2008 AMC 10B Problem 25

Michael and the truck meet whenever the truck's lead over Michael is exactly $0$. So follow that lead through time.

One cycle of the truck.

1. The truck drives $200$ feet at $10$ feet per second, taking $20$ seconds, then stops for $30$ seconds. So its pattern repeats every $50$ seconds.
2. While the truck drives ($20$ seconds), it goes $10$ feet per second and Michael goes $5$. The truck pulls ahead by $5$ feet each second: $5 \times 20 = 100$ feet.
3. While the truck is stopped ($30$ seconds), Michael closes in by $5$ feet each second: $5 \times 30 = 150$ feet.
4. At the start the truck is just leaving the next pail, so its lead is $200$ feet.

The lead, cycle by cycle. Record it at the end of each drive and each stop.

A negative lead means Michael is ahead.

Counting the meetings. Each time the lead reaches or passes through $0$ is a meeting.

1. Cycle 4, stop: the lead falls from $150$ to exactly $0$. Michael reaches the truck just as it starts again. Meeting 1. The truck then drives off ahead.
2. Cycle 5, stop: the lead falls from $100$ past $0$ to $-50$. Michael passes the parked truck. Meeting 2.
3. Cycle 6, drive: the lead rises from $-50$ past $0$ to $50$. The truck passes Michael. Meeting 3.
4. Cycle 6, stop: the lead falls from $50$ past $0$ to $-100$. Michael passes it again. Meeting 4.
5. Cycle 7, drive: the lead rises from $-100$ to exactly $0$. The truck catches Michael just as it reaches a pail and stops. Meeting 5. Then Michael walks on ahead.
6. After that, each drive raises the lead by only $100$ feet, but the lead starts each drive at $-150$ or lower. It never gets back to $0$.

They meet $5$ times.

The answer is $\boxed{\textbf{(B)}\ 5}$.

Two moving objects meet exactly when the distance between them is zero, so follow that one number instead of both positions. The truck's lead goes up and down in a repeating pattern but drifts down by $50$ feet each cycle. Checking the lead at every turning point catches every meeting, including the two where the lead just touches $0$ at a turning point without crossing. Once even the highest point of a cycle stays below $0$, there are no more meetings.

Draw a graph of distance against time. Michael is a straight line rising $5$ feet per second. The truck is a staircase: steep pieces (rising $10$ feet per second for $20$ seconds) joined by flat pieces ($30$ seconds each). Count the places where Michael's line meets the staircase, including two places where it just touches a corner. There are $5$.

Let $g(t)$ be the truck's lead over Michael, in feet. At $t = 0$, $g = 200$.

The truck's motion repeats every $50$ seconds: it drives $200$ feet in $20$ seconds, then stops for $30$ seconds. While driving, the truck gains $5$ ft/s on Michael, so $g$ rises by $100$; while stopped, Michael closes at $5$ ft/s, so $g$ falls by $150$. Net change per cycle: $-50$.

Record $g$ at the end of each drive and each stop:

Meetings are the moments $g = 0$. In cycle 4, $g$ reaches $0$ exactly as the stop ends (meeting 1). In cycle 5, $g$ goes from $100$ down through $0$ to $-50$ (meeting 2). In cycle 6, the drive takes $g$ from $-50$ up through $0$ to $50$ (meeting 3), and the stop takes it back down through $0$ to $-100$ (meeting 4). In cycle 7, the drive brings $g$ from $-100$ up to exactly $0$ (meeting 5), after which the truck stops and Michael pulls ahead. From then on, each drive raises $g$ by only $100$ while $g$ starts at $-150$ or lower, so $g$ stays negative forever.

They meet $5$ times. The answer is $\boxed{\textbf{(B)}\ 5}$.