2008 AMC 12B Problems/Problem 20

The following problem is from both the 2008 AMC 12B #20 and 2008 AMC 10B #25, so both problems redirect to this page.

Problem

Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck intersect?

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$

Solution 1

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.

$[asy] import graph; size(400,300,IgnoreAspect); real[] xt = new real[21]; real[] yt = new real[21]; for (int i=0; i<11; ++i) xt[2*i]=50*i; for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20; for (int i=0; i<11; ++i) yt[2*i]=200*(i+1); for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2); real[] xm={0,500}; real[] ym={0,2500}; draw(graph(xt,yt),red); draw(graph(xm,ym),blue); xaxis("$\mathrm{time}$",Bottom,LeftTicks); yaxis("$\mathrm{position}$",Left,LeftTicks); [/asy]$

Solution 2

Number the trash pails $0, 1, 2, 3, \ldots$ , with 0 being the pail where the truck is at the beginning. For $t \in (50k, 50k+20)$ the truck is traveling between the $k$ th and $(k+1)$ th trash pails at 10 ft/sec. Let $\Delta = t-50k \in (0,20)$ . The truck's position at time $t$ (relative to Michael's starting point) is given by $d = 200 + 200k + 10\Delta$ . Michael's position at this time is of course $5t=5\Delta + 250k$ . If Michael passes the truck at this time then we must have $200 + 200k + 10\Delta = 5\Delta + 250k$ . Solving for $\Delta$ , we get $\Delta = 10k - 40$ . Since $\Delta$ must lie in the interval $(0,20)$ we get $k =5$ .

For $t \in (50k+20, 50(k+1))$ the truck stops at the $(k+1)$ th trash pail and its position is thus given by $200+200(k+1)$ . Let $\Delta = t-50k-20 \in [0,30]$ . If Michael passes the truck at this time then we must have $400 + 200k = 5t = 5\Delta + 250k + 100$ . Solving for $\Delta$ , we get $\Delta = 10(6-k)$ . Since $\Delta$ must lie in the interval $[0,30]$ we get $k \in \{3, 4, 5, 6\}$ .

Thus Michael intersects with the truck 5 times, which is option $\boxed{\textbf{(B)}}$ .

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 12 Problems and Solutions

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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2008 AMC 12B Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See also