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Homework Help: Problem with vectors

  1. Feb 8, 2007 #1
    The velocity of a particle moving in the xy plane is given by = (5.6 t - 4.5 t2) + 8.2, with in meters per second and t (> 0) in seconds. (a) What is the acceleration when t = 3.3 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) does the speed equal 10 m/s?

    I got the first 2 but when I went to find c I can't seem to get it.

    I took

    10m/s = sqrt((5.6t - 4.5t^2)^2 + (8.2)^2
    10^2 = (sqrt((5.6t - 4.5t^2)^2 + (8.2)^2)^2
    100 = (5.6t - 4.5t^2)^2 + 67.24)
    32.76 = (5.6t - 4.5t^2)^2
    sqrt(32.76) = sqrt((5.6t - 4.5t^2)^2)
    5.72 = 5.6t - 4.5t^2
    0 = -4.5t^2 + 5.6t - 5.72
    then I used quadratic to find out the two values...
    but I get a negative in the sqrt on the quadratic and I can't do that... any suggestions?
     
  2. jcsd
  3. Feb 8, 2007 #2

    D H

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    You implicitly assumed [itex]t<=56/45[/itex]
    when you wrote
    [tex]\sqrt{(5.6t-4.5t^2)^2} = 5.6t-4.5t^2[/tex]
    What happens if [itex]t>56/45[/itex]?
     
  4. Feb 8, 2007 #3
    so how do you figure it out? I'm not sure how you would go about doing it.

    Could you take

    5.6t-4.5t^2
    t(5.6-4.5t)
    t = 5.6/4.5 ?

    if so what happens with the 5.72 on the other side of the equation?
     
  5. Feb 8, 2007 #4

    D H

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    My hint wasn't good enough.

    It is not valid to write [itex]\sqrt{(5.6t-4.5t^2)^2} = 5.6t-4.5t^2[/itex] when [itex]t>56/45[/itex] because [itex]5.6t-4.5t^2[/itex] is negative.

    For example, try [itex]t=2[/itex]. The velocity vector at [itex]t=2[/itex] is [itex][ 5.6*2 - 4.5*2^2, 8.2 ] = [ -6.8, 8.2 ] [/itex]. Note that the [itex]x[/itex] component of the velocity is negative. [itex]\sqrt{(-6.8)^2} = 6.8,\text{\ not\ } -6.8[/itex].
     
    Last edited: Feb 8, 2007
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