The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is$150\mathrm{km}/\mathrm{h}.$ The air is moving in a wind at $30.0\mathrm{km}/\mathrm{h}$ toward the north. Find the velocity of the airplane relative to the ground.

Key Concepts: Relative Velocity, Vector Addition Explanation: To find the velocity of the airplane relative to the ground, we need to use vector addition. We first find the velocity of the airplane relative to the air, then add it to the velocity of the air relative to the ground to get the velocity of the airplane relative to the ground. Step by Step Solution: Step 1. Draw a diagram of the situation, with vectors representing the velocities of the airplane relative to the air, the air relative to the ground, and the airplane relative to the ground. Step 2. Resolve the velocity of the airplane relative to the air into its components. Since the airplane is heading due west, its velocity relative to the air has no component in the north direction and a magnitude of $150\mathrm{km}/\mathrm{h}$ in the west direction. Step 3. Add the velocity of the air relative to the ground, which is moving in a wind at $30.0\mathrm{km}/\mathrm{h}$ toward the north, to the component of the airplane's velocity relative to the ground that is in the north direction. This gives a northward component of $30.0\mathrm{km}/\mathrm{h}$, and a westward component of $150\mathrm{km}/\mathrm{h}$. Use the Pythagorean theorem to find the magnitude of the resulting velocity. Step 4. The magnitude of the velocity of the airplane relative to the ground is $\sqrt{(150\mathrm{km}/\mathrm{h}{)}^2+(30.0\mathrm{km}/\mathrm{h}{)}^2}=152.8\mathrm{km}/\mathrm{h}$. The direction of the velocity is given by the inverse tangent of the ratio of the northward component to the westward component, which is ${\tan }^{-1}(30.0\mathrm{km}/\mathrm{h}/150\mathrm{km}/\mathrm{h})=11.3^{\circ }$ North of West. Final Answer: The velocity of the airplane relative to the ground is $152.8\mathrm{km}/\mathrm{h}$, $11.3^{\circ }$ North of West