Asked by T mann

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?
Answered by drwls
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.
Answered by T mann
What exactly is the angle. I got 100.55 degrees and it says it is wrong
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