Difference between revisions of "2008 AMC 12B Problems/Problem 20"
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<math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8</math> | <math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8</math> | ||
− | The | + | ==Solution== |
+ | |||
+ | Pick a coordinate system where Michael's starting pail is <math>0</math> and the one where the truck starts is <math>200</math>. | ||
+ | Let <math>M(t)</math> and <math>T(t)</math> be the coordinates of Michael and the truck after <math>t</math> seconds. | ||
+ | Let <math>D(t)=T(t)-M(t)</math> be their (signed) distance after <math>t</math> seconds. | ||
+ | Meetings occur whenever <math>D(t)=0</math>. | ||
+ | We have <math>D(0)=200</math>. | ||
+ | |||
+ | The truck always moves for <math>20</math> seconds, then stands still for <math>30</math>. During the first <math>20</math> seconds of the cycle the truck moves by <math>200</math> meters and Michael by <math>100</math>, hence during the first <math>20</math> seconds of the cycle <math>D(t)</math> increases by <math>100</math>. | ||
+ | During the remaining <math>30</math> seconds <math>D(t)</math> decreases by <math>150</math>. | ||
+ | |||
+ | From this observation it is obvious that after four full cycles, i.e. at <math>t=200</math>, we will have <math>D(t)=0</math> for the first time. | ||
+ | |||
+ | During the fifth cycle, <math>D(t)</math> will first grow from <math>0</math> to <math>100</math>, then fall from <math>100</math> to <math>-50</math>. Hence Michael overtakes the truck while it is standing at the pail. | ||
+ | |||
+ | During the sixth cycle, <math>D(t)</math> will first grow from <math>-50</math> to <math>50</math>, then fall from <math>50</math> to <math>-100</math>. 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, <math>D(t)</math> will first grow from <math>-100</math> to <math>0</math>, then fall from <math>0</math> to <math>-150</math>. Hence the truck meets Michael at the moment when it arrives to the next pail. | ||
+ | |||
+ | Obviously, from this point on <math>D(t)</math> will always be negative, meaning that Michael is already too far ahead. Hence we found all <math>\boxed{5}</math> 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("$time$",Bottom,LeftTicks); | ||
+ | yaxis("$position$",Left,LeftTicks); | ||
+ | </asy> | ||
+ | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2008|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2008|ab=B|num-b=19|num-a=21}} | ||
{{AMC10 box|year=2008|ab=B|num-b=24|after=Last Question}} | {{AMC10 box|year=2008|ab=B|num-b=24|after=Last Question}} |
Revision as of 00:16, 27 November 2009
- The following problem is from both the 2008 AMC 12B #20 and 2008 AMC 10B #25, so both problems redirect to this page.
Problem
Your mom 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 travels at
feet per second in the same direction as your mom and stops for
seconds at each pail. As your mom passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will your mom and the truck meet?
Solution
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
meters 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.
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 Question | |
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 |