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Homework Help: Homework, Particle Question

  1. Nov 16, 2006 #1
    A particle passes through the origin with an initial speed of 5.6 m/s in the positive y direction. The particle accelerates in the negative x direction at 5.2 m/s2.
    (a) What are its x and y positions after 4.0 s?
    m (x position)
    m (y position)
    (b) What are vx and vy at this time?
    m/s (vx)
    m/s (vy)



    This question seems pretty hard to me, I have tried about 4 times to get the answer to no avail, heres what I have done.

    To try and find out where its x and y positions were after 4.0 seconds, I plugged it into the Position Versus Time Equation, xf=xi+vixt+.5axt^2

    Since I have the givens

    time = 4.0
    Vi=5.6
    a=-5.2^2

    I plug it into the equation.

    xf=xi+vixt+.5axt^2

    xf=0+5.6(4)+.5(-5.2)^2(4.0)^2

    xf=238.72

    however, this answer for part A. was not correct.

    Can anyone tell me what I am doing wrong here?
     
  2. jcsd
  3. Nov 16, 2006 #2
    You have to treat each direction separately.
    you will have two equations
    eq 1: xf=xi+vixt+.5axt^2
    eq 2: yf=yi+viyt+.5ayt^2
    think about what the question says and in which direction each velocity and acceleration is in.
     
  4. Nov 16, 2006 #3
    Aaahh.. I see, so if it is going in the x direction at -5.2m/s^2 then after 4 seconds, it would be xf=0+0+.5(-5.2)^2(4)^2?

    and if it is going in the y direction

    yf=0+5.6(4)+.5(?)(4)^2

    So, how do I figure out the y if I am not given the acceleration?
     
  5. Nov 16, 2006 #4
    If you're not given an acceleration, I think you can assume that there is none and the particle is moving at a constant velocity in the y direction.
     
  6. Nov 16, 2006 #5
    I got the final position of the y correct, but for x I am still lost.

    if X is accelerating at -5.2m/s^2, where will it be in 4 seconds?

    I plugged it into the position versus time equation

    xf=0+0+.5(-5.2)^2(4)^2
    xf=216.32

    The above answer was false, I am completely lost on this, thanks for your help on solving the y position though conquer!
     
  7. Nov 16, 2006 #6
    xf=xi +vixt +.5a(t^2)
    compare that to your equation
    "xf=0+0+.5(-5.2)^2(4)^2"
    you have something I don't...
     
  8. Nov 16, 2006 #7
    what am I missing? I still dont see what I have that you dont
     
  9. Nov 16, 2006 #8
    heh, take a look at the acceleration...after everythign cancels, you have
    xf=.5(a^2)(t^2)
    I have
    xf=.5(a)(t^2)
     
  10. Nov 16, 2006 #9
    But.. its 5.2 meters a second squared, so dosnt that make it 5.2^2? or do I take the square root of 5.2 and put it into a?

    xf=.5(-5.2^2)(4^2)

    or...

    xf=.5(2.28)(4^2)?
     
  11. Nov 16, 2006 #10
    okay, here's the units:
    acceleration = m/s^2
    t^2 = s^2
    so
    a x t^2 = m/s^2 x s^2 = m

    so it's just xf = .5(-5.2)(4^2), this will give you your answer in meters.

    a is already in m/s^2, if you square it again, your units would be m^2/s^4
     
  12. Nov 16, 2006 #11
    Ah thanks, that was correct! I still dont quite understand why you dont leave it m/s^2 though
     
  13. Nov 16, 2006 #12
    hmm... I'm not sure what you mean by this.
    You ARE leaving the acceleration in m/s^2 because by definition, acceleration is in m/s^2. Doing anything to that acceleration would change the formula.
    You don't want the final answer in m/s^2 because you have a distance, x, which should be in meters, on the other side of the equation.

    maybe there's someone else in the forums who can explain this more clearly... anybody care to give it a try?
     
  14. Nov 16, 2006 #13
    I think I understand it now, thanks for all your help!
     
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