Kenneth S. answered • 10/02/16
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They each cover a part of the distance 450 cm.
Slow snail goes at x cm/min, so faster one goes at x+22 cm/min. Each goes a particular distance governed by
distance = rate multiplied by time. So we compute their distances, add them, and we have this equation:
x(25) + (x+22)25 = 450.
You can solve for x, now.
Kara L.
That equation gave me X=-2?
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10/03/16
David W.
To elaborate: At 22 cm/min, a snail would crawl 550 cm in 25 minutes (the other snail crawls 0). That's too much!
Perhaps the rates differ by 2 cm/min [this is much more reasonable bevsudr it gives a reasonable answer].
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10/03/16
David W.
8 and 10
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10/03/16
Kara L.
THANK YOU, DAVID W!!! Bless your heart
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10/03/16
Kenneth S.
A very fine example of a problem that has a numerical answer, but it's unexpected. Still, it works, in the sense that one is actually traveling at a negative speed, i.e. going backwards slowly. They were FACING one another. They met 50 meters behind the original position of the backward-moving snail. It's a word problem--results have to be interpreted and checked with reality.
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10/03/16
David W.
But the problem clearly says, "They crawl toward each other...".
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10/03/16
Kenneth S.
Well, then, David, the answer HAS TO BE THIS IS AN IMPOSSIBLE PROBLEM. (If we believe the towards one another aspect, then we must also believe the 22 cm/min difference in rates.)
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10/04/16
David W.
10/03/16