physics
posted by Megan .
This is the question to one of my physic problem.
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

westbound 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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