Let's break down the problem step by step. ## **Given:** - **Airplane's airspeed:** \(205\) km/h (**bearing N60°E**) - **Wind speed:** \(75\) km/h (**from bearing 120°**) - **Required:** Ground velocity (magnitude and direction) of the airplane --- ## **Step 1: Draw Vector Diagram** Let's start by visualizing the directions: - **N60°E:** From North, 60° towards East. - **Bearing 120°:** Bearings are measured clockwise from North. So, 120° from North is 30° South of East (i.e., 60° East of South). ### **Coordinate System** Let: - Positive y-axis = North - Positive x-axis = East --- ## **Step 2: Resolve Vectors into Components** ### **Airspeed Vector (\(\vec{V}_{PA}\))** - Magnitude: \(205\) km/h - Direction: \(N60^\circ E\) (so, \(60^\circ\) east of north). Resolve into x (east) and y (north): \[ V_{PA,x} = 205\cos(60^\circ) = 205 \times 0.5 = 102.5\, \text{km/h} \\ V_{PA,y} = 205\sin(60^\circ) = 205 \times 0.866 = 177.53\, \text{km/h} \] --- ### **Wind Vector (\(\vec{V}_{W}\))** - Magnitude: \(75\) km/h - Direction: **FROM** bearing \(120^\circ\) (but wind direction is given as the direction it comes from; so its vector points towards bearing \(300^\circ\)). - **Bearing 120°:** 120° clockwise from north. - Wind blows **FROM** 120°, so it heads **TOWARD** 300°. Convert to standard position: 300° is 60° south of west. So, the wind is **towards 300°**. Let's resolve this: - \(300^\circ = 360^\circ - 60^\circ\), so it's 60° below the positive x-axis (east). So, vector makes an angle of \(60^\circ\) below the x-axis (east), or \(30^\circ\) west of south. \[ V_{W,x} = 75\cos(300^\circ) = 75 \times 0.5 = 37.5\, \text{km/h} \] \[ V_{W,y} = 75\sin(300^\circ) = 75 \times (-0.866) = -64.95\, \text{km/h} \] --- ## **Step 3: Add the Vectors** Ground velocity = Airspeed vector + Wind vector \[ V_{G,x} = V_{PA,x} + V_{W,x} = 102.5 + 37.5 = 140\, \text{km/h} \] \[ V_{G,y} = V_{PA,y} + V_{W,y} = 177.53 + (-64.95) = 112.58\, \text{km/h} \] --- ## **Step 4: Calculate Magnitude and Direction** ### **Magnitude:** \[ |\vec{V}_G| = \sqrt{V_{G,x}^2 + V_{G,y}^2} \] \[ |\vec{V}_G| = \sqrt{140^2 + 112.58^2} = \sqrt{19600 + 12674.2564} = \sqrt{32274.2564} \approx 179.6\, \text{km/h} \] --- ### **Direction (angle east of north):** \[ \theta = \tan^{-1}\left(\frac{V_{G,x}}{V_{G,y}}\right) \] \[ \theta = \tan^{-1}\left(\frac{140}{112.58}\right) = \tan^{-1}(1.2438) \approx 51.5^\circ \] So, the direction is **N 51.5° E**. --- ## **Step 5: Vector Diagram** Here’s a sketch: ``` ^ | / | / | / | / |/ --------------------> ``` - The vertical vector is the airspeed (N60°E). - The wind vector goes roughly east-southeast. - The resultant (ground velocity) is the vector sum. --- ## **Final Answers** ### **a) Ground velocity magnitude:** **≈ 179.6 km/h** ### **b) Direction of flight:** **N 51.5° E (or 51.5° east of north)** --- ### **Summary Table** | Quantity | Value | |------------------|------------------------| | Ground velocity | 179.6 km/h | | Direction | N 51.5° E | --- **If you need a cleaner vector diagram, let me know!**