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

1 answer

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
Similar Questions
  1. 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
    1. answers icon 3 answers
  2. Question content area topPart 1 Suppose that two people standing 3 miles apart both see the burst from a fireworks display.
    1. answers icon 1 answer
  3. Question content area topPart 1 Suppose that two people standing 3 miles apart both see the burst from a fireworks display.
    1. answers icon 1 answer
    1. answers icon 2 answers
more similar questions