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 September 13th, 2012, 12:35 PM #1 Newbie   Joined: Sep 2012 From: Springfield, OR Posts: 4 Thanks: 0 This Distance Word Problem Is Confusing "A tortoise crawling at a rate of 0.1mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at a rate of 5mi/h. How many feet must the hare run to catch the tortoise?" I know that you have to figure out the distance the tortoise travels which is 0.05 miles but then you need to figure out the EXTRA distance it travels once the hare starts running. I now know the answer is 533.544 feet. The only problem is that I don't understand how to get the extra distance. Could you please explain it to me? I'm also the kind of person that only understands stuff if you include detail and explain what the variables mean. And I also love this forum so far!
 September 13th, 2012, 12:38 PM #2 Newbie   Joined: Sep 2012 From: Springfield, OR Posts: 4 Thanks: 0 Re: This Distance Word Problem Is Confusing Oh no. I think I got the answer wrong. Well, I already submitted the lesson to my teacher so oh well.
 September 13th, 2012, 01:31 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond Re: This Distance Word Problem Is Confusing 0.05 * 5280 = 264.
September 13th, 2012, 01:37 PM   #4
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Re: This Distance Word Problem Is Confusing

Quote:
 Originally Posted by greg1313 0.05 * 5280 = 264.
That's just the distance that the hare needs to travel to where the tortoise was when the hare started running. When the hare runs the tortoise is still traveling so you need to find that extra distance.

 September 13th, 2012, 02:16 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond Re: This Distance Word Problem Is Confusing (1/2 + t) * 0.1 = 5t, t = 1/98, (1/2 + 1/9 * 0.1 * 5280 ? 269.3878.
September 13th, 2012, 03:34 PM   #6
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Re: This Distance Word Problem Is Confusing

Hello, kensclark15!

Another approach . . .

Quote:
 A tortoise crawling at a rate of 0.1 mph passes a resting hare. The hare rests another 30 minutes before chasing the tortoise at a rate of 5 mph. How many feet must the hare run to catch the tortoise?

$\text{The tortoise moves at }0.1\text{ mph.}

$\text{The difference of their speeds is: }\,5\,-\,0.1 \:=\:4.9\text{ mph.}
\text{It is as if the tortoise has }stopped\text{ and the hare is approaching at 4.9 mph.}$

$\text{How long does it take the hare to cover 0.05 mile at 4.9 mph?}
\text{It will take: }\,\frac{0.05}{4.9} \:=\:0.001020408\text{ hours.}$

$\text{At 5 mph, the hare runs: }5\,\times\,0.001020408 \:=\:0.005102041\text{ miles.}
\text{That is: }\:0.005102041\,\times\,5280 \:=\:269.38775\text{ feet.}$

September 13th, 2012, 05:26 PM   #7
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Re: This Distance Word Problem Is Confusing

Quote:
 Originally Posted by kensclark15 "A tortoise crawling at a rate of 0.1mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at a rate of 5mi/h. How many feet must the hare run to catch the tortoise?"
Sometimes, to "see" how it all works, good idea to make up a simple similar case, like:
Pam Anderson, travelling at a rate of 20mi/h, passes Ken. Ken wants to rest another 6 hours
before chasing Pam, at a rate of 60mi/h. How many miles must Ken travel to catch Pam?

Work out that one, then apply same steps to your problem...get my drift?

September 17th, 2012, 11:30 AM   #8
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Re: This Distance Word Problem Is Confusing

Quote:
Originally Posted by soroban
Hello, kensclark15!

Another approach . . .

Quote:
 A tortoise crawling at a rate of 0.1 mph passes a resting hare. The hare rests another 30 minutes before chasing the tortoise at a rate of 5 mph. How many feet must the hare run to catch the tortoise?

$\text{The tortoise moves at }0.1\text{ mph.}

$\text{The difference of their speeds is: }\,5\,-\,0.1 \:=\:4.9\text{ mph.}
\text{It is as if the tortoise has }stopped\text{ and the hare is approaching at 4.9 mph.}$

$\text{How long does it take the hare to cover 0.05 mile at 4.9 mph?}
\text{It will take: }\,\frac{0.05}{4.9} \:=\:0.001020408\text{ hours.}$

$\text{At 5 mph, the hare runs: }5\,\times\,0.001020408 \:=\:0.005102041\text{ miles.}
\text{That is: }\:0.005102041\,\times\,5280 \:=\:269.38775\text{ feet.}$

Thanks! I get it better from your explanation but there's one thing I don't get: Can you explain why you need to find out how long it will take the hare to run 0.05 miles at 4.9 miles per hour? And then why do you multiply that time by 5mph?

 September 17th, 2012, 01:32 PM #9 Senior Member   Joined: Jun 2011 Posts: 164 Thanks: 0 Re: This Distance Word Problem Is Confusing How about using variables? The distance of lead of the tortoise + its speed * time = speed of the hare * time t=time it took for hare to catch up to tortoise we know the lead of the tortoise because we can use its speed time the extra 30 minutes + 0.1mph * t = 5t 0.05mi+0.1mph*t=5t 0.05mi=4.9t t is about 0.010204081632 <------- (Is this some sort of coincidence? 01,02,04,08,16,32?) now we just have to find the distance, which is speed x time speed=5mph time=0.010204081632 distance is about 0.05102 miles, which is about 269.3877ft
September 17th, 2012, 04:22 PM   #10
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Quote:
 Originally Posted by wuzhe Is this some sort of coincidence? 01,02,04,08,16,32?
No. 0.010204081632... = 0.01/(1 - 0.02) = 0.05/4.9. The hare makes up the tortoise's lead of 0.05 miles at 4.9 mph, which is 49 times the speed of the tortoise, so the tortoise moves another 0.05/49 miles before being caught up.

Hence the hare has run 5280(0.05 + 0.05/49) feet = 264 (1 + 1/49) feet = $^{^{269\frac{19}{49}}}$ feet.

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# A tortoise crawling at the rate of 0.1 mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at the rate of 5 mi/h how many feet must the hare run to catch the tortoise

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