Difference between revisions of "2008 AMC 10B Problems/Problem 25"
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==Solution== | ==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. | ||
| + | |||
| + | 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{4}</math> more meetings: | ||
| + | * 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. | ||
| + | |||
| + | <asy> | ||
| + | import graph; | ||
| + | |||
| + | size(400,300,IgnoreAspect); | ||
| + | |||
| + | real[] xt={0,20,50,70,100,120,150,170,200}; | ||
| + | real[] yt={0,200,200,400,400,600,600,800,800}; | ||
| + | |||
| + | real[] xm={0,200}; | ||
| + | real[] ym={0,1000}; | ||
| + | |||
| + | draw(graph(xt,yt),red); | ||
| + | draw(graph(xm,ym),blue); | ||
| + | |||
| + | xaxis("$time$",Bottom,LeftTicks); | ||
| + | yaxis("$position$",Left,LeftTicks); | ||
| + | </asy> | ||
==See also== | ==See also== | ||
{{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 00:48, 30 January 2009
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 travels at 
feet per second in the same direction as Michael and stops for 
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?
Solution
The truck always moves for 
seconds, then stands still for . In these 
seconds, the truck will drive 
meters. In those seconds Michael will walk 
meters. So ultimately Michael will be way too far ahead of the truck for any more meetings to happen.
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 
more meetings:
- 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.
![[asy] import graph; size(400,300,IgnoreAspect); real[] xt={0,20,50,70,100,120,150,170,200}; real[] yt={0,200,200,400,400,600,600,800,800}; real[] xm={0,200}; real[] ym={0,1000}; draw(graph(xt,yt),red); draw(graph(xm,ym),blue); xaxis("$time$",Bottom,LeftTicks); yaxis("$position$",Left,LeftTicks); [/asy]](http://latex.artofproblemsolving.com/f/2/d/f2da6b0b884bec32df905352db6eefbc0c82262c.png)
See also
| 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 | ||
