Question

A helicopter flies horizontally with constant velocity in a direction $30^\circ$ east of north between two points $A$ and $B$ at a distance $3\sqrt{2}$ km apart. Wind is blowing from south with a constant speed of 20 m/s. The speed of the helicopter is 1.5 times the speed of air, the helicopter flies from $A$ to $B$ and then returns from $B$ to $A$ with same speed relative to air in same wind. Find the total time for the journey in minutes assuming that the helicopter requires negligible time to stop or reverse its direction.

Answer 1

8

Answer 2

The helicopter flies from point A to point B and then returns from B to A. The distance between A and B is $d = 3\sqrt{2}$ km. The direction from A to B is $30^\circ$ east of north. Let's set up a coordinate system with the positive x-axis pointing east and the positive y-axis pointing north. The direction from A to B is represented by a unit vector $\hat{u}_{AB} = \sin(30^\circ) \hat{i} + \cos(30^\circ) \hat{j} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}$. The direction from B to A is $\hat{u}_{BA} = -\hat{u}_{AB} = -\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}$. The distance $d = 3\sqrt{2}$ km $= 3\sqrt{2} \times 1000$ m.

The wind is blowing from the south with a constant speed of $v_w = 20$ m/s. This means the wind velocity is directed towards the north, so $\vec{v}_w = 20 \hat{j}$ m/s. The speed of the helicopter relative to the air is $v_h = 1.5 \times v_w = 1.5 \times 20 = 30$ m/s.

Let $\vec{v}$ be the velocity of the helicopter relative to the ground. The relationship between the velocities is $\vec{v} = \vec{v}_h + \vec{v}_w$.

Journey from A to B: The helicopter flies with a constant velocity $\vec{v}_{AB}$ relative to the ground in the direction $\hat{u}_{AB}$. Let the magnitude of this velocity be $v_{AB}$. So, $\vec{v}_{AB} = v_{AB} \hat{u}_{AB} = v_{AB} (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})$. Let the velocity of the helicopter relative to the air be $\vec{v}_{h,AB}$ with magnitude $v_h = 30$ m/s. Let $\vec{v}_{h,AB} = v_h (\cos\theta \hat{i} + \sin\theta \hat{j})$. $\vec{v}_{AB} = \vec{v}_{h,AB} + \vec{v}_w$ $v_{AB} (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = 30 \cos\theta \hat{i} + 30 \sin\theta \hat{j} + 20 \hat{j}$ $v_{AB} (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = 30 \cos\theta \hat{i} + (30 \sin\theta + 20) \hat{j}$ Comparing components: $\frac{v_{AB}}{2} = 30 \cos\theta \implies \cos\theta = \frac{v_{AB}}{60}$ $\frac{v_{AB}\sqrt{3}}{2} = 30 \sin\theta + 20 \implies \sin\theta = \frac{v_{AB}\sqrt{3} - 40}{60}$ Using $\cos^2\theta + \sin^2\theta = 1$: $(\frac{v_{AB}}{60})^2 + (\frac{v_{AB}\sqrt{3} - 40}{60})^2 = 1$ $v_{AB}^2 + (v_{AB}\sqrt{3} - 40)^2 = 3600$ $v_{AB}^2 + 3v_{AB}^2 - 80\sqrt{3}v_{AB} + 1600 = 3600$ $4v_{AB}^2 - 80\sqrt{3}v_{AB} - 2000 = 0$ $v_{AB}^2 - 20\sqrt{3}v_{AB} - 500 = 0$ Solving the quadratic equation for $v_{AB}$: $v_{AB} = \frac{20\sqrt{3} \pm \sqrt{(-20\sqrt{3})^2 - 4(1)(-500)}}{2} = \frac{20\sqrt{3} \pm \sqrt{1200 + 2000}}{2} = \frac{20\sqrt{3} \pm 40\sqrt{2}}{2} = 10\sqrt{3} \pm 20\sqrt{2}$. Since $v_{AB} > 0$, we take the positive root: $v_{AB} = 10\sqrt{3} + 20\sqrt{2}$ m/s. Time taken for A to B: $t_{AB} = \frac{d}{v_{AB}} = \frac{3000\sqrt{2}}{10\sqrt{3} + 20\sqrt{2}} = \frac{300\sqrt{2}}{\sqrt{3} + 2\sqrt{2}} = \frac{300\sqrt{2}(\sqrt{3} - 2\sqrt{2})}{(\sqrt{3})^2 - (2\sqrt{2})^2} = \frac{300(\sqrt{6} - 4)}{3 - 8} = \frac{300(\sqrt{6} - 4)}{-5} = 60(4 - \sqrt{6})$ seconds.

