Amy and Allison are both running toward the soccer ball during a game.Amy's path is represented by the parametric equations x(t)=40 + 2/3t, y(t) = 6 + 1/6t, where t is on the interval [0,50] and t is measured in tenths of seconds.Allison's path is represented by the parametric equations x(t)=12 + 2t, y(t)= 5+ 1/4t, where t is on the interval [0,50] and t is measured in tenths of seconds.

[tex]\begin{gathered} Amy\begin{cases}x(t)=40+\frac{2}{3}t \\ y(t)={6+\frac{1}{6}}t\end{cases} \\ Allison\begin{cases}x(t)={12+2t} \\ y(t)={5+\frac{1}{4}t,}\end{cases} \end{gathered}[/tex]

If the two girsl collide at a certain t, then

[tex]\begin{gathered} x_{Amy}(t)=x_{Allison}(t) \\ y_{Amy}(t)=y_{All\imaginaryI son}(t) \end{gathered}[/tex]

Then, we can equal each pair of euqations to find if there is any time t when their positions are the same.

Then:

[tex]x_{Amy}(t)=x_{All\imaginaryI son}(t)\Rightarrow40+\frac{2}{3}t={12+2t}[/tex]

And solve for t:

[tex]\begin{gathered} 40+\frac{2}{3}t=12+2t \\ 40-12=2t-\frac{2}{3}t \\ 28=\frac{4}{3}t \\ t=28÷\frac{4}{3}=28\cdot\frac{3}{4} \\ t=21s \end{gathered}[/tex]

Now, if the girls collide, at t = 21s they will be at the same place in in y(t)

Let's see:

[tex]y_{Amy}(21s)={6+\frac{1}{6}}\cdot21[/tex]

And solve:

[tex]y_{Amy}(21s)=6+\frac{7}{2}=9.5[/tex]

Now, for Allison:

[tex]y_{All\mathrm{i}son}(21)=5+\frac{1}{4}\cdot21[/tex]

And solve:

[tex]y_{All\imaginaryI son}(21)=5+5.25=10.25[/tex]

Then, we have:

At t = 21s the two girls are at the same position in the x axis. But, at 21s, in the y axis, Amy is at 9.5 and Allison at 10.25

To collide, they have to be in the same position in the x and y axis at the same time.

Thus, they never collide.