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March 24, 2017

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A train at a constant 41.0 km/h moves east for 38 min, then in a direction 44.0° east of due north for 25.0 min, and then west for 59.0 min. What are the (a) magnitude (in km/h) and (b) angle (relative to north, with east of north positive and west of north negative) of its average velocity during this trip?

  • physics - ,

    First travel segment is 41*38/60 = 25.97 km east.

    Second travel segment is 41*25/60 = 17.08 km, 44 degrees east of north.
    That has components of 17.08 sin44 = 11.87 km east and 17.08 cos44 = 12.29 kn north.

    The third displacement is 40.32 km weast (-40.32 km west).

    Add up the total resultant diplacements north and east. Divide them by 38 + 25 + 59 = 122 minutes = 2.067 hours to get average (resultant) velocity components in km/h.

    The angle east of north is arctan Rx/Ry, where Rx and Ry are the resultant displacements east and north.

  • physics - ,

    What exactly is the angle. I got 100.55 degrees and it says it is wrong

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