Question

Montréal is 510 km from Toronto, 12 degrees north of east. At an altitude of 9,000 meters, the windspeed is 80 km/h out of the north. For the entire 510 km, the aircraft flies at 9,000 meters at an airspeed of 300 knots. Draw a triangle whose sides represent the velocity vectors that correspond to the groundspeed, airspeed, and windspeed. Determine:

(a) the aircraft heading (direction in which the nose of the aircraft points), in degrees from north.

(b) the groundspeed of the aircraft, in km/h.

(c) the flight time, in hours.

Answer 1

Ground speed is the horizontal speed of an aircraft relative to the ground. It is the vector sum of airspeed and windspeed. Airspeed is the aircraft's speed relative to the air mass.

Here , V_g = groundspeed , V_a = airspeed = 300 knots = 555.6 Km/h 1 knot = 1.852 Km/h and  V_w = windspeed = 80 Km/h

a)

Because v_g is the vector sum of the v_w and v_a, its x– and y-components are the sums of the x– and y-components of the airspeed and windspeed.

Components along X-axis(East),

V_g cos(12) = V_a cos() + V_w cos(90)

V_g cos(12) = 555.6 cos() +0

V_g  = 568 cos() ....(1)

Components along Y-axis(North),

V_g sin(12) = V_a sin() -  V_w

V_g sin(12) = 555.6sin() - 80

V_g = 2672.3 sin() - 384.8 .....(2)

From Eq(1) and Eq(2)

568 cos() + 384.8 = 2672.3 sin()

Square both side and use the identity sin²(x)=1 -  cos²(x) to obtain an quadratic equation in cos()

then put cos() = u to solve quadratic

-6993118.96 + 437132.8u+ 7463814 u²=0

Solving it we get u = 0.939 or u = -0.997

Taking positive value

cos() = 0.939

So, = cos^-1 (0.939) = 20^o from East Or 70^o from North

b )

From Eq (1)

V_g  = 568 cos() = 568 x 0.939 = 533.35 Km/h

c)

Flight time = Distance / V_g = 510 / 533.35 = 0.956 hr