@Dr.Peterson @JeffM please reply sir to my above post . I will be grateful to u
Average speed= AM
- Thread starter Saumyojit
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@Dr.Peterson @JeffM please confirm me have i interpretted correctly or understood my doubt . What do u conclude after reading post #20. I am waiting for ur reply please
Dr.Peterson
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Yes, you got the correct average speed over the entire trip, and correctly verified that at this constant speed, the trip would have taken the same time as the actual trip; that is what "average speed" means, so you clearly understand that.
@Dr.Peterson AND this note part have i interpretted correctly of what jeffM was trying to say with this line "Is it not quite clear that traveling even 4 hours at 24 km/hr you will go 96 km. If you travel 3 hours at 45 km/hr and one tenth of an hour at 3 km/hr, you will travel 135.3 km"
@Dr.Peterson please reply sir...i want to end my doubt regarding this by today. please spare some time sir
Dr.Peterson
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Yes, you are right. Is that what you need? You should start trusting your own reasoning rather than depending on other people who may not be available.
Your extra details are unnecessary. Just the fact that 135.5 km is far more than it would take to drive for a longer time at 24 km/h makes it clear. That was his point (as far as I can tell -- I left it for JeffM to explain what JeffM means).
Your extra details are unnecessary. Just the fact that 135.5 km is far more than it would take to drive for a longer time at 24 km/h makes it clear. That was his point (as far as I can tell -- I left it for JeffM to explain what JeffM means).
@JeffM @Dr.Peterson I was not sure with my own reasoning thats why i ask teachers.One thing i did not know until our discussion happened that avg speed is a speed if a car travels with it ;it has to take the same time as the actual journey took otherwise if it is taking less time then it cannot be the average speed.WHY SO? thats bcoz avg speed=total distance/ total time ; so it has to cover that amount of distance in that exact amt of time . No more no less .
conclusion: defination of what i understood of AVG speed : IT is just generalizing all the different speeds into one speed but it has to cover distance in that exact time .
For me asking questions then going through the process clears and opens up many new things. thank u
@Dr.Peterson
Yes it is matching with the AVG speed formula. So the weight that we need to take is the Time average speed each lasted.
@Jomo
Yes that's a nice way to solve for average speed per trip . By recording the speeds at a "particular time" we can list out the speeds and it's frequency.Making it discrete is a good way to solve
Also by choosing a specific recording time we can jot down each speed "Consistently" wrt to that time and it makes the calculation of Arithmetic mean of speeds much easier especially when the duration of trip are diff.
But that recording time has to be at least the hcf of each of the trip duration .
Suppose one trip is for 6 Min (S1= 3km/hr)and another trip is 105 minutes(S2= 2km/hr) so if I need to calculate average speed by Mean method , I need to choose a recording value of time .....which will be the hcf of 105,6 i.e 3 minutes.
By recording every 3 minutes , my data comes out to be Two "3km/hr" and thirty five " 2km/hr"
so Weighted arithmetic mean of the speeds = {3+3+ 35*(2km/hr)} divided by 37 Which is the count of all the speeds and this equals AVG speed by formula method!
Note, We could also do this by Individually taking average of each data(unweighted mean) but that would take more space and more time.
Which brings me to full circle that Dr p told
One thing to note is that in the case of marks - student case as it is discrete , so we will get the individual items(marks) if marks of each student is given or we can evaluate avg of all classes by taking average of averages using weight .
Student 1: 2marks
Student 2: 4 marks
Average marks of class a = 3
Student 1: 6marks
Student 2: 3 marks
Student 3: 3marks
Average marks of class b = 4
If I find out wrongly the AVG marks of both the classes by "averaging the averages" then Avg marks of both the classes= (4+3)/2= 3.5
This is wrong when we check with the method of finding AVG marks individually of each all students from both the classes=>(2+4+6+3+3)/5=3.6 is avg value. Which also equals if I did this way
"Averaging the averages using weight"= (4*3+3*2)/(3+2)=>3.6 Which finally tells us that Weighted average or "Averaging the averages using weight" should be equivalent to taking average of the individual data.!
And in the case of speed distance problem First we need to make the problem Discrete and then we can jot down the individual speeds and then check that finding the avg speed by unweighted method (Which was shown at the start)
In marks - students case I get the logic of considering the weight as the " frequency of the same marks in the data set" or " the no of students getting the same marks" but in the speed - distance problemn we consider the "time each AVG speed lasted" as the weight coz Avg speed= (D1+D2)/(T1+T2) => (S1*T1 + S2*T2)/(T1+T2)
We know that Weighted average formula looks like this=>(w1x1+ w2x2+..)/(w1+w2)
So comparing these two formulas (S1*T1 + S2*T2)/(T1+T2) &(w1x1+ w2x2)/(w1+w2), We find that weights are time in AVG formula case . So is there other logical reason of considering "time" as the weight?
I know that if time is different the avg speed in the longer time journey will contribute more to the entire trip avg speed value .the entire trip avg speed value will be closer to more weighted avg speed .So this is the reason right?
