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A train moving at a constant speed of 68.0 km/h moves east for 33.0 min, then in a direction 40.0° east of due north for 30.0 min, and then west for 59.0 min. What is the average velocity of the train during this run?

  • physics -

    The key is finding the final position.
    1) It goes 68*33/60 km east
    2) it goes 68*30/60 40 deg E of N. Converting that to E and N displacements..
    it goes 60(30/60)sin40 East
    it goes 60(30/60)cos40 North
    3) it goes E for 68(-59/60) Notice the negative sign which changes West to East.

    Now add up the East components.
    That gives you the net East.
    Now add up the N components.

    Now use Pythogerean theorum
    displacement=sqrt(N^2 +E^2)

    Magnitude avg velocity=displacement/timetotal.

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