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3. An airplane traveling north at 220. meters per second encounters a 50.0 -meters-per-second crosswind from west to east, as represented in the diagram below. What is the resultant speed of the plane? A) 170.m/s C) 226m/s . co th B) 214m/s D) 270.m/s

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3. An airplane traveling north at 220. meters per second encounters a 50.0 -meters-per-second crosswind from west to east, as represented in the diagram below. What is the resultant speed of the plane? A) 170.m/s C) 226m/s . co th B) 214m/s D) 270.m/s

3. An airplane traveling north at 220. meters per second encounters a 50.0 -meters-per-second crosswind from west to east, as represented in the diagram below. What is the resultant speed of the plane? A) 170.m/s C) 226m/s . co th B) 214m/s D) 270.m/s

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VaughnElite · Tutor for 8 years

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To find the resultant speed of the plane, we need to consider the vector addition of the northward velocity of the plane and the eastward velocity of the crosswind. Since these two velocities are perpendicular to each other, we can use the Pythagorean theorem to find the resultant velocity.Let's denote the northward velocity of the plane as \(V_p = 220 \, \mathrm{m/s}\) and the eastward velocity of the crosswind as \(V_w = 50.0 \, \mathrm{m/s}\).Step 1: Write down the Pythagorean theorem for the velocities:\[V_r^2 = V_p^2 + V_w^2\]Step 2: Substitute the given values into the equation:\[V_r^2 = (220 \, \mathrm{m/s})^2 + (50.0 \, \mathrm{m/s})^2\]\[V_r^2 = 48400 \, \mathrm{(m/s)^2} + 2500 \, \mathrm{(m/s)^2}\]\[V_r^2 = 50900 \, \mathrm{(m/s)^2}\]Step 3: Take the square root of both sides to find the resultant velocity:\[V_r = \sqrt{50900 \, \mathrm{(m/s)^2}}\]\[V_r \approx 225.6 \, \mathrm{m/s}\]The resultant speed of the plane, rounded to the nearest whole number, is approximately 226 m/s.The correct answer is:C) \(226 \mathrm{~m} / \mathrm{s}\)
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