Question

An airplane trip involves three trips with two stopovers from its base. the first trip is 670 Km west the second one is 510 km northeast and the third one is 620 Km (37 degree) west of south .
a- what is the plane total displacement (magnitude and direction)
b-what is the magnitude and direction of a fourth trip to bring the plane back to its base

Answer 1

We have next picture:

And we have:

$$A B = 6 7 0$$ $$B C = 5 1 0$$ $$C D = 6 2 0$$ $$\angle A B C = 4 5 {}^ {\circ}$$ $$\angle B C D = 8 {}^ {\circ}$$ $$A C = \sqrt {6 7 0 ^ {2} + 5 1 0 ^ {2} - 2 \cdot 6 7 0 \cdot 5 1 0 \cdot \cos 4 5 {}^ {\circ}} \approx 4 7 5. 1 5$$ $$\angle B O C = 1 2 7 {}^ {\circ}$$ $$\frac {6 7 0}{\sin \angle A C B} = \frac {4 7 5 . 1 5}{\sin \angle A B C} = \frac {4 7 5 . 1 5}{\sin 4 5 {}^ {\circ}} \Rightarrow \sin \angle A C B = \frac {6 7 0 \cdot \sin 4 5 {}^ {\circ}}{4 7 5 . 1 5} \Rightarrow$$ $$\Rightarrow \angle A C B = \arcsin \left(\frac {6 7 0 \cdot \sin 4 5 {}^ {\circ}}{4 7 5 . 1 5}\right) = 8 5. 6 {}^ {\circ}$$ $$\angle A C D = \angle A C B - 8 {}^ {\circ} = 8 5. 6 {}^ {\circ} - 8 {}^ {\circ} = 7 7. 6 {}^ {\circ}$$ $$A D = \sqrt {6 2 0 ^ {2} + 4 7 5 . 1 5 ^ {2} - 2 \cdot 6 2 0 \cdot 4 7 5 . 1 5 \cdot \cos 7 7 . 6 {}^ {\circ}} = 6 9 5. 6;$$ $$\frac {6 9 5 . 6}{\sin 7 7 . 6 {}^ {\circ}} = \frac {4 7 5 . 1 5}{\sin \angle A D C} \Rightarrow$$ $$\Rightarrow \angle A D C = \arcsin \left(\frac {4 7 5 . 1 5 \cdot \sin 7 7 . 6 {}^ {\circ}}{6 9 5 . 6}\right) = 4 1. 8 {}^ {\circ}$$ $$\angle B A D = 1 8 0 {}^ {\circ} - 1 2 7 {}^ {\circ} - 4 1. 8 {}^ {\circ} = 1 1. 2 {}^ {\circ}$$

So, we can say that total displacement of the plane is: magnitude = 695.6 km and direction = 11.2 degree south of west.

And the magnitude and the direction of a fourth trip are: magnitude = 695.6 km and direction = 11.2 degree north of east.

Answer: a. magnitude = 695.6 km and direction = 11.2 degree south of west.

b. magnitude = 695.6 km and direction = 11.2 degree north of east.