how to set this up? 3mph walking away; 6mph running back; 30 min total time; find d

allegansveritatem

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Here is the problem:



29. A person walked from home at the rate of three miles per hour (mph), and then ran home along the same route at 6 mph. If the round trip took thirty minutes, how far did the person walk?



I tried to do this thus: let x = time walked at 3 mph let 0.5x = time at 6 mph

3x + 6(0.5x) = 30

3x+ 3x= 30

x=5

?
I don't see how this can be right. I mean if someone walks for 5 minutes at 3 miles an hour, he will have walked 1/12th of a mile....What is wrong with this other than plenty?
 

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Here is the problem:



29. A person walked from home at the rate of three miles per hour (mph), and then ran home along the same route at 6 mph. If the round trip took thirty minutes, how far did the person walk?



I tried to do this thus: let x = time walked at 3 mph let 0.5x = time at 6 mph

3x + 6(0.5x) = 30

3x+ 3x= 30

x=5

?
I don't see how this can be right. I mean if someone walks for 5 minutes at 3 miles an hour, he will have walked 1/12th of a mile....What is wrong with this other than plenty?
''Just a fun example. It's not quite done. My point is for you to notice that you can ALWAYS be more organized and formal...

Distance = Rate * Time -- Every time. Never anything else.


Trip Out Formula: D1 = R1 * T1
Trip Back Formula: D2 = R2 * T2

Given #1: D1 = D2
Given #2: T1 + T2 = 30 min = ½ hr è T1 = ½ hr – T2
Given #3: D1 = 3 mph * T1
Given #4: D2 = 6 mph * T2

Addition: D1 + D2 = 3 mph * T1 + 6 mph * T2
Distributive Property: D1 + D2 = 3 mph (T1 + 2*T2)
Substitution From Given #1: D1 + D2 = D2 + D1 = 2*D2 = 3 mph (T1 + 2*T2)


Substitution From Given #2 Conclusion: 2*D2 = 3 mph (T1 + 2*T2) = 3 mph (½ hr – T2 + 2*T2) = 3 mph (½ hr + T2)
Algebra: D2 = 3 mph (¼ hr + ½*T2)
Equality Principle from Given #4: 6 mph * T2 = 3 mph (¼ hr + ½*T2)
Division: Equality Principle from Given #4: 2 * T2 = ¼ hr + ½*T2
Time to Solve: 8*T2 = 1 hr + 2*T2 ==> 6*T2 = 1 hr ==> T2 = 1/6 hr = 10 min
From Given #2: T1 = 30 min – T2 = 30 min – 10 min = 20 min
 
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Here is the problem:



29. A person walked from home at the rate of three miles per hour (mph), and then ran home along the same route at 6 mph. If the round trip took thirty minutes, how far did the person walk?



I tried to do this thus: let x = time walked at 3 mph let 0.5x = time at 6 mph

3x + 6(0.5x) = 30

3x+ 3x= 30

x=5

?
I don't see how this can be right. I mean if someone walks for 5 minutes at 3 miles an hour, he will have walked 1/12th of a mile....What is wrong with this other than plenty?

More useful answer:
1) What are the units on 'x'?
2) What are the units on '30'?
 
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I mean if someone walks for 5 minutes at 3 miles an hour, he will have walked 1/12th of a mile....
I understand that 5 minutes is 1/12 of an hour but you seem to be saying that regardless of your speed you will go 1/12 of a mile in the 5 minutes. Does this really make sense? 60 miles per hour is 1 mile per minute so in 5 minutes you traveled 1/12 of a mile?
 
More useful answer:
1) What are the units on 'x'?
2) What are the units on '30'?

they should be minutes, no? 3 miles for so many minutes. 6 miles for so many minutes. And 30 means thirty minutes. The reason I used .05x is that the guy waling 6 mph is going to take half the time to go the same distance as the 3 mph guy. Does this make sense?
 
I understand that 5 minutes is 1/12 of an hour but you seem to be saying that regardless of your speed you will go 1/12 of a mile in the 5 minutes. Does this really make sense? 60 miles per hour is 1 mile per minute so in 5 minutes you traveled 1/12 of a mile?

Well, it means that the guy walked 3 miles an hour for 1/12th of an hour--accoding to the working out of the equation. So he walked, what? 1/4th of a mile? The book says the answer is 1 mile.
 
''Just a fun example. It's not quite done. My point is for you to notice that you can ALWAYS be more organized and formal...

Distance = Rate * Time -- Every time. Never anything else.