Journey from B to A: The helicopter flies with a constant velocity $\vec{v}_{BA}$ relative to the ground in the direction $\hat{u}_{BA}$. Let the magnitude of this velocity be $v_{BA}$. So, $\vec{v}_{BA} = v_{BA} \hat{u}_{BA} = v_{BA} (-\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})$. Let the velocity of the helicopter relative to the air be $\vec{v}_{h,BA}$ with magnitude $v_h = 30$ m/s. Let $\vec{v}_{h,BA} = 30 (\cos\phi \hat{i} + \sin\phi \hat{j})$. $\vec{v}_{BA} = \vec{v}_{h,BA} + \vec{v}_w$ $v_{BA} (-\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) = 30 \cos\phi \hat{i} + 30 \sin\phi \hat{j} + 20 \hat{j}$ $-v_{BA} (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = 30 \cos\phi \hat{i} + (30 \sin\phi + 20) \hat{j}$ Comparing components: $-\frac{v_{BA}}{2} = 30 \cos\phi \implies \cos\phi = -\frac{v_{BA}}{60}$ $-\frac{v_{BA}\sqrt{3}}{2} = 30 \sin\phi + 20 \implies \sin\phi = \frac{-v_{BA}\sqrt{3} - 40}{60}$ Using $\cos^2\phi + \sin^2\phi = 1$: $(-\frac{v_{BA}}{60})^2 + (\frac{-v_{BA}\sqrt{3} - 40}{60})^2 = 1$ $v_{BA}^2 + (v_{BA}\sqrt{3} + 40)^2 = 3600$ $v_{BA}^2 + 3v_{BA}^2 + 80\sqrt{3}v_{BA} + 1600 = 3600$ $4v_{BA}^2 + 80\sqrt{3}v_{BA} - 2000 = 0$ $v_{BA}^2 + 20\sqrt{3}v_{BA} - 500 = 0$ Solving the quadratic equation for $v_{BA}$: $v_{BA} = \frac{-20\sqrt{3} \pm \sqrt{(20\sqrt{3})^2 - 4(1)(-500)}}{2} = \frac{-20\sqrt{3} \pm \sqrt{1200 + 2000}}{2} = \frac{-20\sqrt{3} \pm 40\sqrt{2}}{2} = -10\sqrt{3} \pm 20\sqrt{2}$. Since $v_{BA} > 0$, we take the positive root: $v_{BA} = -10\sqrt{3} + 20\sqrt{2} = 20\sqrt{2} - 10\sqrt{3}$ m/s. Time taken for B to A: $t_{BA} = \frac{d}{v_{BA}} = \frac{3000\sqrt{2}}{20\sqrt{2} - 10\sqrt{3}} = \frac{300\sqrt{2}}{2\sqrt{2} - \sqrt{3}} = \frac{300\sqrt{2}(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2})^2 - (\sqrt{3})^2} = \frac{300(4 + \sqrt{6})}{8 - 3} = \frac{300(4 + \sqrt{6})}{5} = 60(4 + \sqrt{6})$ seconds.

Total time for the journey $T = t_{AB} + t_{BA} = (240 - 60\sqrt{6}) + (240 + 60\sqrt{6}) = 480$ seconds. The question asks for the total time in minutes. $T = 480 \text{ seconds} = \frac{480}{60} \text{ minutes} = 8 \text{ minutes}$.