Another thing I want to add using jomo's method of recording speed at a particular time,
This eg : suppose
T1= 1hr S1=2km/hr
T2= 3hr S2= 3km/hr
Recording speed at every 1hr , we get the data :
2,3,3,3. km/hr
Avg speed=Sum of the speeds(data)/ no of the speeds(data)= {2+3*(3)}/4
Here I get a subtle intuition of why we consider the weight as "duration each avg speed lasted" : in jomo's method i can think of the weight as how many times one of the two speeds repeat wrt to that "recording time"
I never thought this way.
Yes it is matching with the AVG speed formula. So the weight that we need to take is the Time average speed each lasted.
@Jomo
Yes that's a nice way to solve for average speed per trip . By recording the speeds at a "particular time" we can list out the speeds and it's frequency.Making it discrete is a good way to solve
Also by choosing a specific recording time we can jot down each speed "Consistently" wrt to that time and it makes the calculation of Arithmetic mean of speeds much easier especially when the duration of trip are diff.
But that recording time has to be at least the hcf of each of the trip duration .
Suppose one trip is for 6 Min (S1= 3km/hr)and another trip is 105 minutes(S2= 2km/hr) so if I need to calculate average speed by Mean method , I need to choose a recording value of time .....which will be the hcf of 105,6 i.e 3 minutes.
By recording every 3 minutes , my data comes out to be Two "3km/hr" and thirty five " 2km/hr"
so Weighted arithmetic mean of the speeds = {3+3+ 35*(2km/hr)} divided by 37 Which is the count of all the speeds and this equals AVG speed by formula method!
Note, We could also do this by Individually taking average of each data(unweighted mean) but that would take more space and more time.
Which brings me to full circle that Dr p told
Yes that's why there is a problem of not keeping an account when we unconciously just average the averages but their weight is different so leading to wrong avg value.
One thing to note is that in the case of marks - student case as it is discrete , so we will get the individual items(marks) if marks of each student is given or we can evaluate avg of all classes by taking average of averages using weight .
Student 1: 2marks
Student 2: 4 marks
Average marks of class a = 3
Student 1: 6marks
Student 2: 3 marks
Student 3: 3marks
Average marks of class b = 4
If I find out wrongly the AVG marks of both the classes by "averaging the averages" then Avg marks of both the classes= (4+3)/2= 3.5
This is wrong when we check with the method of finding AVG marks individually of each all students from both the classes=>(2+4+6+3+3)/5=3.6 is avg value. Which also equals if I did this way
"Averaging the averages using weight"= (4*3+3*2)/(3+2)=>3.6 Which finally tells us that Weighted average or "Averaging the averages using weight" should be equivalent to taking average of the individual data.!
And in the case of speed distance problem First we need to make the problem Discrete and then we can jot down the individual speeds and then check that finding the avg speed by unweighted method (Which was shown at the start)
does not equal "discrete method" where discrete method gives us always the correct avg Speed
In marks - students case I get the logic of considering the weight as the " frequency of the same marks in the data set" or " the no of students getting the same marks" but in the speed - distance problemn we consider the "time each AVG speed lasted" as the weight coz Avg speed= (D1+D2)/(T1+T2) => (S1*T1 + S2*T2)/(T1+T2)
We know that Weighted average formula looks like this=>(w1x1+ w2x2+..)/(w1+w2)
So comparing these two formulas (S1*T1 + S2*T2)/(T1+T2) &(w1x1+ w2x2)/(w1+w2), We find that weights are time in AVG formula case . So is there other logical reason of considering "time" as the weight?
I know that if time is different the avg speed in the longer time journey will contribute more to the entire trip avg speed value .the entire trip avg speed value will be closer to more weighted avg speed .So this is the reason right?
Another thing I want to add using jomo's method of recording speed at a particular time,
This eg : suppose
T1= 1hr S1=2km/hr
T2= 3hr S2= 3km/hr
Recording speed at every 1hr , we get the data :
2,3,3,3. km/hr
Avg speed=Sum of the speeds(data)/ no of the speeds(data)= {2+3*(3)}/4
Here I get a subtle intuition of why we consider the weight as "duration each avg speed lasted" : in jomo's method i can think of the weight as how many times one of the two speeds repeat wrt to that "recording time"
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It is not fair to either me or to Dr. Peterson to ask him to comment on my remarks. He may believe that I am wrong. He may believe that I am correct but believe that my language is not illuminating.
It is not fair to anyone to wait six months to ask a long rambling set of questions about an already very long thread.
It is not fair to anyone to ask people to go back to post #9 WHEN YOU COULD QUOTE IT IN ITS ENTIRETY. You have this habit of giving partial quotes and thereby leaving context unclear.
Why do you turn exact fractions into inexact decimals?
And all of this is to ask people to confirm the following:
Because average speed is defined as total distance divided by total time, the simplest way to way to compute it CORRECTLY is to APPLY THE DEFINITION. If the speed is not uniform during the entire time, taking the arithmetic mean of the average speeds during various intervals of time will give a wrong answer unless each interval of time is of equal duration.
