Question

A jet aircraft is aimed south and travelling 851km/h. A wind blows the plane from 40 degrees N of E at 36km/h. What is the plane's resultant velocity?

Answer 1

A jet aircraft is aimed south and travelling $851\mathrm{km/h}$ . A wind blows the plane from 40 degrees N of E at $36\mathrm{km/h}$ . What is the plane's resultant velocity?

Solution.

$v_{a}$ - the velocity of the aircraft.

$v_{w}$ - the velocity of the wind.

$v$ - the resultant velocity of the aircraft.

A wind blows the plane from 40 degrees N of E as on the diagram.

By the diagram:

An angle $\varphi = 90{}^{\circ} - 40{}^{\circ} = 50{}^{\circ}$

$\varphi_{1} = \varphi = 50{}^{\circ}$ , as the angles between mutually parallel lines.

An angle $\alpha = 180{}^{\circ} - \varphi_{1} = 180{}^{\circ} - 50{}^{\circ} = 130{}^{\circ}$ .

$\alpha = 130{}^{\circ}$

By the diagram there is a triangle with sides: $v_{a}, v_{w}, v$ .

By Law of cosines:

$$v ^ {2} = v _ {a} ^ {2} + v _ {w} ^ {2} - 2 v _ {a} v _ {w} \cos \alpha .$$

The resultant velocity of the aircraft is:

$$v = \sqrt {v _ {a} ^ {2} + v _ {w} ^ {2} - 2 v _ {a} v _ {w} \cos \alpha}.$$ $$v = \sqrt {\left(851 \frac {k m}{h}\right) ^ {2} + \left(36 \frac {k m}{h}\right) ^ {2} - 2 \cdot 851 \frac {k m}{h} \cdot 36 \frac {k m}{h} \cdot \cos (130 {}^ {\circ})} = 874.6 \frac {k m}{h}.$$

Answer: The resultant velocity of the aircraft is $v = 874.6 \frac{km}{h}$ .