A vector A has a magnitude of 45.0 m and points in a direction 20.0° below the positive x axis.?
A vector A has a magnitude of 45.0 m and points in a direction 20.0° below the positive x axis. A second vector, B, has a magnitude of 70.0 m and points in a direction 49.0° above the positive x axis. Find the magnitude of the vector D = A - B.
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A = <45.0cos20.0°, -45.0sin20.0°>
B = <70.0cos49.0°, 70.0sin49.0°>
D = A - B
D = <45.0cos20.0° - 70.0cos49.0°, -45.0sin20.0° - 70.0sin49.0°>
D ≈ <-3.638, -68.22> m
||D|| ≈ sqrt[(-3.638)² + (-68.22)²]
D ≈ 68.32 m