Find their respective speeds

YES.

Now I shall show you I would do this problem if I found it difficult. (People like lev, Subhotosh, and me, who have been doing algebra problems for decades, do not find this a difficult problem and so take shortcuts, but when we find a problem to be hard, we do something like this.) I like Subhotosh's numbering.

(Step 1) Read the problem carefully and make sure you understand it well. This may allow you to take shortcuts. If it doesn't

(Step 2) Assign a distinct letter to each potentially relevant quantity and write them down. Try to use the notation to help your memory.

[MATH]d_1 = \text {distance travelled by car 1.}[/MATH]
[MATH]r_1 = \text {average rate of car 1.}[/MATH]
[MATH]t_1 = \text {time traveled by car 1.}[/MATH]
[MATH]d_2= \text {distance travelled by car 2.}[/MATH]
[MATH]r_2 = \text {average rate of car 2.}[/MATH]
[MATH]t_2 = \text {time traveled by car 2.}[/MATH]
[MATH]d_3 = \text {final distance between the cars.}[/MATH]
Notice that the distances are all subscripted d, all the rates are subscripted r, and all the times are subscripted t. It puts the least stress on my aged memory.

(Step 3): Look for clues, translate them into mathematical notation, and write the translations down as you find them. The whole point of this writing down is to relieve your memory while you are trying to think. This step is where most of the thinking comes in, and you do not want to be trying to remember what everything means while you read and think. Here is a tip: look for the easy stuff first.

[MATH]d_3 = 40.[/MATH] Easy, the problem explicitly tells you that in words.

[MATH]t_1 = 2 \text { and } t_2 = 2.[/MATH] Easy, the problem implicitly tells you in words

[MATH]r_1 = r_2 + 4.[/MATH] Easy, the problem explicitly tells you that in words.

We need to find three more clues and translate them into mathematical notation. If you have drawn a diagram after noting the word "perpendicular," you can actually see physically that the Pythagorean Theorem provides a big clue.

[MATH]d_1^2 + d_2^2 = d_3^2.[/MATH]
We still need two more clues. But we know a general formula relating rates, times, and distances, which gives us those two extra two clues.

[MATH]r * t = d \implies r_1 * t_1 = d_1 \text { and } r_2 * t_2 = d_2.[/MATH]
This step is where the brain work occurs, but you have a goal. Seven relevant pronumerals requires seven independent equations, which means seven clues to find and translate. You are not working entirely in the dark.

When you are done, you should be looking at

[MATH]d_3 = 40,\\ t_1 = 2\\ t_2 = 2,\\ r_1 = r_2 + 4,\\ d_1^2 + d_2^2 = d_3^2,\\ r_1 * t_1 = d_1 \text { and,}\\ r_2 * t_2 = d_2.[/MATH](Step 4): Now it is mechanical. Just carefully solve the system equations by substitution

It is trivial to get to

[MATH]d_1^2 + d_2^2 = d_3^2 = 40^2 = 1600,\\ 2r_1 = d_1,\\ 2r_2 = d_2,\\,[/MATH]Now it is easy to get to

[MATH](2r_1)^2 + (2r_2^2) = 1600 \implies 4r_1^2 + 4r_2^2 = 1600 \implies r_1^2 + r_2^2 = 400.[/MATH]
Just two unknowns left. Do we know how they are related? We do so

[MATH](r_2 + 4)^2 + r_2^2 = 400 \implies r_2^2 + 8r_2 + 16 + r_2^2 = 400 \implies\\ 2r_2^2 + 8r_2 - 384 = 0 \implies r_2^2 + 4r_2 - 192 = 0.[/MATH]Can you solve that equation? If so, you can say what every relevant quantity is.

Then go on to step 5: check that the answers work.

It is not a formula. It is a procedure that requires persistence, thought, and care. It is a way of thinking step by step through complicated problems.
 
Last edited by a moderator:
JeffM said:
YES.

