Saturday

August 30, 2014

August 30, 2014

Posted by **victoria** on Sunday, January 2, 2011 at 1:06pm.

I have figured out that after the 2 seconds when the fast runner passes the slow runner, there are 4.19047 feet between the runners. I got this by:

5280 ft/360 sec = x ft/2 sec and

5280 ft/420 sec = x ft/2 sec then finding the difference in the distance the runners had travelled over those 2 seconds. However, I can't figure out what to do now.

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