A baseball is hit with a bat and, as a result, its direction is completely reversed and its speed is doubled. If the actual contact with the bat lasts 0.31 s , what is the ratio of the acceleration to the original velocity??
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Before the impact, let the velocity of the baseball was `v` m/s.
After being hit by the bat its velocity is `-2v`
So, change in velocity, `Deltav=v-(-2v)=3v`
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.
Therefore, `a/v=3/0.31=9.7 s^-1`
So, the ratio of acceleration of the baseball to its original velocity is 9.7.
Let the initial velocity be x m/s
Then the final velocity, v = - 2u m/s
(the minus sign indicates that the direction was reversed and doubled)
Acceleration = (v - u) / t = - 3u / t
Original velocity / Acceleration = - t / 3 = - 0.31 / 3
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