In this challenge, we were seeking to understand where two carts, going at different speeds would collide starting from 3 meters apart.
In order to solve the challenge, we measured the rate of acceleration (5 trials averaged) for both carts and then inserted the following data into a formula we had learnt, X = vot + (1/2)at^2, and set the x's equal to each other.
Our first cart had an acceleration of .0909 m/s^2, and the second cart had an acceleration of .137 m/s^2.
Thus,when you set the formula's equal to each other:
1/2(0.909)t^2 = 1/2 (-0.137)t^2=3
Which ends up being:
0.4545t^2+ 0.0685t^2=3
Which ends up with 3 meters ( our displacement) equaling 0.4545t^2 + 0.685t^2, which ends up being 0.11395^2.
Thus, 3=0.11395t^2, divided by 0.11395 on both sides, we find t^2=25.9, wherein the square root of both sides is t=5.08s. Thus, when we let both of them run for 5.08 seconds, our prediction results in an expected collision of 1/2(0.909)(5.08^2)= collision.
Thus our collision should occur 1.2 meters from the top, 80 cm from the bottom.
When we tested it, our cart ran 1.23 meters, coming in at 77 centimeters from the bottom.
This difference, of 3.75 degrees of error, is much less than 10 percent error.
