A private airplane left the airport at 1pm with speed 150 mph and its direction

Question

A private airplane left the airport at 1pm with speed 150 mph and its direction was 30 degrees east of north. At 3pm, the airplane reached point A and then changed its speed to 100 mph heading west. At 7pm, it reached point B. The pilot decided to change its direction to 20 degrees east of south, without changing its speed. At 9pm, the plane reached point C. Use OA = ax i + ay j to represent the displacement of the plane from 1pm to 3pm. Use AB = bx i + by j to represent the displacement of the plane from 7pm to 9pm. Use OC = dx i + dy j to represent the position of the plane at 9pm, i.e., the displacement from 1pm to 9pm, measured from the airport. We take East as positive x-direction and North as positive y-direction.

Also what are the magnitudes for each interval: OA, AB, BC, OC and what is the direction (theta from +x axis) for OC?

And what would the result look like on a graph?

Explanation / Answer

the magnitude of OA is 2*150 = 300 miles ax = 300 sin(30deg) = 150 miles ay = 300 cos(30deg) = 259.8 miles the magnitude of AB is 4*100 = 400 miles west is -x direction so, bx = -400 miles by = 0 miles the magnitude of BC is 2*100 = 200 miles cx = 200 sin(20deg) = 68.4 miles cy = 200 cos(20deg) = 187.9 miles southward = -187.9 miles now for OC: dx = ax + bx + cx = 150 -400 + 68.4 = -181.6 miles dy = ay + by + cy = 259.8 + 0 - 187.9 = 71.9 miles and the magnitude of OC can be found using the Pythagorean theorem |OC| = sqrt( dx^2 + dy^2) = sqrt( 32978.56 + 5169.61) = 195.3 miles if you need more help, let me know

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