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acceleration of a superball

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A 50.0g superball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50ms, what is the magnitude of the average acceleration of the ball during this time interval? (note 1 ms= 10^-3 s.) Textbook answer to this problem is 1.34*10^4 m/s^2

now i know that the average acceleration is = delta V/delta T= (Vf-Vi)/(Tf-Ti).

so i tried (22-25)/(.0035)=-857 i came up with this wrong answer

then i said to my self since the ball started at 25.0 m/s then hits the wall, it's velocity should be 0. therfore i reason that 0 should be the intial velocity and 22m/s is the final

so i tried (22-0)/(.0035)=6285 wrong answer again

Please explain to me why i am off and help me understand acceleration so i wont make this mistake on a exam.

thank you very much

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Solution Summary

It helps the student understanding the concept of the acceleration.

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Hi, I think you used the correct formula: the average acceleration is = delta V/delta T= (Vf-Vi)/(Tf-Ti).

The only thing you did not correctly is that the velocity of the ball has changed the direction before and after ...

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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