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A model airplane is flying horizontally due north at $20 \mathrm{mi} / \mathrm{hr}\( when it encounters a horizontal crosswind blowing east at \)20 \mathrm{mi} / \mathrm{hr}\( and a downdraft blowing vertically downward at \)10 \mathrm{mi} / \mathrm{hr}$. a. Find the position vector that represents the velocity of the plane relative to the ground. b. Find the speed of the plane relative to the ground.

Short Answer

Expert verified
#Answer: The overall velocity of the airplane relative to the ground is represented by the position vector <20, 20, -10>, and its speed is 30 mi/hr.

Step by step solution

01

Define The Given Vectors

We are given the following three vectors - airplane's velocity (v_plane), crosswind (v_wind), and downdraft (v_downdraft). Represent them as: - v_plane = 20 mi/hr north - v_wind = 20 mi/hr east - v_downdraft = 10 mi/hr downward Let's represent them as 3D vectors: - v_plane = <0, 20, 0> - v_wind = <20, 0, 0> - v_downdraft = <0, 0, -10>
02

Find The Position Vector

To find the position vector representing the velocity of the plane relative to the ground, add the three given vectors. v_total = v_plane + v_wind + v_downdraft v_total = <0, 20, 0> + <20, 0, 0> + <0, 0, -10> Combine the elements: v_total = <0 + 20 + 0, 20 + 0 + 0, 0 + 0 + (-10)> v_total = <20, 20, -10> The position vector representing the velocity of the plane relative to the ground is <20, 20, -10>.
03

Calculate The Speed Of The Plane

To find the speed of the plane relative to the ground, calculate the magnitude of the velocity vector (v_total) that we found in step 2. Magnitude formula for 3D vectors: |v_total| = √((20)^2 + (20)^2 + (-10)^2) |v_total| = √(400 + 400 + 100) |v_total| = √(900) The speed of the plane relative to the ground is 30 mi/hr.

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