function of t, subject to the initial condition?

    Chet
     
  7. Jul 30, 2015 #6

    rcgldr

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    Why twice, you only need to integrate dv once to end up with v, and you're solving for v = 0.

    I missed that the question was about distance traveled, not time, so you will need to integrate twice.

    So that would be

    $$\frac{dv}{v^{\frac{1}{2}}} = -2.5 \ dt$$
     
    Last edited: Jul 30, 2015
  8. Jul 30, 2015 #7
    I would have had the -2.5 associated with the dt.
     
  9. Jul 30, 2015 #8
    I thought I needed to s end up with an expression for position / distance
     
  10. Jul 30, 2015 #9

    rcgldr

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    You're correct. I lost a update to my prior post; and it's fixed now. So yes, you can start with dv/(v^(1/2) = -2.5 dt, and noting that v = ds/dt, this results in some function f(ds/dt) + 5 = t. (5 being the constant of integration, in this case the initial velocity). Again you can separate the equation so ds and the s terms are on one side and dt and the t terms are on the other and integrate again.
     
    Last edited: Jul 31, 2015
  11. Jul 31, 2015 #10

    ehild

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    You can use the distance travelled as independent variable instead of t. dv/dt=(dv/ds)(dv/dt)(ds/dt)=0.5 d(v2)/ds.You need to integrate with respect to s.
     
    Last edited: Jul 31, 2015
  12. Jul 31, 2015 #11

    rcgldr

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    If using chain rule, shouldn't that be dv/dt = (dv/ds)(ds/dt) = v dv/ds ?
     
  13. Jul 31, 2015 #12

    ehild

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    Yes, it was a typo, but the last formula is true. a=dv/dt= 0.5d(v2)/ds.
     
  14. Jul 31, 2015 #13

    rcgldr

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    OK, but is it any simpler than just continuing with the chain rule result:

    $$v \ \frac{dv}{ds} = -2.5 v^{\frac{1}{2}}$$

    $$v^{\frac{1}{2}} \ dv = -2.5 \ ds $$

    Which only requires a single integration, while the previously mentioned equation

    $$\frac{dv}{v^{\frac{1}{2}}} = -2.5 \ dt$$

    requires two integrations.

    As for the constant(s) of integration, the problem states that v = 5 m/s at t = 0 and s = 0, and that the initial position is s = 0, which allows you to solve for the constant(s) of integration with either approach. I got the same result for s using both methods.
     
    Last edited: Jul 31, 2015
  15. Aug 25, 2015 #14
    Ok it took long for me to get back to this thread but I did get it eventually.
     
  16. Aug 25, 2015 #15

    rcgldr

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    If anyone is curious the distance is

    $$\frac{4}{3} \ \sqrt{5} \approx 2.981424 $$
     
    Last edited: Aug 25, 2015
  17. Aug 25, 2015 #16
    Yep.

    Thanks everyone for your help.
     
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