? asked in Science & MathematicsPhysics · 9 years ago

Need help ASAP!!! Determine the MAGNITUDE of the bird watcher's (a) displacement and (b) average velocity.?

A bird watcher meanders through the woods, walking 0.219 km due east, 1.42 km due south, and 4.76 km in a direction 75.4 ° north of west. The time required for this trip is 1.616 h. Determine the MAGNITUDE of the bird watcher's (a) displacement and (b) average velocity.

I've tried this a couple of times but i seem to be getting the wrong answer.. i think it is to do with the magnitude ('+ve or -ve') .. thanks for any help!!

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  • 9 years ago
    Best Answer

    Displacement and velocity are both vectors so only the initial and final points count, not distance traveled.

    Here is what you have:

    +0.219 km right (positive x axis)

    -1.42 km south (negative y axis)

    And 4.76 km north of west which must be broken down into x and y components:

    The x component is cosine of the angle (75.4) multiplied by 4.76, while the y component is sine of the angle multiplied by 4.76

    Since it is northwest the x component will be negative and y will be positive.

    Now you need to add your x components together and your y components together and draw it on a piece of paper (be careful with signs).

    Now using pythagoreans theorem (x^2 + y^2 = z^2) to find the total displacement.

    Velocity=displacement/time

    If you have to be in SI units (meters and seconds) divide displacement by 1000 and multiply time 1.616 by 3600).

    Then just solve for velocity

    Source(s): If anything is unclear, edit your question and let me know... I don't give solutions onY! Answers, I just provide the required inforation to allow them to be found.
  • 4 years ago

    West component of 3rd. leg = (cos 44) x 2.2, = 1.5825km. North component = (sin 44) x 2.2, = 1.52825km. (1.5825 - 0.3) = 1.2825km. west of origin, and (1.52825 - 0.75) = 0.77825km. north of origin. Displacement = sqrt. (0.77825^2 + 1.2825^2) = 1.5km., and angle = arctan (0.77825/1.2825) = 31.25 deg. N of W.

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