Find their respective speeds

The problem says that both “start at the same time” and “after two hours.” Therefore the two times are the same and both equal 2 hours.

Perhaps I was wrong to say that the two equations are obvious. I have been thinking like this for a lifetime almost, and it is new to you.

I start by thinking about relevant quantities when faced with a quantitative problem and I assign letters to them in writing so I can think and talk about them without getting mixed up about what exactly I am talking about. Some call that naming the variables; others talk about labelling. I don’t care what you call it; it is a way to guide you through the initial mental fog that a hard problem represents. That to me is the first step. It is not hard. You just ask yourself what quantities may be relevant. Then you see which ones are known. This tells you how many equations you need.

The second step is where you do most of the hard thinking. I call it translation, but I have had students who prefer to think of it as a mystery story. You need to find a certain number of clues in English that you then must translate into mathematic notation. Your job is to search out the clues and then do the translation. In this case the clue to the fourth equation is that they start from the same place but go in perpendicular directions. If you draw a diagram, you get a right triangle and then you can begin to search for helpful mathematical ideas relevant to right triangles.

Once you have your system of equations, your third step is to just follow the mechanical steps to get a solution. No deep thought required; just careful attention to details.

Your fourth step is to check your work.

That method will let you solve not only word problems but real life quantitative problems.
 
The problem says that both “start at the same time” and “after two hours.” Therefore the two times are the same and both equal 2 hours.

Perhaps I was wrong to say that the two equations are obvious. I have been thinking like this for a lifetime almost, and it is new to you.

I start by thinking about relevant quantities when faced with a quantitative problem and I assign letters to them in writing so I can think and talk about them without getting mixed up about what exactly I am talking about. Some call that naming the variables; others talk about labelling. I don’t care what you call it; it is a way to guide you through the initial mental fog that a hard problem represents. That to me is the first step. It is not hard. You just ask yourself what quantities may be relevant. Then you see which ones are known. This tells you how many equations you need.

The second step is where you do most of the hard thinking. I call it translation, but I have had students who prefer to think of it as a mystery story. You need to find a certain number of clues in English that you then must translate into mathematic notation. Your job is to search out the clues and then do the translation. In this case the clue to the fourth equation is that they start from the same place but go in perpendicular directions. If you draw a diagram, you get a right triangle and then you can begin to search for helpful mathematical ideas relevant to right triangles.

Once you have your system of equations, your third step is to just follow the mechanical steps to get a solution. No deep thought required; just careful attention to details.

Your fourth step is to check your work.

That method will let you solve not only word problems but real life quantitative problems.
I would add one more step - the step 0 :

Read and re-read the problem very carefully.

Rewrite the problem on paper without looking back! Then check for accuracy.
 
The problem says that both “start at the same time” and “after two hours.” Therefore the two times are the same and both equal 2 hours.

Perhaps I was wrong to say that the two equations are obvious. I have been thinking like this for a lifetime almost, and it is new to you.

I start by thinking about relevant quantities when faced with a quantitative problem and I assign letters to them in writing so I can think and talk about them without getting mixed up about what exactly I am talking about. Some call that naming the variables; others talk about labelling. I don’t care what you call it; it is a way to guide you through the initial mental fog that a hard problem represents. That to me is the first step. It is not hard. You just ask yourself what quantities may be relevant. Then you see which ones are known. This tells you how many equations you need.

The second step is where you do most of the hard thinking. I call it translation, but I have had students who prefer to think of it as a mystery story. You need to find a certain number of clues in English that you then must translate into mathematic notation. Your job is to search out the clues and then do the translation. In this case the clue to the fourth equation is that they start from the same place but go in perpendicular directions. If you draw a diagram, you get a right triangle and then you can begin to search for helpful mathematical ideas relevant to right triangles.

Once you have your system of equations, your third step is to just follow the mechanical steps to get a solution. No deep thought required; just careful attention to details.

