Question

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

Answer 1

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.