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?

Answer 1

s1=v •t1=44•42/60 =30.8 km

s2 =v•t2 = 44•17/60 = 12.5 km
s3 = v•t3 = 44•51/60=37.4 km
Cosine Law:
displacement d= sqrt[(s2² +(s3-s1)²-2•s2•(s3-s1) •cos135] = 17.8 km
ave velocity = d/time taken = 17.8• 60/(42+17+51) =9.7 m/s

Sine Law:
(s3-s1)/sinα =d/sin135
sinα = sin135(s3-s1)/d = 0.707•6.6/17.8 =0.26.
α =15.07°
β=135-90=45°
θ= α+ β =15.07°+45° =60.07°