Question

The runner traveled 2km 22 degrees North East from y axis, 4km 25 degrees North West from x axis, 7km 70 degrees South West from y axis, 6km 50 degrees South East from y axis. what is the resultant displacement of the runner using resolution of vectors to its components

Answer 1

if you can resolve one vector, you can do four. Then just add them up...

<u>2km 22 degrees North East from y axis</u>
makes no sense. I assume you meant
2km @ N22°E = (2sin22°,2cos22°) = (0.75,1.85)
Resolve the others similarly, and then just add them up for a final (x,y) value.
Then the displacement is the usual √(x^2+y^2)

Answer 2

All angles are measured CW from +Y-axis,
Diisp. = 2km[22o] + 4km[295o] + 7km[250o] + 6km[130o],
X = 2*sin22+4*sin295+7*sin250+6*sin130 = -4.86 km,
Y = 2*Cos22+4*Cos295+7*Cos250+6*Cos130 = -2.71 km,

Disp. = Sqrt(X^2+Y^2),
Tan A = X/Y,
A = ?.