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Resultant velocity from easterly and northerly components

  1. Nov 5, 2007 #1
    The easterly and northerly components of a car’s velocity are 24 m/s and 30 m/s, respectively. In what direction and with what speed is the car moving? In other words, what is car’s velocity. Hint: this question requires input from Trigonometry for the right angle triangle.
    Thus, the car was moving with the velocity of 38 m/s [E510N] (or, to the north of due east).

    I have no clue with this one ... can someone please help me?..
  2. jcsd
  3. Nov 5, 2007 #2
    Velocity is a vector and we're given the components of the velocity vector. How do we find the magnitude of a vector?
  4. Nov 5, 2007 #3
    I don't know T.T
  5. Nov 5, 2007 #4
    Ok, forget about velocity for a second. Say we move 24m east, then move 30m north what's the total distance we have gone? On a coordinate system say we started at (0,0) now we're at (24,30) right? How do we find the total distance?
  6. Nov 5, 2007 #5
    what about thinking it as pathagorean theorem. Hypotenuse?
  7. Nov 5, 2007 #6
    Is there like a formula or something...? because its really confusing..
  8. Nov 5, 2007 #7
    The magnitude of a vector is the square root of the sum of its components squared. I'm sure you're confused by now, I know I'd be, so here's a usefull formula you should remember:

    [tex]\vec v = \sqrt{v_x^2 + v_y^2}[/tex]
  9. Nov 5, 2007 #8
    We can use the distance formula which (kinda what antineutron is saying) will give us the distance.

    Magnitude: [tex]|\vec v| = \sqrt{v_x^2+v_y^2}[/tex]

    Direction: [tex]\theta = \arctan{\frac{v_y}{v_x}}[/tex]
    Last edited: Nov 5, 2007
  10. Nov 5, 2007 #9
    ok i thinkk i get it ... Thank youvery much everyone. ^^
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