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
3 answers
I used the equation above since the initial velocity was 0 same goes for the initial distance. For some reason it is wrong.
They 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.
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.
post it.
1 answer