Difference between revisions of "2008 AMC 12B Problems/Problem 20"
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<math>\begin{tabular}{c|c|c}Time (s) &Distance of Michael&Distance of Garbage\\ \hline | <math>\begin{tabular}{c|c|c}Time (s) &Distance of Michael&Distance of Garbage\\ \hline |
Revision as of 13:58, 7 June 2021
- 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 feet per second on a long straight path. Trash pails are located every
feet along the path. A garbage truck traveling at
feet per second in the same direction as Michael stops for
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?
Solution 1
Pick a coordinate system where Michael's starting pail is and the one where the truck starts is
.
Let
and
be the coordinates of Michael and the truck after
seconds.
Let
be their (signed) distance after
seconds.
Meetings occur whenever
.
We have
.
The truck always moves for seconds, then stands still for
. During the first
seconds of the cycle the truck moves by
feet and Michael by
, hence during the first
seconds of the cycle
increases by
.
During the remaining
seconds
decreases by
.
From this observation it is obvious that after four full cycles, i.e. at , we will have
for the first time.
During the fifth cycle, will first grow from
to
, then fall from
to
. Hence Michael overtakes the truck while it is standing at the pail.
During the sixth cycle, will first grow from
to
, then fall from
to
. 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, will first grow from
to
, then fall from
to
. Hence the truck meets Michael at the moment when it arrives to the next pail.
Obviously, from this point on will always be negative, meaning that Michael is already too far ahead. Hence we found all
meetings.
The movement of Michael and the truck is plotted below: Michael in blue, the truck in red.
Solution 2
The truck takes seconds to go from one pail to the next and then stops for
seconds at the new pail. Thus it sets off from a pail every 50 sec. Let
denote the time elapsed and write
, where
. In this time Michael has traveled
feet. What about the truck? In the first
seconds the truck covers
pails, i.e.
feet so it moves
feet from Michael's starting point. Then we have two cases:
(a) if , then the truck travels an additional
feet. For them to intersect we must have
. Solving, we get
. Since
must lie in the interval
we get
.
(b) if , then the truck travels an additional
feet. For them to intersect we must have
. Solving, we get
. Since
must lie in the interval
we get
.
Thus Michael intersects with the truck times, which is option
.
Solution 3
We make a chart by seconds in increments of ten.
Notice at 200, 240, 260, 280, and 320 seconds, Michael and the garbage truck meet. It is clear that they met, and will meet no more. Thus the answer is .
~superagh
See also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.