Question

An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

Answer 1

Given:
Distance between points A and B = 500 km
B from A is 30˚ east of north.
Speed of wind due north, v_w = 20 m/s
Airspeed of the plane, v_a = 150 m/s
Let $\overrightarrow{R}$ be the resultant direction of the plane to reach point B.

(a) Using the sine formula in ∆ACB, we get:
$$\begin{gathered} \frac{20}{\sin A}=\frac{150}{\sin 30°} \\ \Rightarrow \sin A=\frac{20}{150}\sin 30° \\ =\frac{20}{150}\times \frac{1}{2}=\frac{1}{15} \\ \Rightarrow A={\sin }^{-1}\left(\frac{1}{15}\right) \end{gathered}$$

Direction of the aeroplane is ${\sin }^{-1}\left(\frac{1}{15}\right)$ east of line AB.

(b) Time taken by the plane to reach point B from point A:
${\sin }^{-1}\left(\frac{1}{15}\right)=3°48\text{'}$
⇒ 30° + 3°48' = 33°48

$R=\sqrt{A^2+B^2+2AB\cos \theta }$

$$\begin{gathered} R=\sqrt{150+20+2\left(150\right)\left(20\right)\cos \left(33°48\text{'}\right)} \\ =\sqrt{27886}=167\mathrm{m}/\mathrm{s} \\ \mathrm{Time}=\frac{\mathit{s}}{\mathit{v}}=\frac{500000}{167} \\ =2994\mathrm{s}=49.0\approx 50\mathrm{minutes} \end{gathered}$$