An airplane undergoes the following displacements: First,it flies 66 km in a direction 30° east of north.Next,it flies 49 km due south.Finally,it flies 100 km 30° north of west.Using vector components,determine how far the airplane ends up from its starting point.
A) 79 km
B) 81 km
C) 82 km
D) 78 km
E) 76 km

Question

Quizplus · Accepted Answer

To solve this, we break each displacement into its north-south and east-west components and then sum these components separately. 1. For the first displacement: - North component = 66 km * cos(30°) = 57.2 km- East component = 66 km * sin(30°) = 33 km2. The second displacement is directly south, so:- North component = -49 km (since it's south, it's negative)- East component = 0 km3. For the third displacement:- North component = 100 km * cos(30°) = 86.6 km- West component = 100 km * sin(30°) = -50 km (since it's west, it's negative)Summing up the components:- Total north-south = 57.2 km (north) - 49 km (south) + 86.6 km (north) = 94.8 km (north)- Total east-west = 33 km (east) - 50 km (west) = -17 km (west)The total displacement is the hypotenuse of the right triangle formed by these components, which can be found using the Pythagorean theorem:$$ \sqrt{(94.8)^2 + (-17)^2} = \sqrt{8990.24 + 289} = \sqrt{9279.24} \approx 79 \, \text{km} $$

An Airplane Undergoes the Following Displacements: First,it Flies 66 Km