Your fourth step is to check your work.

That method will let you solve not only word problems but real life quantitative problems.
Good morning, and can't thank you enough for such a explanatory reply. And I am most thankful to you, and to the others, because in the last few posts and previous problems, you started giving me advice on how to initiate, or how best set up a problem. You are not only helping me solve problems. You are teaching me how to think. I can't be more thankful for that. I'll start working on the problem right now.
 
First law of solving "word problem":

What is/are the find/s: \(\displaystyle \ \ \to \ \ \ \) find their respective speeds.

Assign names to those "finds": \(\displaystyle \ \ \to \ \ \ \) Speed of person 1 = N and Speed of person 2 = T

Find Givens:

speed of one of them is 4 km/h more than the other\(\displaystyle \ \ \to \ \ \ \) O = T + 4 .............. and

but the next given:

after two hours are 40 km away ...... ← ...... does not make sense

Please check the original problem and post corrected version.
Morning, what does the O in O= T+4 stand for?>
 
So, I know I should get the hypothenuse and then apply the speed formula.
One question, though.
Which should I get first , and why?
Khan said: get the hypothenuse and the find speed.
My question is, why is that order?. ( I feel this problem is going to teach me a lot so I do not want to miss a thing)
 
So, I know I should get the hypothenuse and then apply the speed formula.
One question, though.
Which should I get first , and why?
Khan said: get the hypothenuse and the find speed.
My question is, why is that order?. ( I feel this problem is going to teach me a lot so I do not want to miss a thing)
Your question about the order tells me you don't understand what's going on at all. The problem is not that you don't understand it. The problem is that you don't admit it and try to move forward - not a good approach.
So, given that you _know_ what the length of the hypotenuse is, why are you still talking about finding it?
 
Your question about the order tells me you don't understand what's going on at all. The problem is not that you don't understand it. The problem is that you don't admit it and try to move forward - not a good approach.
So, given that you _know_ what the length of the hypotenuse is, why are you still talking about finding it?
So the hypothenuse is 40.
I will work the rate of speed now.
 
So the hypothenuse is 40.
I will work the rate of speed now.
So the hypothenuse is 40.
I will work the rate of speed now.
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?.
 
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?.
Looks good.
 
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?.
Since the two distances are NOT equal - you should call those d1 and d2.

But - what was your "find"?
 
Well, I think now it is when the Pythagoras theorem comes in, right?.
 
I know the hypothenuse. That was a given
40^2=
I am stuck here. What should do in place of a^2 and b^2. Any clue?.
I have two more knowns now.
I have that the rate of car 1 is 2x, and that the rate of the car 2 is 2x+8.
But I don't want to substitute without thought. I need to understand why I am using the distances found. It is not clear for me.
 
So, I know I should get the hypothenuse and then apply the speed formula.
One question, though.
Which should I get first , and why?
Khan said: get the hypothenuse and the find speed.
My question is, why is that order?. ( I feel this problem is going to teach me a lot so I do not want to miss a thing)
The length of the legs of this triangle are the distances that the two cars traveled and are named d1 and d2.This is why writing things down is so important.
 
I know the hypothenuse. That was a given
40^2=
I am stuck here. What should do in place of a^2 and b^2. Any clue?.
I have two more knowns now.
I have that the rate of car 1 is 2x, and that the rate of the car 2 is 2x+8.
But I don't want to substitute without thought. I need to understand why I am using the distances found. It is not clear for me.
jeffM said: The length of the legs of this triangle are the distances that the two cars traveled and are named d1 and d2. Yes, I tumbled in it!. I t was pretty evidnet. From now I am writing everything down.
Hence,
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?.
 
jeffM said: The length of the legs of this triangle are the distances that the two cars traveled and are named d1 and d2. Yes, I tumbled in it!. I t was pretty evidnet. From now I am writing everything down.
Hence,
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?.
You may want to review the square of a sum formula.
 
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