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A pilot diverted a fighter jet 20.0 km [S 38.0° E] to avoid thunderstorm, and then flew 60.0 km [S 55.0° W] to get back on trac...Asked by nicholas do
A pilot diverted a fighter jet 20.0 km [S 38.0° E] to avoid thunderstorm, and then flew 60.0 km [S 55.0° W]
to get back on track. If the diversion lasted a total of 2.00 minutes, what was the average velocity of the
jet during the diversion?
to get back on track. If the diversion lasted a total of 2.00 minutes, what was the average velocity of the
jet during the diversion?
Answers
Answered by
oobleck
starting at (0,0) final location is at
20cis308° + 38cis215° = 42 km S26W
so the average velocity is 21km/min in the direction
20cis308° + 38cis215° = 42 km S26W
so the average velocity is 21km/min in the direction