Physics help needed right now

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EditThe following data table gives the position of an object during freefall. The time interval between adjacent points is 1/60 seconds.  

Point Position (cm) 
1 94.30 
2 91.53 
3 88.48 
4 85.17 
5 81.58 
6 77.72 
7 73.58 
8 69.17 
9 64.49 
10 59.54 

What is the instantaneous velocity (in cm/s) of the object when it is at position 3?

Sep 28th, 2015

Thank you for the opportunity to help you with your question!

position 3 is in 88.48cm

position 2 is in 91.53cm

so the object has moved 91.53-88.48=3.05 cm in 1/60 of a second as the intervals are 1/60 of asecond

so velocity is 3.05cm divided by 1/60s ( Velocity is distance divided by time) so velocity is 60 x 3.05=183 cm/s

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 28th, 2015

Hello thanks for your help! I thought thats what the answer was as well but sadly the computer keeps saying no. Any other ideas?

Sep 28th, 2015



Ok harder than we thought. It is possible that your computer wants you to do position 3 and 4 instead of 2 and 3 or both and get an average of it but attached find calculating the exact velocity. Please let me know if you have any question. I hope it is not late.



Sep 28th, 2015

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