Friday
March 24, 2017

Post a New Question

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

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

  • physics - , Thursday, August 30, 2012 at 4:37pm

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question