# Use the $x(t)$ functions describing the motion of the two cars to predict where they will meet.

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:

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

• Hi, welcome to MSE. Please visit this tutorial – SarGe May 19 at 15:48
• I don't think the question has anything to do with graph theory; anyway, assume that at a time interval T the position function of Car A equals that of Car B. You'll obtain a polynomial equation in terms of T, solving for which gives possible values of T. Substitute T in the position function of either of the cars to get the answer. – Manan May 19 at 15:48
• Further, which physical parameters do A and B represent? – Manan May 19 at 15:51
• This is the answer I got, but I think it's wrong. One fast and one slow; two meters apart. 1.374*10^2 -2.719*10^2= 6.473*10^2t- 4.402*10^2 (6.473*10^2 -1.374*10^2)t= 4.402*10^2- 2.719*10^2 t= 4.402*10^2-2.719*10^2/ 6.473*10^2- 1.374*10^2= 0.33 s Using any of equations x= 1.374*10^2(0.33)-2.719*10^2 x= -226.55 m APART – student May 19 at 15:53
• hello! I just saw this: math.meta.stackexchange.com/q/31368/532409 , maybe it is relevant since this question looks like something that should have the tag "homework" (not certainly "graph theory"). – Quillo May 19 at 15:58