Posted by **T mann** on Monday, February 6, 2012 at 7:00am.

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 -
**drwls**, Monday, February 6, 2012 at 8:20am
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 -
**T mann**, Monday, February 6, 2012 at 8:50am
What exactly is the angle. I got 100.55 degrees and it says it is wrong

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