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September 20, 2014

September 20, 2014

Posted by **Megan** on Thursday, August 30, 2012 at 1:56pm.

A person standing outside their house is passed by a runner heading West at a steady pace of 3.3 m/s. Right when the runner passes by, the person accelerates towards the south at a rate of 0.36 m/s2 for 1.2 seconds. Determine how far apart the people are when t = 1.2 seconds.

I tried distance = 1/2at^2 but that did not work. Any help would be greatly appreaciated

- physics -
**Steve**, Thursday, August 30, 2012 at 4:37pmwestbound runner travels 3.3*1.2 = 3.96

southbound person travels 1/2 (.36)(1.2^2) = .26

d = √(3.96^2+.26^2) = 3.97

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