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A student runs 9.0m [W] and then turns and runs 12m [N]. If he is able to do this in 5 seconds, what is his average velocity?

Respuesta :

AL2006
His displacement is √(9² + 12²) = √(81 + 144) = √(225) = 15 m
Displacement/time = 15m/5s = 3 m/s

The direction from start-point to end-point is

90° + tan⁻¹(-3/4) = 90° + tan⁻¹(-3/4) = 126.7°


The student's average velocity is 

 3 m/s in the direction 36.7° west of north.

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