AMC12 2008 B

Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$. Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds. Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds. Meetings occur whenever $D(t)=0$. We have $D(0)=200$. The truck always moves for $20$ seconds, then stands still for $30$. During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$, hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$. During the remaining $30$ seconds $D(t)$ decreases by $150$. From this observation it is obvious that after four full cycles, i.e. at $t=200$, we will have $D(t)=0$ for the first time. During the fifth cycle, $D(t)$ will first grow from $0$ to $100$, then fall from $100$ to $-50$. Hence Michael overtakes the truck while it is standing at the pail. During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$, then fall from $50$ to $-100$. Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail. During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$, then fall from $0$ to $-150$. Hence the truck meets Michael at the moment when it arrives to the next pail. Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5 \Longrightarrow B}$ meetings. The movement of Michael and the truck is plotted below: Michael in blue, the truck in red.