Magnitude of the average acceleration

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SUMMARY

The average acceleration of a 49.0-g Super Ball rebounding off a brick wall can be calculated using the formula \( a = \frac{v_f - v_i}{t} \). With an initial velocity (\( v_i \)) of 27.0 m/s and a final velocity (\( v_f \)) of -19.0 m/s (considering the direction change), and a contact time of 3.65 ms (0.00365 s), the average acceleration is determined to be approximately -12,876.71 m/s². This calculation highlights the importance of correctly applying the signs for velocity when determining acceleration.

PREREQUISITES
  • Understanding of basic physics concepts, specifically Newton's laws of motion.
  • Familiarity with the formula for average acceleration.
  • Knowledge of unit conversions, particularly milliseconds to seconds.
  • Ability to interpret velocity direction in physics problems.
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  • Study the principles of momentum and impulse in physics.
  • Learn about the effects of collision dynamics on objects.
  • Explore advanced topics in kinematics, including projectile motion.
  • Investigate the use of high-speed cameras in motion analysis.
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A 49.0-g Super Ball traveling at 27.0 m/s bounces off a brick wall and rebounds at 19.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.65 ms, what is the magnitude of the average acceleration of the ball during this time interval?


I converted 3.65 ms to 0.00365 s and I know I need to use (v2-v1/t2-t1) but everytime I try to figure this problem out I am getting lost and coming up with the wrong answer somehow.
 
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nevermind I figured it out :)
 

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