Answer
a. $t=\frac{\pi}{3}, \frac{4\pi}{3}$
b. $t=\frac{5\pi}{6}, \frac{11\pi}{6}$, $1m$
c. $t=\frac{\pi}{3}, \frac{4\pi}{3}$, $1m/sec$
Work Step by Step
a. When the two particles meet, they should have the same position $s_1=s_2$ or $sin(t)=sin(t+\frac{\pi}{3})$. Using the Sum to Product Formula, we have $sin(t+\frac{\pi}{3})-sin(t)=2sin(\frac{\pi}{6})cos(t+\frac{\pi}{6})=0$ which gives solutions at $t=\frac{\pi}{3}sec, \frac{4\pi}{3}sec$ over the domain of $[0,2\pi]$.
b. We need to find the extrema of $D=s_2-s_1=sin(t+\frac{\pi}{3})-sin(t)$ and then choose the largest absolute value. Take the derivative and set to zero to get $D'=cos(t+\frac{\pi}{3})-cos(t)=0$ or $sin(\frac{\pi}{6})sin(t+\frac{\pi}{6})=0$ which gives solutions at $t=\frac{5\pi}{6}, \frac{11\pi}{6}$ corresponding to $|D1|=1$ and $|D2|=1$. Thus the farthest apart that the particles ever get is $1m$.
c. The change of distance between the particles can be represented by $D'$; we need to find the maximum value of $D'$ (again compare absolute values). Take its derivative to get $D''=-sin(t+\frac{\pi}{3})+sin(t)=-2sin(\frac{\pi}{6})cos(t+\frac{\pi}{6})$. Let $D''=0$; we can get $t=\frac{\pi}{3}sec, \frac{4\pi}{3}sec$ over the domain of $[0,2\pi]$. $|D'_1|=|D'_2|=1m/sec$ and we can check the sign changes of $D''$ across these critical points to see that they give a local maximum.