Now I shall show you I would do this problem if I found it difficult. (People like lev, Subhotosh, and me, who have been doing algebra problems for decades, do not find this a difficult problem and so take shortcuts, but when we find a problem to be hard, we do something like this.) I like Subhotosh's numbering.

(Step 1) Read the problem carefully and make sure you understand it well. This may allow you to take shortcuts. If it doesn't

(Step 2) Assign a distinct letter to each potentially relevant quantity and write them down. Try to use the notation to help your memory.

[MATH]d_1 = \text {distance travelled by car 1.}[/MATH]
[MATH]r_1 = \text {average rate of car 1.}[/MATH]
[MATH]t_1 = \text {time traveled by car 1.}[/MATH]
[MATH]d_2= \text {distance travelled by car 2.}[/MATH]
[MATH]r_2 = \text {average rate of car 2.}[/MATH]
[MATH]t_2 = \text {time traveled by car 2.}[/MATH]
[MATH]d_3 = \text {final distance between the cars.}[/MATH]
Notice that the distances are all subscripted d, all the rates are subscripted r, and all the times are subscripted t. It puts the least stress on my aged memory.

(Step 3): Look for clues, translate them into mathematical notation, and write the translations down as you find them. The whole point of this writing down is to relieve your memory while you are trying to think. This step is where most of the thinking comes in, and you do not want to be trying to remember what everything means while you read and think. Here is a tip: look for the easy stuff first.

[MATH]d_3 = 40.[/MATH] Easy, the problem explicitly tells you that in words.

[MATH]t_1 = 2 \text { and } t_2 = 2.[/MATH] Easy, the problem implicitly tells you in words

[MATH]r_1 = r_2 + 4.[/MATH] Easy, the problem explicitly tells you that in words.

We need to find three more clues and translate them into mathematical notation. If you have drawn a diagram after noting the word "perpendicular," you can actually see physically that the Pythagorean Theorem provides a big clue.

[MATH]d_1^2 + d_2^2 = d_3^2.[/MATH]
We still need two more clues. But we know a general formula relating rates, times, and distances, which gives us those two extra two clues.

[MATH]r * t = d \implies r_1 * t_1 = d_1 \text { and } r_2 * t_2 = d_2.[/MATH]
This step is where the brain work occurs, but you have a goal. Seven relevant pronumerals requires seven independent equations, which means seven clues to find and translate. You are not working entirely in the dark.

When you are done, you should be looking at

[MATH]d_3 = 40,\\ t_1 = 2\\ t_2 = 2,\\ r_1 = r_2 + 4,\\ d_1^2 + d_2^2 = d_3^2,\\ r_1 * t_1 = d_1 \text { and,}\\ r_2 * t_2 = d_2.[/MATH](Step 4): Now it is mechanical. Just carefully solve the system equations by substitution

It is trivial to get to

[MATH]d_1^2 + d_2^2 = d_3^2 = 40^2 = 1600,\\ 2r_1 = d_1,\\ 2r_2 = d_2,\\,[/MATH]Now it is easy to get to

[MATH](2r_1)^2 + (2r_2^2) = 1600 \implies 4r_1^2 + 4r_2^2 = 1600 \implies r_1^2 + r_2^2 = 400.[/MATH]
Just two unknowns left. Do we know how they are related? We do so

[MATH](r_2 + 4)^2 + r_2^2 = 400 \implies r_2^2 + 8r_2 + 16 + r_2^2 = 400 \implies\\ 2r_2^2 + 8r_2 - 384 = 0 \implies r_2^2 + 4r_2 - 192 = 0.[/MATH]Can you solve that equation? If so, you can say what every relevant quantity is.

Then go on to step 5: check that the answers work.

It is not a formula. It is a procedure that requires persistence, thought, and care. It is a way of thinking step by step through complicated problems.
Click to expand...
 
