1. A new pilot forgets to compensate for the wind. He believes he is heading \( \left[\mathrm{N} 20^{\circ} \mathrm{W}\right] \) at \( 250 \mathrm{~km} / \mathrm{h} \) but the he's in a north wind with a velocity of 60.0 \( \mathrm{~km} / \mathrm{h} \). He flies for 30 minutes.

Let’s begin by defining our coordinate system. We choose positive \( x \) to be east and positive \( y \) to be north. 1. **Determine the airspeed components.** The pilot thinks he is flying at a speed of \( 250\,\mathrm{km/h} \) in the direction \(\mathrm{N}20^\circ\mathrm{W}\). This means his motion is \(20^\circ\) west of north. Hence, the components of his airspeed are: - Westward (negative \( x \)) component: \[ v_{x,\text{air}} = -250\sin 20^\circ \] - Northward (positive \( y \)) component: \[ v_{y,\text{air}} = 250\cos 20^\circ \] 2. **Write the wind vector.** The problem states that there is “a north wind” at \( 60.0\,\mathrm{km/h} \). A north wind comes from the north and blows toward the south, so its vector is: \[ \vec{v}_\text{wind} = \langle 0,\,-60.0 \rangle\,\mathrm{km/h}. \] 3. **Find the ground speed vector.** The actual ground speed is the vector sum of the airspeed and the wind speed. Thus: \[ v_x = v_{x,\text{air}} + 0 = -250\sin 20^\circ, \] \[ v_y = v_{y,\text{air}} - 60.0 = 250\cos 20^\circ - 60.0. \] 4. **Compute the displacement over 30 minutes.** Since the pilot flies for 30 minutes (\( 0.5\,\mathrm{h} \)), the displacement components are: \[ \Delta x = v_x \times 0.5 = -125\sin 20^\circ, \] \[ \Delta y = v_y \times 0.5 = 125\cos 20^\circ - 30.0. \] 5. **Evaluate the numerical values.** Using approximate values: - \(\sin 20^\circ \approx 0.342\), - \(\cos 20^\circ \approx 0.940\). We then have: \[ \Delta x \approx -125 \times 0.342 \approx -42.75\,\mathrm{km}, \] \[ \Delta y \approx 125 \times 0.940 - 30.0 \approx 117.5 - 30.0 \approx 87.5\,\mathrm{km}. \] 6. **Find the distance from the starting point and the heading.** The total displacement (distance) is: \[ d = \sqrt{(\Delta x)^2 + (\Delta y)^2} \approx \sqrt{(-42.75)^2 + (87.5)^2} \approx \sqrt{1827 + 7656} \approx \sqrt{9483} \approx 97.4\,\mathrm{km}. \] The direction relative to north is obtained by: \[ \theta = \arctan\left(\frac{|\Delta x|}{\Delta y}\right) \approx \arctan\left(\frac{42.75}{87.5}\right) \approx \arctan(0.489) \approx 26^\circ. \] Since the \( x \)-component is negative (westward) and the \( y \)-component is positive (northward), the direction is \(\mathrm{N}26^\circ\mathrm{W}\). Thus, after 30 minutes, the pilot’s actual displacement is approximately \(97.4\,\mathrm{km}\) from the starting point on a bearing of \(\mathrm{N}26^\circ\mathrm{W}\).