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Homework Help: Vectors of position, velocity, and acceleration.

  1. Aug 18, 2007 #1
    A particle initially located at the origin has an acceleration of a = 3.00j m/s^2 and an initial velocity of v[initial] = 500i m/s. Find (a) the vector position and velocity at any time t and (b) the coordinates and speed of the particle at t = 2.00s.

    (a) I looked at the examples given in the chapter, and to find the velocity at any time seemed to be the final velocity, which was found through the equation of v[f] = v + at. That means that the final velocity is equal to it's components added, and the same goes for the initial velocity and acceleration. So:

    ax = 0 ay = 3.00 vx = 500 vy = 0

    Working each component separately, I got 500i + (3.00t)j.

    For the position, I, again, separately worked each component, using the equation r[f] + r + vt + .5at^2.

    The initial position is (0,0), so my answer is:

    r[f] = (500t)i + (1.5(t^2))j

    (b) To find the coordinates of position, I took my last equation from above, and broke it down into it's components, where x[f] = x + vx + .5(ax)(t^2) and y[f] = y + vy + .5(ay)(t^2), all while plugging in 2 for time. x = 1000 and y = 3.

    For speed, my previous findings of 500i + (3.00t)j, I got 500i + 6.00j when plugging in 2. That would be the velocity, and taking the absolute value would equal speed. But, I don't think speed is in it's components, and so I drew the hypotenuse of the triangle the vectors formed, and got 500.036 m/s.

    ...for some reason, I'm not sure if that is right. I kind of guess on taking the hypotenuse, but I don't know if I should do that, if I've already plugged into an equation specifically for velocity. What I'm wondering seems stupid, but, well, X/

    Thank you in advance.
  2. jcsd
  3. Aug 18, 2007 #2
    (a) is correct

    for (b), y=3 is incorrect, check your calculations

    the speed of 500.036 m/s is correct. Remember, velocity is a vector, so it must have components. Speed is a scalar, hence doesnt have components. It is the magnitude of velocity
  4. Aug 18, 2007 #3
    ah! I forgot to square the time before multiplying it by .5 and the vertical component of acceleration.

    Thank you for your help! :D
    Last edited: Aug 18, 2007
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