Let the xaxis point in the East direction and the yaxis poin

Let the +x-axis point in the East direction and the +y-axis point in the North direction. An airplane pilot wants to fly her airplane from Richmond, Virginia to Beckley, West Virginia. Beckley is 200 miles due west of Richmond. A strong 100mi/hr wind is blowing from the North-Northwest in a direction 30 degree East of South. Compute the components v_wind, y of the vector V_wind where V_wind = V_wind, x i + v_wind, y j Assume that the airspeed of the airplane is 150 mi/hr (V_airspeed| = 150 mi/hr). Compute the compontens v_airspeed, x, V_airspeed, y of the vector wherer V_airspeed = v_airspeed, x i + V_airspeed, y J Compute theta_airspeed, the direction of the vector V_airspeed, where theta_airspeed is an angle measured counterclockwise from the +x-axis. Compute |v|, the speed of the airplane relative to the ground. How long does it take the airplane to fly from Richmond to Beckley? (compute the time in minutes)

Solution

Here, it needs to be understood that the net velocity of the plane should be along the X axis only that is, along the west. For that to happen the Y component of the velocity of the aircraft must balance the counter velocity being provided by the air along the Y axis.

We will assume that the aircraft\'s velocity makes an angle of \'a\' with negative x axis and it flies at this angle so as to make the resultant velocity along the X axis..

a.) Wind blows at an angle of 30 degrees with the vertical.

Therfore, Vwind,x = V sin30 = 100/2 = 50 mi/hr

Vwind,y = - Vcos30 = -86.603 mi/hr

b.) Now, the Y component of the airspeed must balance Vwind,y

That is 150 sina = 86.603

Sina = 0.57735

That is \'a\' = 35.265

Hence the Vair,x = -150 cos 35.265 = -122.4736 mi/hr

Vair,y = 86.603 mi/hr

c.)The direction of this airspeed with +x axis counterclockwise = 180 - 35.265 = 144.735

d.) Speed of the airplane relative to the ground will be the magnitude of its velocity along the x axis

That is, |V| = 122.4736 mi/hr

e.) The time taken to reach Beckley would be:

Time = Distance / speed = 200 / 122.4736 = 1.633 hours

NOTE: For such problems you need to try and visualise the direction of the objects at hand, That greatly simplifies the approach. For instance, in the above problem you need to realise that as the wind is blowing from North-NorthWest the countering velocity must be against that.