Trip Out Formula: D1 = R1 * T1
Trip Back Formula: D2 = R2 * T2

Given #1: D1 = D2
Given #2: T1 + T2 = 30 min = ½ hr è T1 = ½ hr – T2
Given #3: D1 = 3 mph * T1
Given #4: D2 = 6 mph * T2

Addition: D1 + D2 = 3 mph * T1 + 6 mph * T2
Distributive Property: D1 + D2 = 3 mph (T1 + 2*T2)
Substitution From Given #1: D1 + D2 = D2 + D1 = 2*D2 = 3 mph (T1 + 2*T2)


Substitution From Given #2 Conclusion: 2*D2 = 3 mph (T1 + 2*T2) = 3 mph (½ hr – T2 + 2*T2) = 3 mph (½ hr + T2)
Algebra: D2 = 3 mph (¼ hr + ½*T2)
Equality Principle from Given #4: 6 mph * T2 = 3 mph (¼ hr + ½*T2)
Division: Equality Principle from Given #4: 2 * T2 = ¼ hr + ½*T2
Time to Solve: 8*T2 = 1 hr + 2*T2 ==> 6*T2 = 1 hr ==> T2 = 1/6 hr = 10 min
From Given #2: T1 = 30 min – T2 = 30 min – 10 min = 20 min

It is 1 oclock in the morning and I am not up to understanding this right now. I will have to look at it again in the morning. I can say right now that I don't know what the è means at all.
 
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It is 1 oclock in the morning and I am not up to understanding this right now. I will have to look at it again in the morning. I can say right now that I don't know what the è means at all.
It is repaired. No more e-accent.

Minutes, hours? It makes no difference. Just be consistent. Get the units right and a lot of things just take care of themselves.
 
It is repaired. No more e-accent.

Minutes, hours? It makes no difference. Just be consistent. Get the units right and a lot of things just take care of themselves.

I wet this again today. Yes. The way I have set up the problem doesn't make sense. I am looking for distance, not m minutes.So the 30 shouldn't even be there
 
Al, the point tkhunny was making is that setting things up is easier if you think in small steps.

First, I haul out my handy little formula of d = r * t TWICE, once for walking and once for running. Note the distance is the same in both cases, but also note that the formula only works with consistent units.

\(\displaystyle d = \dfrac{3}{60} * t_1 \text { and } d = \dfrac{6}{60} * t_2 \implies \dfrac{3t_1}{60} = \dfrac{6t_2}{60} \implies 3t_1= 6t_2 \implies t_1 = 2t_2.\)

One equation but two unknowns? Do I have any other numeric information about time? Yes.

\(\displaystyle t_1 + t_2 = 30 \text { but } t_1 = 2t_2 \implies 2t_2 + t_2 = 30 \implies 3t_2 = 30 \implies t_2 = 10.\)

\(\displaystyle \therefore d = \dfrac{6}{60} * 10 = 1.\)

EDIT: It is easier of course to adjust the time units than the rate units. Then we get

\(\displaystyle d = 3t_1 \text { and } d = 6t_2 \implies t_1 = 2t_2.\)

But using hours rather than minutes we have

\(\displaystyle t_1 + t_2 = \dfrac{1}{2} \implies 2t_2 + t_2 = \dfrac{1}{2} \implies 3t_2 = \dfrac{1}{2} \implies t_2 = \dfrac{1}{6}.\)

\(\displaystyle \therefore d = 6 * \dfrac{1}{6} = 1.\)

It makes no difference whether we use hours or minutes so long as we are consistent.
 
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My additional point, though unspoken, is to be able to be so thorough and so complete and so organized that you NEVER have to say, "I have no idea where to start." or "I just got all confused." or "I don't see where I went wrong."
 
Darn it Al, just solve this for d:
d/3 + d/6 = 1/2

1st step: multiply the equation by 6

I have doen this thus:

d/3 + d/6 = 1/2

2d +d = 3

d= 1

But...I have a question: How do I connect this answer, which according to the book is correct, with the walker and not the runner?
 
Al, the point tkhunny was making is that setting things up is easier if you think in small steps.

First, I haul out my handy little formula of d = r * t TWICE, once for walking and once for running. Note the distance is the same in both cases, but also note that the formula only works with consistent units.

\(\displaystyle d = \dfrac{3}{60} * t_1 \text { and } d = \dfrac{6}{60} * t_2 \implies \dfrac{3t_1}{60} = \dfrac{6t_2}{60} \implies 3t_1= 6t_2 \implies t_1 = 2t_2.\)

One equation but two unknowns? Do I have any other numeric information about time? Yes.

\(\displaystyle t_1 + t_2 = 30 \text { but } t_1 = 2t_2 \implies 2t_2 + t_2 = 30 \implies 3t_2 = 30 \implies t_2 = 10.\)

\(\displaystyle \therefore d = \dfrac{6}{60} * 10 = 1.\)

EDIT: It is easier of course to adjust the time units than the rate units. Then we get

\(\displaystyle d = 3t_1 \text { and } d = 6t_2 \implies t_1 = 2t_2.\)

But using hours rather than minutes we have

\(\displaystyle t_1 + t_2 = \dfrac{1}{2} \implies 2t_2 + t_2 = \dfrac{1}{2} \implies 3t_2 = \dfrac{1}{2} \implies t_2 = \dfrac{1}{6}.\)

\(\displaystyle \therefore d = 6 * \dfrac{1}{6} = 1.\)

It makes no difference whether we use hours or minutes so long as we are consistent.

I confess this is puzzling to me. Where do you get the 60 from in:

. . .d = \dfrac{3}{60} * t_1 \text { and } d = \dfrac{6}{60} * t_2 \implies \dfrac{3t_1}{60} = \dfrac{6t_2}{60} \implies 3t_1= 6t_2 \implies t_1 = 2t_2.

and what does "text" mean? I guess dfrac means something like defractionate.
 
