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Chapter 04, Problem 005 A train at a constant 77.0 km/h moves east for 31.0 min, then in a direction 54.0° east of due north for 22.0 min, and then west for 41.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip? (a) Number Units (b) Number Units

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Answer #1

east disp: x = 77km/h(31 + 22*sin54º - 41)h/60 = 10.00 km

north disp: y = 77km/h(0 + 22cos54º + 0)h/60 = 16.60 km

displacement s = √(x² + y²) = 19.38 km

total time t = (31 + 22 + 41)/60 h = 1.57 h

(a) Vavg = s / t = 12.37 km/h

(b) Θ = arctan(16/10) = 58º N of E

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