Thanks, jeffM. Awesome!. I will give it a try.
I don't understand something
I know d is distance, but can you explain more this:
you said:
We need to find three more clues and translate them into mathematical notation. If you have drawn a diagram after noting the word "perpendicular," you can actually see physically that the Pythagorean Theorem provides a big clue.

and then you write the three d's squared and subscripted with numbers 1,2=3.
I didn't get that?.
Or can the variables in the hypothenuse formula be changed? .Like, instead of ,a,b,c you wrote the d's?. Is that so?.
 
eddy2017 said:
Thanks, jeffM. Awesome!. I will give it a try.
I don't understand something
I know d is distance, but can you explain more this:
you said:
We need to find three more clues and translate them into mathematical notation. If you have drawn a diagram after noting the word "perpendicular," you can actually see physically that the Pythagorean Theorem provides a big clue.

and then you write the three d's squared and subscripted with numbers 1,2=3.
I didn't get that?.
Or can the variables in the hypothenuse formula be changed? .Like, instead of ,a,b,c you wrote the d's?. Is that so?.
Click to expand...
As long as the variables correspond to the lengths of the sides of a right triangle you can name them anything from a, b, c to Jim, Bob and Marry. Although using meaningful names is better for all involved. If d1, etc are the variables denoting the 2 distances traveled by the cars and one between them, then that's the variables which should be used in the Pythagorean Theorem formula.

Your post #59 was going in the right direction, you just made a mistake when squaring the sum.
 
You need to ask questions about anything that a helper says and that you do not fully understand.

In my very first post, I said that you should try to use the notation to help you.

The assignment of letters to quantities is completely arbitrary. It is up to you.

There is a guideline that says propositions true for a wide class of numbers should be shown in letters at the start of the alphabet. That guideline then says that, because the sum of the square of the length of one of the short sides of a RIGHT triangle and the square of the other short side are ALWAYS equal to the square of the longest side, called the hypotenuse, then it is advisable to use the first three letters of the alphabet for those lengths when talking in general terms. But that is a general guideline, not a rule, and we are discussing a specific problem rather than a general proposition. There is nothing whatsoever to stop you from substituting d1 for a, d2 for b, and d3 for c. The naming that is useful for you in a given problem is the best naming.
 
lev888 said:
As long as the variables correspond to the lengths of the sides of a right triangle you can name them anything from a, b, c to Jim, Bob and Marry. Although using meaningful names is better for all involved. If d1, etc are the variables denoting the 2 distances traveled by the cars and one between them, then that's the variables which should be used in the Pythagorean Theorem formula.

Your post #59 was going in the right direction, you just made a mistake when squaring the sum.
Click to expand...
What mistake?. I did it slowly, so where is the mistake?. I was understanding it very well.
 
JeffM said:
I think at some point lev slipped a cog; it happens to every one of us, especially me, which is why you must always check your work.

The relevant formula is

[MATH]d_1^2 + d_2^2 = d_3^2[/MATH] using my notation.

But it is not generally true that [MATH](d_1 + d_2)^2 = d_1^2 + d_2^2.[/MATH]
I want to reiterate: word problems CAN teach a systematic way of thinking about a certain type of quantitative problem. While doing that, it is always possible to slip up: the phone rings, your sister has a hissy fit in your ear, etc. That's why we check.
Click to expand...
I was referring to a mistake in post 59:
40^2==(2x)^2+ (2x+8)^2
1600=4x^2+ 4x^2 +64
 
JeffM said:
You need to ask questions about anything that a helper says and that you do not fully understand.

In my very first post, I said that you should try to use the notation to help you.

The assignment of letters to quantities is completely arbitrary. It is up to you.

There is a guideline that says propositions true for a wide class of numbers should be shown in letters at the start of the alphabet. That guideline then says the the square of the length of one of the short sides of a RIGHT triangle and the square of the other short side are ALWAYS equal to the square of the longest side, called the hypotenuse, then it is advisable to use the first three letters of the alphabet when talking in general terms. But that is a general rule, and we are discussing a specific problem. There is nothing whatsoever to stop you from substituting d1 for a, d2 for b, and d3 for c. The naming that is useful for you in a given problem is the best naming.
Click to expand...
You need to ask questions about anything that a helper says and that you do not fully understand.
that is exactly what i am doing. You
lev888 said:
I was referring to a mistake in post 59:
40^2==(2x)^2+ (2x+8)^2
1600=4x^2+ 4x^2 +64
Click to expand...
Please, lev, can you explain where the mistake is?.
I plugged in the correct values in the theorem. Or, did I not?.
you said: 'Please expand (a+b)^2.
I don't understand that.
 
