physics
posted by tom on .
A car is driven east for a distance of 49 km, then north for 21 km, and then in a direction 30° east of north for 27 km. Determine (a) the magnitude (in km) of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.

30deg E of N = 60deg CCW.
X = hor = 49 + 27*cos60,
X=49 + 13.5=62.5km=hor comp. of vector.
Y = ver = 21 + 27*sin60,
Y = 21 + 23.38 = 44.38km.
a. D = sqrt(x^2 + y^2),
D = sqrt(3906.25+1969.58),
D = 76.7km.
b. tanA = Y/X = 44.38 / 62.5 = 0.7101.
A = 35.4 deg = Direction.