An airplane must fly due north and cover s = 300 km in t = 1…

An airplane must fly due north and cover $s = 300\ \mathrm{km}$ in $t = 1\ \mathrm{h}$. A north-east wind blows with speed $v = 35\ \mathrm{km/h}$ at an angle $\alpha = 40^\circ$ to the meridian. With what speed (relative to the air) must the airplane fly, and what course must it hold?

Explanation of Method

Flying north against a crosswind.

Step 1: Which way does the wind blow?

Winds are named after the direction they come from. A north-east wind comes from the north-east and blows toward the south-west, at $40^\circ$ to the north–south line. Take north and east as positive directions. The wind's components are then

\[ w_N = -v\cos 40^\circ = -35 \cdot 0.766 = -26.81\ \mathrm{km/h}, \]

\[ w_E = -v\sin 40^\circ = -35 \cdot 0.643 = -22.50\ \mathrm{km/h}. \]

Step 2: Required ground velocity

To cover 300 km north in 1 h, the airplane must move over the ground at 300 km/h due north: $u_N = 300\ \mathrm{km/h}$, $u_E = 0$.

Step 3: Velocity relative to the air

Airplane velocity (relative to the air) + wind = ground velocity, so the airplane's own velocity is ground velocity minus wind: \[v_{aN} = u_N - w_N = 300 + 26.81 = 326.81\ \mathrm{km/h},\]

\[ v_{aE} = u_E - w_E = 0 + 22.50 = 22.50\ \mathrm{km/h}. \]

$\vec v_{\text{ground}} = 300\ \mathrm{km/h}\ \text{(north)}$ $\vec v_a\ \text{(airplane in air)}$ $\vec w\ \text{(wind)}$ $\theta$ $N$ $E$ $\text{wind from NE}$

required ground velocity
airplane relative to air
wind

Airplane velocity + wind = ground velocity. The wind blows toward the south-west, so the airplane must point slightly east of north and fly a little faster than 300 km/h. (The wind arrow is drawn 3× longer for clarity, so the angle looks bigger than it is.)

Step 4: Speed and course

\[ v_a = \sqrt{v_{aN}^2 + v_{aE}^2} = \sqrt{326.81^2 + 22.50^2} \approx 327.6\ \mathrm{km/h}, \]

\[ \tan\theta = \frac{v_{aE}}{v_{aN}} = \frac{22.50}{326.81} \approx 0.0688 \;\Longrightarrow\; \theta \approx 3.9^\circ. \]

\[\begin{gathered} v_a \approx 328\ \mathrm{km/h} \\ \text{course } \approx 3.9^\circ \text{ east of north} \end{gathered}\]

Check

North: $ 326.81 - 26.81 = 300\ \mathrm{km/h}\ \checkmark $

East: $ 22.50 - 22.50 = 0\ \checkmark $ (no sideways drift)

Result

\[\boxed{\,\begin{gathered} v_a \approx 328\ \mathrm{km/h} \\ \text{course } \approx 3.9^\circ \text{ east of north} \end{gathered}\,}\]

Remark

About the original solution. It treated the wind as blowing toward the north-east. With that reading the airplane would get a tailwind: $v_{aN} = 300 - 26.81 = 273.19$, $v_{aE} = -22.50$, giving $v_a = \sqrt{273.19^2 + 22.50^2} \approx 274.1\ \mathrm{km/h}$ and a course $4.7^\circ$ west of north. (Its value 274.9 km/h was also a small arithmetic slip.) With the standard meaning of "north-east wind" the answer is 328 km/h, 3.9° east of north.