Taking north as the positive direction, the displacement of the car = 10m - 5m = 5m
average velocity = displacement/total time = 5/(9 + 8) = 5/17 = 0.294m/s
Hadassah J.
asked • 02/18/20What is the magnitude of the average velocity of the car during this time interval?
A car travels 10.0 meters north in 9.0 seconds. The car then turns around and travels 5.0 meters south in 8.0 seconds. What is the magnitude of the average velocity of the car during this time interval?
Kinematic Equation which may help with this problem.
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