2008 AMC 10B Problem 25

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 $\boxed{\textbf{(B)}\ 5}$.

Two movers meet whenever their separation is zero, so the problem is about the sign changes of a piecewise-linear function with a repeating pattern and a steady downward drift. Bookkeeping the gap at the turning points of each cycle captures every crossing, including the two "grazing" meetings where the gap touches zero at a cycle boundary. Once the peaks of $g$ drop below zero, no further meetings are possible.

Plot distance versus time: Michael is a line of slope $5$; the truck is a staircase of slope-$10$ segments and horizontal rests. Counting the intersections of the line with the staircase, including tangential touches at corners, gives $5$.

• Assuming that once Michael catches the truck he stays ahead, or missing the two meetings where the gap only touches zero at the end of a stop or a drive.
• Counting the drift of $50$ feet per cycle and answering $200/50 = 4$, which ignores the back-and-forth within each cycle.