# Simple Dynamics Problem Requiring Explanation

## Homework Statement

Boxes are placed on a slope at uniform intervals of time trelease and slide down the slope with uniform acceleration. Knowing that as a box B is released, the preceding box A has slid 6 meters down the slope and that 1 second later they are 10 meters apart, determine the following:

(a) The value of trelease

(b) The acceleration of the boxes.

ΔV = a⋅Δt
Δx=Vi⋅t+½⋅a⋅t2
2⋅a⋅Δx=Vf2-Vi2

## The Attempt at a Solution

This problem isn't actually homework, I'm just studying. I'm posting the solution to the problem because I simply don't understand how exactly they arrive at the equations that they do. Here it is:

http://i.imgur.com/oSyIFsm.png

For example, equation (2) doesn't make sense to me as t = tr + 1, yet in the problem t = 1. Also, how come in eqn. (3) they don't employ the 10 m @ t=tr+1? Why is Δx simply xA@tr+1?

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When using the equations of motion listed you need to put the time interval that the particle has been under that constant acceleration.
When Box B is released: Box A has already been travelling for "tr" time. Box B has travelled for "Zero" time.
After 1 second of releasing Box B: Box A has travelled for "tr time PLUS One extra second". Box B has only travelled for "1 second".

Coming to Eq. 3:
Notice that the diagram isn't drawn properly. After 1 second both the boxes would have travelled a certain distance from their initial position. The distance between them is given to be 10m. So you need to use (xb - xa) = 10 @t=tr+1.

That really clears it up. Thank you, I appreciate that.