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Why isn't this working 2D Kinematics

  1. Sep 20, 2006 #1
    Why isn't this working!! 2D Kinematics

    Here is the problem:

    Relative to the ground, a car has a velocity of 18.2 m/s, directed due north. Relative to this car, a truck has a velocity of 22.9 m/s directed 47.2 ° 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.

    For a) I found the answer as follows (which is correct)

    V of T in relation to G (Vtg) = V of T in relation to C (Vtc) + V of C in relation to G Vcg.

    The x factor looks like this:

    Vtgx = Vtc + Vcg

    22.9 cos 47.2 + 0 = 15.6

    Y factor:

    Vtgy = Vtc + Vcg

    -22.9 sin 47.2 = 1.39

    Vtg = Sqrt (Vtgx^2 + Vtgy^2) = 15.7 <-------the Right answer

    For b) I did

    tan^-1 y/x, but I keep getting the wrong answer. What is going on. I keep getting 5.07, and this is incorrect!
     
  2. jcsd
  3. Sep 20, 2006 #2

    berkeman

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    Staff: Mentor

    Looks reasonable to me. Do you know how to convert vectors in polar notation (like those in your problem) into rectangular coordinates? As a cross-check on your work, you could convert the vectors into rectangular coordinates and just add Vc and deltaVtc to get Vt. Then convert Vt back to polar coordinates to get the angle a different way.
     
  4. Sep 21, 2006 #3
    I reviewed some of my noted but I am unclear about this, could the angle be found as such:

    Theta = Cos^-1 (Vcg/Vct) ? ?

    so...

    Cos^-1 (18.2/22.9) = 37.36

    Thanks!
     
  5. Sep 21, 2006 #4

    berkeman

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    From the problem statement, I don't think that's what they are asking for. Aren't they asking for the travel direction of the truck with respect to the ground?
     
  6. Sep 21, 2006 #5

    Doc Al

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    That method is correct. Recheck your calculations for the velocity components--make sure you didn't make any round-off errors.
     
  7. Sep 22, 2006 #6

    Thanks Doc....but I have checked and rechecked....do you think the part that says, what is the angle relative to Due east, has anything to do with the answer being different than what it is?
     
  8. Sep 22, 2006 #7

    berkeman

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    Did you try it in rectangular coordinates yet?
     
  9. Sep 22, 2006 #8
    I hate to say it, but I have no idea what that means :blushing:
     
  10. Sep 22, 2006 #9

    radou

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    [tex]\vec{v}_{T, G} = \vec{v}_{T, C} + \vec{v}_{C, G}[/tex], where T stands for truck, C for car, and G for ground. Just vector addition. Or am I missing something?
     
  11. Sep 22, 2006 #10

    yes, because first off, that part of the question was answered. And secondly, that would work if the two Vectors we were adding were moving along the same direction, or opposite directions, but in this case they are not.


    What we are trying to solve is the angle of the Truck relative to the Ground.
     
  12. Sep 22, 2006 #11

    Doc Al

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    No, since that's what you are giving. The angle is some positive value with respect to the x-axis.

    I assume you know your answer is "wrong" because some system rejected it? If so, it could be something stupid like roundoff error. Don't round anything off until the very last step, when you are finding the angle. (Doing so I get a slightly different answer than you did.)
     
  13. Sep 22, 2006 #12
    Thanks! I will try that and report back!
     
  14. Sep 22, 2006 #13

    Doc Al

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    Yes, it's just vector addition. But mmiller39 understands that!

    Actually, radou's version is correct as it stands since he wrote the equation in terms of vectors, not components.
     
  15. Sep 22, 2006 #14
    Doc, thanks for your attention to detail, 5.13 was my final angle and it was correct.

    Thanks Again!
     
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