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fast ball with a velocity of 43 m/s south. The batter hits the ball and gives it a velocity of 51 m/s north. What was the acceleration of the ball if it was in contact with the bat for 1 s? What direction did it go?


PLEASE HELP I NEED THIS DONE FOR MY NEXT CLASS

Respuesta :

AL2006

Acceleration = (change in velocity) / (time for the change)

change in velocity = (velocity after) - (velocity before)

change in velocity = (51 m/s north) - (43 m/s south)

that's the same thing as (51 m/s north) + (43 m/s north)

change in velocity = 94 m/s north

Acceleration = (94 m/s north) / (1 second)

Acceleration = 94 m/s² north

The direction of the acceleration is North.

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