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Sagot :
Answer:
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 = [tex]\sqrt{v_x^2 + v_y^2 }[/tex]
v = [tex]\sqrt{9.97^2 + 1.54^2}[/tex]
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
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