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Speed for a 2D problem.

  1. Sep 20, 2003 #1
    A particle starts from the origin at t = 0 with an initial velocity having an x component of 20:7 m/s and a y component of -16:3 m/s. The particle moves in the xy plane with an x component of acceleration only, given by 5:03 m/s2.
    A)Determine the x component of velocity after 9:01 s. Answer in units of m/s.

    B)Find the speed of the particle after 9:01 s.
    Answer in units of m/s.

    I have found the answer to A as 66.02 but for B isn't the answer 66.02-(-16.32)= 82.32?
     
  2. jcsd
  3. Sep 21, 2003 #2

    HallsofIvy

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    The point of the first part of the problem is that you can do the x and y components separately.

    Since the x-component of velocity is initially 20.7 m/s and there is an acceleration of 5.03 m/s2, the x-component of velocity at any time t is given by 20.7+ 5.03t. When t= 9.02, this is 20.7+ 5.03*9.01= 65.8 m/s.
    That's not quite what you got. Is your "20:7" something other than "20.7"??

    More importantly, you cannot add components like you do for B.
    Set up a right triangle with x-component as one leg and y-component as the other. The "speed" is the length of the vector itself- the hypotenuse of your right triangle. Use the Pythagorean theorem to calculate that.
     
  4. Sep 21, 2003 #3
    "never mind I figured that out, but how do I find the displacement after 9.01s?(in the previous post colon was supposed to be a dot sign(:=.)"

    should have read my other thread
     
  5. Sep 21, 2003 #4

    HallsofIvy

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    Just as I said: after you have found the "x-component" and the "y-component" you know the legs of a right triangle having the vector itself as hypotenuse. Use the Pythagorean theorem to find the length of that vector. That's the "displacement".
     
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