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January 28, 2015

January 28, 2015

Posted by **Megan** on Thursday, August 30, 2012 at 1:57pm.

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 -
**Megan Lee**, Thursday, August 30, 2012 at 2:06pmI used the equation above since the initial velocity was 0 same goes for the initial distance. For some reason it is wrong.

- physics -
**bobpursley**, Thursday, August 30, 2012 at 3:17pmThey are traveling at right angles.

distance=sqrt(W^2+S^2) where W, S are the distances west, and south.

distanceW=3.3*1.2

distanceS=1/2 .36*1.2^2

solve for distance from the first relation above.

- physics -
**Henry**, Saturday, September 1, 2012 at 8:02pmpost it.

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