An airplane heads northeast at an airspeed of km/hr, but there is a wind blowing from the west at km/hr. in what direction does the plane end up flying

The direction of the plain would be northeast east. If a wind is blowing from the west, it goes from west to east. The final speed of the plain is the vectorial addition of the wind speed and the relative speed of the plain. Therefore the flight direction of the plain seen from a stationary point in the earth is obtained from the sum of the direction of the wind and the plain.

Resultant velocity = 711.2km/hr

Direction = 41°

(41° North of east)

Complete Question:

An airplane heads northeast at an airspeed of 660km/hr, but there is a wind blowing from the west at 70 km/hr. In what direction does the plane end up flying?

Explanation:

Given;

Airplane velocity = 660km/hr northeast

Wind blowing velocity = 70 km/hr from the west (towards east)

Finding the horizontal and vertical component of the speed.

For vertical component Ry:

Ry = 660sin45

Ry = 466.7km/hr

For the horizontal component Rx:

Rx = 660cos45 + 70

Rx = 536.7km/hr

The Resultant velocity R can be given as;

R = √(Rx^2 +Ry^2)

Substituting Rx and Ry

R = √(536.7^2 + 466.7^2)

R = 711.2 km/hr

The direction can be derived from:

Tan∅ = opposite/adjacent = Ry/Rx

∅ = taninverse (Ry/Rx)

Substituting Ry and Rx

∅ = taninverse (466.7/536.7)

∅ = 41°

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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