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I confess this is puzzling to me. Where do you get the 60 from in:
Where others did everything in hours, JeffM used minutes. The "60" is "60 minutes per hour" so that "3 miles per hour" became "\(\displaystyle \frac{3}{60}= \frac{1}{20}\) miles per minute".

and what does "text" mean? I guess dfrac means something like defractionate.
In Latex "\text" allows you to put actual text words in the math. "\dfrac" is almost identical to "\frac" showing a fraction.

"text": \(\displaystyle \text{This is an exponential function: }f(x)= e^x\)

"\frac": \(\displaystyle \frac{A}{B}\)

"\dfrac": \(\displaystyle \dfrac{A}{B}\)

Click on "Reply with Quote" to see the Latex code.
 
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I confess this is puzzling to me. Where do you get the 60 from in:
\(\displaystyle d = \dfrac{3}{60} * t_1 \text { and } d = \dfrac{6}{60} * t_2 \implies \dfrac{3t_1}{60} = \dfrac{6t_2}{60} \implies 3t_1= 6t_2 \implies t_1 = 2t_2.\)

and what does "text" mean? I guess dfrac means something like defractionate.
I had said that you have to make the units consistent. The problem as stated gives rates in miles per hour but gives time in minutes. The formula of d = r * t only works if the units of measurement are consistent. You can adjust the time so it is in hours, which changes 30 minutes to 1/2 hour. Or you can change the rates so that they are in miles per minute. If it takes an hour to go 3 miles, then it takes a minute to go 3/60 miles. As my edit shows, you get the same answer either way.

\dfrac indicates a fraction in LaTeX and \text indicates plain text. You left the initial bracket off your quotation. Therefore the parser did not work properly. I have fixed it for you.
 
I have doen this thus:

d/3 + d/6 = 1/2

2d +d = 3

d= 1

But...I have a question: How do I connect this answer, which according to the book is correct, with the walker and not the runner?
The problem requires careful attention to what is asked. If we define d as the distance walked, the total distance both ways is 2d. But you are NOT asked to find the total distance both ways. You are asked to find the distance one way even though you are only given the time both ways.

Denis attacked the problem by defining the distance walked as d, which is also the distance run, and then by focusing on the time in hours. The spent walking was d/3. The time spent running was d/6. The total time was the sum of the time spent walking and time spent running giving

\(\displaystyle \dfrac{d}{3} + \dfrac{d}{6} = \dfrac{1}{2} \implies 2d + d = 3 \implies d = 1.\)

Now you could define d as the total distance. In which case the distance walked is 0.5d, and the distance run is 0.5d as well. Using Denis's method of focusing on time but a different definition of d, we get

\(\displaystyle \dfrac{0.5d}{3} + \dfrac{0.5d}{6} = \dfrac{1}{2} \implies d + 0.5d = 3 \implies 1.5d = 3 \implies d = 2.\)

But that gives the total distance, not the distance walked. So the answer 0.5 * 2 = 1.

There are numerous correct ways to set the problem up. The reason that some advocate ALWAYS defining your variables is that it helps you to set things up correctly.
 
WHY are you asking that question? Makes no sense...
Look at it this way:

@3mph...........d............>@6mph...........d.............>total of 1/2 hour

Since time = distance / speed:
d/3 + d/6 = 1/2
Multiply by 6:
2d + d = 3
3d = 3
d = 1
Over and out!
Yes, I get that but I don't understand why, when the equation presents both runner and walker, how to connect this d with the walker and thus answer the question. I can solve this problem in other ways than algebra, but I don't seem to be able to understand the algebra solution.
 
The problem requires careful attention to what is asked. If we define d as the distance walked, the total distance both ways is 2d. But you are NOT asked to find the total distance both ways. You are asked to find the distance one way even though you are only given the time both ways.

Denis attacked the problem by defining the distance walked as d, which is also the distance run, and then by focusing on the time in hours. The spent walking was d/3. The time spent running was d/6. The total time was the sum of the time spent walking and time spent running giving

\(\displaystyle \dfrac{d}{3} + \dfrac{d}{6} = \dfrac{1}{2} \implies 2d + d = 3 \implies d = 1.\)

Now you could define d as the total distance. In which case the distance walked is 0.5d, and the distance run is 0.5d as well. Using Denis's method of focusing on time but a different definition of d, we get

\(\displaystyle \dfrac{0.5d}{3} + \dfrac{0.5d}{6} = \dfrac{1}{2} \implies d + 0.5d = 3 \implies 1.5d = 3 \implies d = 2.\)

But that gives the total distance, not the distance walked. So the answer 0.5 * 2 = 1.


There are numerous correct ways to set the problem up. The reason that some advocate ALWAYS defining your variables is that it helps you to set things up correctly.

I am beginning to see the light...but I will have to copy this explanation and read it tomorrow in the morning when my head is on straight...it is going on what? 2 am? I feel the answer is in this but I need to puzzle it out. Thanks.
 
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