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2D Kinematics I tried so hard

  1. Sep 11, 2008 #1
    2D Kinematics Please Help! I tried so hard

    Relative to the ground, a car has a velocity of 17.3 m/s, directed due north. Relative to this car, a truck has a velocity of 20.5 m/s directed 48.5 ° south of east. Find the (a) magnitude and (b) direction of the truck's velocity relative to the ground. Give the directional angle relative to due east.

    I got:

    Vcg = 0x + 17.3y
    Vtg = 20.5 cos (45) - 20.5 sin(45)

    The resultant is: 14.496x + 2.804y
    So, by the Pythagorean theorem, Vtg = 14.765

    (But that's wrong)

    And the direction would be: Inverse tangent of (2.804/14.496)

    so it's 10.948 degrees

    And that's wrong too.

    I also tried it with adding the 20.5 sin(45) on the Vtg part, but those answers are wrong too. I dont know what to do!
     
  2. jcsd
  3. Sep 12, 2008 #2

    alphysicist

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    Re: 2D Kinematics Please Help! I tried so hard

    Hi lmbiango,

    I don't see why are you using 45 degrees here; they give a different angle in the problem.
     
  4. Sep 12, 2008 #3
    Re: 2D Kinematics Please Help! I tried so hard

    you're right, that was my first dumb mistake... but even after getting the correct number for Vtg, which I now have 13.584, the angle is still incorrect. If I do, inverse tangent of (2.804/13.723) = 11.548, that is wrong for the angle and I only have one try left.
     
  5. Sep 12, 2008 #4

    alphysicist

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    Re: 2D Kinematics Please Help! I tried so hard


    Looking at your numbers, I think you might be mixing up a few numbers.

    You say the answer for Vtg was 13.584, but in your inverse tangent equation you have 13.723 in the denominator. These should be the same number and I think the 13.723 is correct, so you might want to check why you have different numbers.

    In the numerator of your inverse tangent you use 2.804, which was the y component you found when using the incorrect angle of 45 degrees. You need the new total y component from using 48.5 degrees.
     
  6. Sep 14, 2008 #5
    Re: 2D Kinematics Please Help! I tried so hard

    you're right, thanks, I got it now!
     
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