Question

A train at a constant 64.0 km/h moves east for 40.0 min, then in a direction 53.0° east of due north for 27.0 min, and then west for 50.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip?

Answer 1

consider the east-west direction along X-axis and north-south direction along y-axis

AB = (64i + 0j) (40/60) = 42.7 i + 0 j

BC = ((64 Sin53) i + (64 Cos53) j) (27/60) = (51.1 i + 38.5 j) (27/60) = 23 i + 17.3 j

CD = (- 64i + 0j) (50/60) = - 53.3 i + 0 j

displacement is given as

D = AB + BC + CD = 42.7 i + 0 j + 23 i + 17.3 j - 53.3 i + 0 j = 12.4 i + 17.3 j

t = total time = 40 + 27 + 50 = 1.95 h

average velocity is given as

V_avg = D/t = (12.4 i + 17.3 j)/1.95 = 6.4 i + 8.8 j

magnitude of average velocity is given as

|V_avg | = sqrt((6.4)² + (8.8)²) = 10.88 km/h

b)

direction : = tan^-1(8.8/6.4) = 54 degree north of east