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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: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  Re: This Distance Word Problem Is Confusing
0.05 * 5280 = 264.

September 13th, 2012, 01:37 PM  #4  
Newbie Joined: Sep 2012 From: Springfield, OR Posts: 4 Thanks: 0  Re: This Distance Word Problem Is Confusing Quote:
 
September 13th, 2012, 02:16 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 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  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: This Distance Word Problem Is Confusing Hello, kensclark15! Another approach . . . Quote:
 
September 13th, 2012, 05:26 PM  #7  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: This Distance Word Problem Is Confusing Quote:
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  
Newbie Joined: Sep 2012 From: Springfield, OR Posts: 4 Thanks: 0  Re: This Distance Word Problem Is Confusing Quote:
 
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  
Global Moderator Joined: Dec 2006 Posts: 20,927 Thanks: 2205  Quote:
Hence the hare has run 5280(0.05 + 0.05/49) feet = 264 (1 + 1/49) feet = feet.  

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