An athlete running the velocity 3m/s due east is confronted with two trade winds. One wind travelling at 10m/s in a direction north 65degrees east and another wind travelling at 8m/s in a direction south 70degrees east. Find the resultant velocity and direction of the athlete

v = 10.09 m / s,   8.78 North of East

Explanation:

The easiest way to solve this exercise is to decompose the velocities into a coordinate system, where the x-axis coincides with the West-East direction and the y-axis coincides with the South-North direction.

athlete's velocity v₁ₓ = 3 m / s

wind speed 2 v₂ = 10 m / s with direction 65 north of east

          sin 65 = v_{2y} / v₂

          cos 65 = v₂ₓ / v₂

          v_{2y} = v₂ sin 65

          v₂ₓ = v₂ cos 65

          v_{2y} = 10 sin 65 = 9.06 m / s

          v₂ₓ = 10 cos 65 = 4.23 m / s

wind speed 3 v₃ = 8 m / s with direction 70 south of east

This angle measured from the positive side of the x-axis is

          θ = 360 - 70

          θ = 290

          sin 290 = v_{3y} / v₃

          cos 290 = v₃ₓ / v₃

          v_{3y} = v₃ sin 290

          v₃ₓ = v₃ cos 290

           v_{3y} = 8 sin 290 = -7.52 m / s

          v₃ₓ = 8 cos 290 = 2.74 m / s

now we can find each component of the velocity

X axis

        vₓ = v₁ + v₂ₓ + v₃ₓ

        vₓ = 3 + 4.23 + 2.74

        vₓ = 9.97 m / s

Y axis  

        v_y = v_{2y} + v_{3y}

        v_y = 9.06 - 7.52

         v_y = 1.54 m / s

to find the modulus let's use the Pythagorean theorem

         v =

         v =

         v = 10.09 m / s

let's use trigonometry for the direction

         tan θ = v_y / vₓ

        θ = tan⁻¹ v_y / vₓ

        θ = tan-1 1.54 / 9/97

         θ = 8.78

the address is 8.78 North of East