Question

a pilot flies for 3 h with an average velocity of 350 km/h n 35° e. in the first part of the trip, the pilot flies for 2.0 h through a displacement of 380 km e 20° n. she then flies directly to her destination. determine the angle of the direction in east of north of displacement for the second part of the flight.

Explanation:

Step1: Calculate total displacement

The total time of flight is $t = 3$ h and average - velocity is $v=350$ km/h. So the total displacement $d = vt=350\times3 = 1050$ km in the direction of $\theta_{total}=35^{\circ}$ east of north.
The first - part displacement $d_1 = 380$ km in the direction of $\theta_1 = 20^{\circ}$ east of north.

Step2: Resolve displacements into components

For the total displacement:
The north - south component $d_{total,y}=d\cos\theta_{total}=1050\cos35^{\circ}\approx1050\times0.819 = 859.95$ km.
The east - west component $d_{total,x}=d\sin\theta_{total}=1050\sin35^{\circ}\approx1050\times0.574 = 602.7$ km.
For the first - part displacement:
The north - south component $d_{1,y}=d_1\cos\theta_1=380\cos20^{\circ}\approx380\times0.940 = 357.2$ km.
The east - west component $d_{1,x}=d_1\sin\theta_1=380\sin20^{\circ}\approx380\times0.342 = 129.96$ km.

Step3: Calculate the components of the second - part displacement

The north - south component of the second - part displacement $d_{2,y}=d_{total,y}-d_{1,y}=859.95 - 357.2=502.75$ km.
The east - west component of the second - part displacement $d_{2,x}=d_{total,x}-d_{1,x}=602.7 - 129.96 = 472.74$ km.

Step4: Calculate the angle of the second - part displacement

Let the angle of the second - part displacement be $\theta_2$ east of north. Then $\tan\theta_2=\frac{d_{2,x}}{d_{2,y}}=\frac{472.74}{502.75}\approx0.940$.
So $\theta_2=\arctan(0.940)\approx43.2^{\circ}$.

Answer:

$43.2^{\circ}$