What Is the Average Acceleration of a Baseball Hit by a Bat?

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SUMMARY

The average acceleration of a baseball hit by a bat, as calculated from the given data, is determined using the formula a = (Vf - Vi) / t. In this case, the final velocity (Vf) is -56 mi/hr, the initial velocity (Vi) is 96 mi/hr, and the contact time (t) is 0.75 milliseconds (7.5 x 10^-4 seconds). The correct calculation yields an average acceleration of -2.03 x 10^-3 mi/hr/s, which is significantly different from the erroneous value of 2.0 x 10^5 mi/hr/s mentioned in the discussion.

PREREQUISITES
  • Understanding of basic kinematics and acceleration calculations
  • Familiarity with unit conversions, specifically between miles per hour and meters per second
  • Knowledge of the formula for average acceleration: a = (Vf - Vi) / t
  • Ability to perform calculations involving scientific notation
NEXT STEPS
  • Study the concept of average acceleration in physics
  • Learn about unit conversion techniques, particularly between mi/hr and m/s
  • Explore kinematic equations and their applications in real-world scenarios
  • Review common mistakes in physics calculations and how to avoid them
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of acceleration calculations in sports contexts.

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Homework Statement

Need assistance with this problem.

Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 96 mi/hr sinking curve ball head on, sending it off his bat in the exact opposite direction at 56 mi/hr. The actually contact between ball and bat lasted for 0.75 milliseconds. Determine the magnitude of the average acceleration of the ball during the contact with the bat. Express your answer in both mi/hr/s and in m/s/s. (Given: 1.00 m/s = 2.24 mi/hr)

Vf=-56mi/hr Vi=96mi/hr
t=7.5×10^4s

Homework Equations


a=v÷t

The Attempt at a Solution


(-56-96)÷7.5×10^4=-2.03×10^-3mi/hr/s

The answer to the problem on the link http://www.physicsclassroom.com/calcpad/1dkin/prob8.cfm says 2.0 x 10^5 mi/hr/s for one of the answers.
 
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I think you simply plugged it in wrong, (-56-96)÷7.5×10^4 = 2.0 x 10^5
 

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