Question

4. A pilot flying from Honolulu to California flies at an airspeed of 300.0 m/s (“airspeed” = airplane’s speed through the surrounding air) with her airplane pointed 70.0˚ to the east of north. However, the air itself is moving with a 50.0-m/s windspeed blowing due east. When the pilot adds together the two vectors of her airspeed and the windspeed, she gets her groundspeed: the airplane’s actual velocity relative to the ground below. (Even though these are called “speeds” by pilots, they are all vector velocities.) You may wish to create a sketch of the vector addition to help you visualize the problem.

a.) Using either the component method or the “triangle” method, find the magnitude and compass bearing (expressed in degrees to the east of north) of the airplane’s groundspeed. Show your work completely.

b.) Suppose the airplane and wind both maintain their constant velocities. How much time is needed for the airplane to travel a distance of 4100 km (roughly 2500 miles) relative to the ground below? Convert your final answer into hours. Show your work.

c.) At what compass bearing should the pilot aim her airplane heading (i.e., direction of her airspeed vector, still with a magnitude of 300.0 m/s) so that her resulting groundspeed ends up at an angle of 70.0˚ east of north? Assume that the wind vector is the same as in the problem above. (Hint: Use the “triangle”/trigonometric method instead of the component method. Another Hint: This is NOT the same triangle as in the problem above!) You do NOT need to show your work for this part.

Answer 1

a)

V_p = velocity of plane = 300 m/s

V_w = velocity of wind = 50 m/s

consider the east-west direction along X-axis and north-south direction along y-axis

net velocity along X-direction is given as

V_x = 300 Cos70 + 50 = 152.6 m/s

net velocity along Y-direction is given as

V_y = 300 Sin70 = 282 m/s

net velocity is given as

V_pg = Velocity of plane relative to ground = sqrt(Vx² + V_y²) = sqrt((152.6)² + (282)²) = 320.6 m/s

= tan^-1(V_y/V_x) = tan^-1(282/152.6) = 61.6 degree north of east.

or 28.4 degree east of north

b)

t = D/ V_pg = 4100 x 1000/320.6 = 12788.52 sec = 3.6 hours

c)

V_x = (300 Cos + 50 )

V_y = 300 Sin

Given that

tan70 = V_x /V_y

tan70 = (300 Cos + 50 )/300 Sin

tan70 = (300 Cos + 50 )/(300 sqrt(1 - Cos²))

Cos= 0.92

= 23.1 degree