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Two Dimension Motion Vectors

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose the positon vector for a particle is given as a function of time by vector r(t)
    =x(t)i + y(t)j with x(t) =at + b and y(t) = ct2 + d, where a=1.00 m/s, b=1.00m,
    c=.125 m/s2, and d=1.00m. (a) Calculate the average velocity from t=2.00s to t=4.00s.
    (b) Determine the velocity and the speed at t=2.00s

    2. Relevant equations
    (a) plug in the values for a, b, c, and d to
    x(t) = at + b and y(t) = ct2 + d
    Vavg = delta r / delta t

    3. The attempt at a solution[/c]
    (a) Ri =
    x(t) = (1.00m/s) 2.00s + 1.00m = 3.00 m/s
    y(t) = (.125m/s2) 2.00s2 + 1.00m = 2.41 m/s
    x(t) = (1.00m/s) 4.00s + 1.00m = 5m/s
    y(t) = (.125m/s) 4.00s2 + 1.00m = 6.66m/s

    Rf -Ri / Tf - Ti

    Rf = 6.66 - 2.41 = 4.25
    Ri = 5 - 3 = 2
    t = 2
    4.25 - 2 /2 = 1.13 Vavg from t = 2 to t=4

    (b) From equation at 2 s
    velocity is
    x(t) = 2.41
    y(t) = 3
    velocity is 5.41 m/s at t=2

    Speed is magnitute of vector square root of (3)2 + (2.41)2 = square root of 14.81
    Speed = 3.85 m/s

    Am I right or close?
    Thanks for the help

  2. jcsd
  3. Sep 11, 2008 #2


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    Homework Helper

    Without checking the actual math, it looks like you have grasped the concepts.

  4. Sep 12, 2008 #3


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    This equation means

    y(t) = (.125m/s^2) (2.00s)^2 + 1.00m
    right? I don't believe the answer is 2.41. It looks like the same things happens for y(t) for the Rf value.
    These last two equations do not match your formula. The value 2 is the change in the x component of the displacement, and the 4.25 (though that number does not look right to me from above) is the change in the y component of the displacement.

    That is (using the value of 4.25 for the moment),

    \Delta \vec R = \vec R_f - \vec R_i = 2 \hat i + 4.25 \hat j

    which is what you would use in your average velocity equation:

    \vec v_{\rm ave} = \frac{\Delta \vec R}{\Delta t}

    The values of 3 and 2.41 were the positions you found at t=2s (but they were the other way around; you had x=3 and y=2.41). But you need to find the components of the velocity at t=2 seconds.

    Based on the x(t) and y(t) equation they give you, what are the equations for [tex]v_x(t)[/itex] and [itex]v_y(t)[/itex]?

    (Once you have the velocity components, you cannot just add the numerical values together to get the velocity, since they are not in the same direction.)

    Once you find the right components of the velocity, this is the way to find the speed.

  5. Sep 12, 2008 #4


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    My apologies. I looked briefly at the method and thought you were substituting values and calculating correctly using the RSS to get your resultant magnitude. I should have actually checked your math more closely now I see.

    Thanks to alphysicist for catching the errors here.
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