# Homework Help: Velocity and Time

1. Mar 22, 2010

1. The problem statement, all variables and given/known data

Boxes are placed on a chute at a uniform rate of time Tr and slide down the chute with uniform acceleration. Knowing that as any box B is released, the preceeding box A is 6m down the chute; 1 second later there is a 10m distance between them. Determine the value of Tr

2. Relevant equations

3. The attempt at a solution

As x = 0.5at2
6 = 0.5a(Tr)2

I found a to be (change in y) / (change in x) = 4 m/sec2

So, $$\sqrt{6/(0.5*0.4)}$$ = 3^(0.5) = Tr

But the answer is plain old 3...