Posted by **D** on Wednesday, October 10, 2012 at 12:53am.

A train at a constant 44.0 km/h moves east for 42 min, then in a direction 45.0° east of due north for 17.0 min, and then west for 51.0 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?

- phyics -
**Henry**, Thursday, October 11, 2012 at 6:25pm
d1 = 44km/h * (42/60)h = 30.80 km @ 0o.

d2 = 44km/h * (17/60)h = 12.47 km @ 45o.

d3 = 44km/h * (51/60)h = 37.4 km @ 180o.

a.X=Hor.=30.80+12.47*cos45+37.40*cos180 =2.22 km.

Y = 12.47*sin45+0 = 8.818 km.

D^2 = X^2 + Y^2 = 4.928 + 77.75 = 82.68

D = 9.1 km. = Total distance.

T = 42/60 + 17/60 + 51/60 = 1.833 h.

V = D/T = 9.1km / 1.833h = 4.965 km/h.

b. tanA = Y/X = 8.818/2.22 = 3.97207

A = 75.9o,CCW = 14.13o East of North.

= Direction.

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