Posted by **Robin** on Sunday, February 9, 2014 at 6:16am.

A train at a constant 41.0 km/h moves east for 28 min, then in a direction 53.0° east of due north for 15.0 min, and then west for 64.0 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 -
**Henry **, Tuesday, February 11, 2014 at 10:13pm
41km/h[0o], 28min.

41km/h[37o], 15min.

41km/h[180o], 64min

a. d1=41km/h[0o] * (28/60)h=19.13km[0o]

d2=41km/h[37o] * (15/60)h=10.25km[37o]=

8.19 + i6.17

d3=41km/h[180o] * (64/60)h=43.73km[180o]

V=(d1+d2+d3)/(t1+t2+t3)=(19.13+8.19+i6.17-43.73)/(28+15+64) =

(-16.41 + i6.17)/107=17.53[159.4o]/107 =

0.16384km/min = 9.83 km/h.

b. Tan Ar = Y/X = 6.17/-16.41 = -0.37599

Ar = -20.61 = Reference angle.

A = -20.6 + 180 = 159.4o, CCW. = 69.4o

West of North.

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