- #1
Dan Feerst
- 12
- 0
Homework Statement
A train at a constant 42.0 km/h moves east for 33 min, then in a direction 64.0° east of due north for 25.0 min, and then west for 69.0 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?
Homework Equations
The way I inturpreted the problem I needed to first find the time in hours and the displacement in kilometers for each vector then I added the displacements over the times. here is the work I did.
The Attempt at a Solution
First I calculated the total time. Since time isn't a vector I added everything.
33'*/60'=.550h
25'*/60'=.417h
69'/60'=1.15h
Sum = 2.12h
Next I found the total distance
42km/h*.550h=23.1km
42km/h*.417h=17.5km
42km/h*1.15h=48.3km
then I added the magnitudes of the three vectors taking the magnitude of the x coordinate for the second
r=(23.1km)+(17.5km *Cos(26))-(48.3km)= -9.47
Finally I divided displacement over time to get an unconvincing and completely wrong answer of -4.47
What did I do wrong.