Answer: B

Solution:

Number the pails so Michael is at pail $0$ and the truck at pail $1$ at time $0.$ Michael reaches pail $n$ at $40n$ seconds. The truck spends $20$ seconds between pails and $30$ stopped, so it leaves pail $n$ at $50(n - 1)$ seconds and (for $n \ge 2$) arrives at $50(n - 1) - 30.$

Michael is at pail $n$ while the truck is there exactly when $50(n-1) - 30 \le 40n$ $\le 50(n-1),$ which simplifies to $5 \le n \le 8.$ So they meet at pail $5$ (at $t = 200,$ as the truck departs), pail $6$ ($t = 240$), pail $7$ ($t = 280$), and pail $8$ ($t = 320,$ as the truck arrives).

Between pails $6$ and $7$ the truck (moving at $10$ ft/s) pulls ahead of and is then overtaken by Michael once more, adding one crossing. In all, they meet $5$ times.

Thus, the correct answer is B.