We have Car A and B, and I have the position of each car:
Car A:
$A= 1.374E2 *10^2$, $B= -2.719E2 *10^2$
$x= 1.374*10^2 -2.719*10^2...(1)$
Car B:
$A=6.473E2= 6.473*10^2, B=-4.402E2= -4.402*10^2 $
$x= 6.473*10^2t -4.402*10^2...(2)$
Use the $x(t)$ functions describing the motion of the two cars to predict where they will meet.
Graphs I drew in the tracker app:
Car A: https://i.stack.imgur.com/a10WA.png
Car B: https://i.stack.imgur.com/ghO7j.png
This is the answer I got, but I think it's wrong. I need to find it in order to find where the two cars will meet.
One fast and one slow; two meters apart.
1.374*10^2 -2.719*10^2= 6.473*10^2t- 4.402*10^2 (1)
(6.473*10^2 -1.374*10^2)t= 4.402*10^2- 2.719*10^2 (2)
t= 4.402*10^2-2.719*10^2/ 6.473*10^2- 1.374*10^2= 0.33 seconds
Using any of equations
x= 1.374*10^2(0.33)-2.719*10^2
x= -226.55 m APART