Given
The path of particle 1 is,
[tex]x(t)=3t-6,\text{ }y(t)=t^2-2t[/tex]
And, the path of second particle is,
[tex]x(t)=\sqrt{t+6},\text{ }y(t)=-3+2t[/tex]
To model the path of the two particles in cartesian form and to find whether, the two particles collide.
Explanation:
It is given that,
The path of the first particle is,
[tex]x(t)=3t-6,\text{ }y(t)=t^2-2t[/tex]
That implies,
[tex]x=2t-6,\text{ }y=t^2-2t[/tex]
Consider,
[tex]\begin{gathered} x=2t-6 \\ 2t=x+6 \\ t=\frac{x+6}{2} \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} y=(\frac{x+6}{2})^2-2(\frac{x+6}{2}) \\ y=\frac{x^2+12x+36}{4}-\frac{2x+12}{2} \\ y=\frac{x^2+12x+36-2(2x+12)}{4} \\ y=\frac{x^2+12x+36-4x-24}{4} \\ y=\frac{x^2+8x+12}{4}\text{ \_\_\_\_\_\_\_\_\_\_\lparen1\rparen} \end{gathered}[/tex]
Also, the path of second particle is,
[tex]x(t)=\sqrt{t+6},\text{ }y(t)=-3+2t[/tex]
That implies,
[tex]x=\sqrt{t+6},\text{ }y=-3+2t[/tex]
Consider,
[tex]\begin{gathered} y=-3+2t \\ 2t=y+3 \\ t=\frac{y+3}{2} \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} x=\sqrt{t+6} \\ \Rightarrow x^2=(t+6) \\ \Rightarrow x^2=(\frac{y+3}{2})+6 \\ \Rightarrow x^2=\frac{y+3+12}{2} \\ \Rightarrow2x^2=y+15 \\ \Rightarrow y=2x^2-15\text{ \_\_\_\_\_\lparen2\rparen} \end{gathered}[/tex]
Hence, y=(x^2+8x+12)/4, y=2x^2-15 are the paths of the two particles respectively.
The graph of the path of the two particles are,
From, this it is clear that the particle collide at the points (-2.686, -0.568) and (3.829, 14.324).
