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−  ==Problem==
 +  #redirect [[2008 AMC 12B Problems/Problem 20]] 
−  Michael walks at the rate of <math>5</math> feet per second on a long straight path. Trash pails are located every <math>200</math> feet along the path. A garbage truck travels at <math>10</math> feet per second in the same direction as Michael and stops for <math>30</math> seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving for the next pail. How many times will Michael and the truck meet?
 
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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>
 
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−  ==Solution==
 
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−  The truck always moves for <math>20</math> seconds, then stands still for <math>30</math>. In these <math>50</math> seconds, the truck will drive <math>200+0=200</math> meters. In those <math>50</math> seconds Michael will walk <math>250</math> meters. So ultimately Michael will be way too far ahead of the truck for any more meetings to happen.
 
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−  The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. We can easily verify that indeed there will be <math>\boxed{5}</math> more meetings: (the graph of the truck should start 200 units higher)
 
−  * Michael will catch and overtake the truck while it is standing at the first pail.
 
−  * The truck will start moving again and on its way to the second pail it will overtake Michael.
 
−  * While the truck is standing at the second pail, Michael will walk past it.
 
−  * The last meeting will occur exactly when both Michael and the truck arrive at the same time to the third pail.
 
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−  <asy>
 
−  import graph;
 
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−  size(400,300,IgnoreAspect);
 
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−  real[] xt={0,20,50,70,100,120,150,170,200};
 
−  real[] yt={0,200,200,400,400,600,600,800,800};
 
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−  real[] xm={0,200};
 
−  real[] ym={0,1000};
 
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−  draw(graph(xt,yt),red);
 
−  draw(graph(xm,ym),blue);
 
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−  xaxis("$time$",Bottom,LeftTicks);
 
−  yaxis("$position$",Left,LeftTicks);
 
−  </asy>
 
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−   
−  ==See also==
 
−  {{AMC10 boxyear=2008ab=Bnumb=24after=Last Question}}
 