My question asks different numbers: A train at a constant 61.0 km/h moves east f

Question

My question asks different numbers: A train at a constant 61.0 km/h moves east for 39 min, then ina direction 38.0° east of due north for 23.0 min, and then westfor 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 northnegative) of its average velocity during this trip?
I got part a right and for some reason I'm getting b wrong! Idon't know why because I'm just supposed to plug it into theinverse tangent.. What am I doing wrong!? I got avg v= -2.94i+9.14jand the magnitude is 9.6 which is correct. For the angle I've tried-72.17, 72.17, and 107.8 and they were all wrong. THANKS :) A train at a constant 61.0 km/h moves east for 39 min, then ina direction 38.0° east of due north for 23.0 min, and then westfor 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 northnegative) of its average velocity during this trip?
I got part a right and for some reason I'm getting b wrong! Idon't know why because I'm just supposed to plug it into theinverse tangent.. What am I doing wrong!? I got avg v= -2.94i+9.14jand the magnitude is 9.6 which is correct. For the angle I've tried-72.17, 72.17, and 107.8 and they were all wrong. THANKS :)

Explanation / Answer

For this problem it is asking for the angle relative tonorth. So when you are finding the angel with the inversetangent you are finding it realtive to west. Which is the72.17. 107.8 is relative to East or the positivex-axis. So you need it relative to the positive y axis ornorth which would be using your numbers -17.83. because itsays that west of north is negative and east of north ispositive. I hope this helps.

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