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November 26, 2014

November 26, 2014

Posted by **Greg** on Tuesday, April 29, 2008 at 11:14pm.

0 m/s squared

20 m/s squared

40.82 m/s squared

need more information

wouldnt it be need more information because isnt acceleration ending velocity minus starting velocity divided by the time, or can it be 20 m/s squared where 400 was over 20 s

- Physics -
**Quidditch**, Tuesday, April 29, 2008 at 11:23pmYou have the right idea. Yes, the AVERAGE acceleration is just as you stated, (ending velocity - starting velocity)/time. The problem says the starting and ending velocities are the same (always 400m/s). (400m/s - 400m/s) is 0. So the acceleration during this time is 0.

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