Example You travel 120 kilometers in two hours and then travel 30 kilometers in the next hour. The distance travelled is 150 kilometers in the course of 3 hours. Therefore, by definition, your average speed is 150 km /3 hr = 50 kilometers per hour. During the first two hours you travelled at 60 km per hour. During the third hour you travelled at 30 km per hour. If you average 60 and 30, you get 45, which is incorrect.
Now you could say that you travelled at 60 km per hour for one hour, 60 km per hour for another hour, and 30 km per hour for still yet another hour. If you average 60, 60, and 30, you do get the correct answer of 50.
What is obscure?
It is not fair to anyone to wait six months to ask a long rambling set of questions about an already very long thread.
It is not fair to anyone to ask people to go back to post #9 WHEN YOU COULD QUOTE IT IN ITS ENTIRETY. You have this habit of giving partial quotes and thereby leaving context unclear.
Why do you turn exact fractions into inexact decimals?
And all of this is to ask people to confirm the following:
Because average speed is defined as total distance divided by total time, the simplest way to way to compute it CORRECTLY is to APPLY THE DEFINITION. If the speed is not uniform during the entire time, taking the arithmetic mean of the average speeds during various intervals of time will give a wrong answer unless each interval of time is of equal duration.
Example You travel 120 kilometers in two hours and then travel 30 kilometers in the next hour. The distance travelled is 150 kilometers in the course of 3 hours. Therefore, by definition, your average speed is 150 km /3 hr = 50 kilometers per hour. During the first two hours you travelled at 60 km per hour. During the third hour you travelled at 30 km per hour. If you average 60 and 30, you get 45, which is incorrect.
Now you could say that you travelled at 60 km per hour for one hour, 60 km per hour for another hour, and 30 km per hour for still yet another hour. If you average 60, 60, and 30, you do get the correct answer of 50.
What is obscure?
yes that is what I discovered on seeing jomo's comment .
Recording speed at every 1hr by choosing a specific recording time we can jot down each speed "Consistently" wrt to that time and it makes the calculation of Arithmetic mean of speeds much easier especially
So is there other logical reason of considering "time" as the weight?
Yes. It goes back to "unit." If we measuring km per HOUR, that means we are talking about the exact distance that would be travelled if we maintained that exact speed for exactly one hour. So when you take the arithmetic mean of average speeds of different durations, your "observations" are not observations of the same type.
Stop thinking in terms of words. 50% of your doubts come from trying to formulate in language things that are obvious in terms of mathematical notation. Think symbolically to understand math.
[MATH]\text {average speed} \equiv \dfrac{\text {total distance}}{\text{total time}}[/MATH] is a definition.
Letting d be distance, t time, and r average speed, we express this algebraically.
[MATH]r \equiv \dfrac{d}{t} \implies d = r * t.[/MATH]
Now if we have two periods of different duration during which average speeds are different, the simple way to compute average speed for the entire time is
[MATH]\text { total distance} = d_1 + d_2.[/MATH]
[MATH]\text { total time} = t_1 + t_2.[/MATH]
Therefore the average speed for the entire time is
[MATH]r = \dfrac{d_1 + d_2}{t_1 + t_2}.[/MATH] SImple.
Of course we can compute the average speed for each period.
[MATH]r_1 = \dfrac{d_1}{t_1}.[/MATH]
[MATH]r_2 = \dfrac{d_2}{t_2}.[/MATH]
And, you can verify numerically that
[MATH]t_1 \ne t_2 \text { and } r_1 \ne r_2 \implies \dfrac{r_1 + r_2}{2} \ne r.[/MATH]
Therefore taking an arithmetic average of speeds is an incorrect way to compute an average speed over multiple periods.
It is as simple as that.
Stop thinking in terms of words. 50% of your doubts come from trying to formulate in language things that are obvious in terms of mathematical notation. Think symbolically to understand math.
[MATH]\text {average speed} \equiv \dfrac{\text {total distance}}{\text{total time}}[/MATH] is a definition.
Letting d be distance, t time, and r average speed, we express this algebraically.
[MATH]r \equiv \dfrac{d}{t} \implies d = r * t.[/MATH]
Now if we have two periods of different duration during which average speeds are different, the simple way to compute average speed for the entire time is
[MATH]\text { total distance} = d_1 + d_2.[/MATH]
[MATH]\text { total time} = t_1 + t_2.[/MATH]
Therefore the average speed for the entire time is
[MATH]r = \dfrac{d_1 + d_2}{t_1 + t_2}.[/MATH] SImple.
Of course we can compute the average speed for each period.
[MATH]r_1 = \dfrac{d_1}{t_1}.[/MATH]
[MATH]r_2 = \dfrac{d_2}{t_2}.[/MATH]
And, you can verify numerically that
[MATH]t_1 \ne t_2 \text { and } r_1 \ne r_2 \implies \dfrac{r_1 + r_2}{2} \ne r.[/MATH]
Therefore taking an arithmetic average of speeds is an incorrect way to compute an average speed over multiple periods.
It is as simple as that.