Answer
a. $t=k\pi+\frac{3\pi}{8}$ where $k\geq0$, max separation $2sin(\frac{\pi}{8})\approx0.765$
b. $t=k\pi-\frac{\pi}{8}$ where $k\geq1$
Work Step by Step
a. Given the position of the two particles as $s_1=cos(t)$ and $s_2=cos(t+\frac{\pi}{4})$, we have $y=s_2-s_1=cos(t+\frac{\pi}{4})-cos(t)=-2sin(t+\frac{\pi}{8})sin(\frac{\pi}{8})$ (addition to product formula). We need to find the extrema of this function and compare their absolute values. Taking the derivative, we have $y'=-2cos(t+\frac{\pi}{8})sin(\frac{\pi}{8})$ . Thus $y'=0$ gives $t+\frac{\pi}{8}=k\pi+\frac{\pi}{2}$ (where $k$ is an integer satisfies the condition of $t\geq0$) and $t=k\pi+\frac{3\pi}{8}$ which gives $|y|=|2sin(k\pi+\frac{3\pi}{8}+\frac{\pi}{8})sin(\frac{\pi}{8})|=2sin(\frac{\pi}{8})\approx0.765$ which gives the maximum distance between the two particles.
b. The two particles collide when they have the same position; we have $s_1=s_2$ and $cos(t)=cos(t+\frac{\pi}{4})$ or $y=cos(t+\frac{\pi}{4})-cos(t)=-2sin(t+\frac{\pi}{8})sin(\frac{\pi}{8})=0$ which leads to $t+\frac{\pi}{8}=k\pi$ thus $t=k\pi-\frac{\pi}{8}$ where $k$ is an integer that satisfies the condition of $t\geq0$.
