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
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 ...
It helps the student understanding the concept of the acceleration.