Science:Math Exam Resources/Courses/MATH100/December 2014/Question 07 (a)/Solution 1

Drawn is the starting position and velocity of the particles.

Let $t$ be time in seconds. The position of the first particle is given by $(0,y)$ where $y=6-2t$ The position of the second particle is $(x,0)$ with $x=t$ The squared-distance between the particles is therefore

${\begin{aligned}d^{2}&=(\mathrm {difference\ in\ } x)^{2}+(\mathrm {difference\ in\ } y)^{2}\\&=t^{2}+(6-2t)^{2}\\&=5t^{2}-24t+36\\d&={\sqrt {5t^{2}-24t+36}}\end{aligned}}$