Question

Please help me figure out what separates these two into x & y to solve the magnitude of the displacement of the planes

Answer 1

Magnitude is 2691.33m 5.32° North of East

Step-by-step explanation:

The plane is flying directly east so the only component is the x component ie + 135 m/s

The wind acceleration is directly northeast. That means 45° N relative to the x-axis(which is the same as saying 45° E relative to the y-axis)

Therefore there are two components to the acceleration due to the wind

The y or North component of acceleration is 2.18 x sin45°

The x or East component of acceleration is 2.18 x cos 45°

The displacement is given by the formula

s = ut + (1/2)at² where a is the acceleration, t is the time, u = initial velocity and v the final velocity

Substituting values for t= 18s, u = 0, v = 135 m/s and a = 2.18 cos 45° we get

Horizontal displacement
Sx = 135 x 18 + (1/2) x (2.18cos(45°) x (18²)
= 2,679.72 m

Vertical displacement in the +ve y-axis is
Sy = (1/2) x 2.18sin(45°) x (18²) = 249.72 m/s

Both components are equal since sin45 = cos 45

The magnitude of the resulting vector displacement

`= $√(2679.72^2 +249.72^2)\\\\\\= 2691.33$ m\\\\Angle is given bytan\theta = $(y-component)/(x-component)$\\or $\theta = tan^(-1)((249.72)/(2679.72))$\\\\\\= 5.32^\circ$ north of east`

The resultant displacement direction is at an angle of 5.32 ° North of East or directly NE