eddy2017 said:
You need to ask questions about anything that a helper says and that you do not fully understand.
that is exactly what i am doing. You

Please, lev, can you explain where the mistake is?.
I plugged in the correct values in the theorem. Or, did I not?.
you said: 'Please expand (a+b)^2.
I don't understand that.
Click to expand...
You wrote previously that from 40^2==(2x)^2+ (2x+8)^2 we get 1600=4x^2+ 4x^2 +64.
Which means (2x+8)^2 = 4x^2 +64.
Can you explain this part?
 
At post# 50 I wrote this:

I will follow the useful advice given.
I'll proceed to label unknowns

d1=distance travelled by car 1.

r1=average rate of car 1.

t=time traveled by car 1.
as time is the same I'd rather leave as t. (not to confuse me labeling with numbers or letters here)

=====================
d2=distance travelled by car 2.

r2=average rate of car 2.

t=time traveled by car 2.

x=final distance between the cars.
given

We know that time is equal to 2 for both cars
We also know that x = 40.

I'll go with car 1 first.
distance that it travels in 2 hrs?
r1=d1
t
d= r1*2 (I am gonna label rate of speed as x)
d=x*t
d=x * 2
d=2x

I'll do the same with car 2
distance that it travels in 2 hrs?
r2= d2
t
d2=r2*t
d2=(x+4)* t (x+4 'cause one of the car's speed is 4kmp more than the other car's speed)
d2=(x+4)* 2 ( t=2) ( I'll distribute the 2 in to get rid of the parentheses, so
d=2(x+4)
d= 2x+ 8
What do you think?.

you said it looked good to you.

Then Khan said that should name the distances 1 and 2 since they were not the same.

Then jeefM said: The length of the legs of this triangle are the distances that the two cars traveled and are named d1 and d2.
I said: Yes, I tumbled in it!. It was pretty evident. From now I am writing everything down.

At post # 59 I wrote the theorem and plugged in the values and asked if there was something missing from the theorem or something wrong with it.

c^2= a^2+b^2
40^2==(2x)2+ (2x+8)^2
1600=4x^2+ 4x^2 +64
this is just to confirm if all the terms are accounted for in the equation above?. If there is nothing missing?



And then lev said this:

You may want to review the square of a sum formula.

I said I did not understand.

And then at 63 lev adds: ‘(2x+8)^2 - square of a sum

I am still in doubt as to what that means,

And then jeffM writes how he would solve it but I was left with the doubt about what you meant by square of a sum. That was never cleared up. I think that was the problem.

Because every thing was going good according to you. I just made a mistake in solving the theorem but it was not explained what mistake it was. Just a square of a sum and it was left at that. I am still in doubt about that,.

I think it is good if we can go back and retake it from my last question and your answer it'd be great: the mistake is in the square of a sum.
Let's take it from there if you don't mind to finish it the way I was doing it, step by step and asking you when i did not understand.
 
eddy2017 said:
I think it is good if we can go back and retake it from my last question and your answer it'd be great: the mistake is in the square of a sum.
Let's take it from there if you don't mind to finish it the way I was doing it, step by step and asking you when i did not understand.
Click to expand...
I am taking it from there.
That's why I asked you how (2x+8)^2 became 4x^2 +64. Can you explain it?
 
Eddy,
Do you know the following equation (formula)?

(a + b)^2 = a^2 + b^2 + 2ab ......?
 
lev888 said:
I am taking it from there.
That's why I asked you how (2x+8)^2 became 4x^2 +64. Can you explain it?
Click to expand...
Sorry, when i git your message i had already gone to bed.
how (2x+8)^2 became 4x^2 +64
2x^2=4x^2+ 8^2
2x^2+ 64
 
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