Average speed= AM

Saumyojit

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BOTH DOUBTS ARE EQUALLY IMPORTANT

DOUBT 1: I discovered that if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes . D1=20km d2=20/3km
Avg speed per trip=40 km/hr

ALSO , Arithmetic mean of 60 and 20 = avg of 2 speeds = (60 km/hr + 20km/hr)/2= sum of observation / no of observation=40km/hr

So avg speed=Arithmetic mean (only when t1=t2)
Is there any intutive explanation why avg speed = AM ? other than doing a sum and then at the last finding out oh yes Avg speed per trip= AM


DOUBT 2: If i take the case of first trip only suppose trip1 was for 20 minutes not 1 hr then the car covered 20km. So the driver says we covered 20 km in 20 minutes . From that info can we say the car was travelling 60km/hr ; though the car has not travlled for 1 hr it has just travlled for 20 minutes covering 20 km , thn how can we say first trip car was travelling for 60 km/hr .

NOTE: IS it because distance and time are directly prorportional . If d1/t1=20 km/20min (from what we travelled) now if we want to measure in terms of 60 minutes that means t1 needs to be tripled so t1 becomes 60 minutes and as d1 is proportion to t1; d1 also triples so d2/t2=60km/60minutes so we can say THE FIRST TRIP WAS TRAVELLING AT A SPEED OF 60km/hr

@Dr.Peterson @JeffM
 
The answer is "Yes."

The definition of "average speed" is total distance divided by total time.

In general, you cannot take the arithmetic mean of average speeds to calculate average speeds. Let's take an example.
A car takes two hours to travel 150 miles on the turnpike for two hours. It then exits the turnpike and takes three hours to travel 105 miles on country roads. During the first two hours, it traveled at an average rate of 150/2 or 75 miles per hour; during the next three hours, it traveled at an average rate of 105/3 or 35 miles per hour. The arithmetic mean of those is
(75 + 35)/2 = 55. That, however, is the wrong answer. The car traveled a total of 150 + 105 = 255 miles. It took 5 hours to do so. So its average speed was 255/5 or 51 miles per hour.

What accounts for this? The car traveled at the lower average speed for a LONGER time. The arithmetic mean ignores this fact. If, however, as you discovered, we take the arithmetic mean of average speeds for equal times, we get the right overall average speed.
 
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BOTH DOUBTS ARE EQUALLY IMPORTANT

DOUBT 1: I discovered that if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes . D1=20km d2=20/3km
Avg speed per trip=40 km/hr

ALSO , Arithmetic mean of 60 and 20 = avg of 2 speeds = (60 km/hr + 20km/hr)/2= sum of observation / no of observation=40km/hr

So avg speed=Arithmetic mean (only when t1=t2)
Is there any intutive explanation why avg speed = AM ? other than doing a sum and then at the last finding out oh yes Avg speed per trip= AM

Speeds are in km/h, that is, d/t. When the times are equal, that means equal denominators, and you can add the fractions directly. That's why the average speed, (d1 + d2)/(t + t), is the same as the average (arithmetic mean) of the speeds, (d1/t + d2/t)/2.

If the distances of two trips are the same, then the numerators are equal, and the reciprocals of the speeds add; that's why in that case the average speed is the harmonic mean of the speeds.

DOUBT 2: If i take the case of first trip only suppose trip1 was for 20 minutes not 1 hr then the car covered 20km. So the driver says we covered 20 km in 20 minutes . From that info can we say the car was travelling 60km/hr ; though the car has not travlled for 1 hr it has just travlled for 20 minutes covering 20 km , thn how can we say first trip car was travelling for 60 km/hr .

NOTE: IS it because distance and time are directly prorportional . If d1/t1=20 km/20min (from what we travelled) now if we want to measure in terms of 60 minutes that means t1 needs to be tripled so t1 becomes 60 minutes and as d1 is proportion to t1; d1 also triples so d2/t2=60km/60minutes so we can say THE FIRST TRIP WAS TRAVELLING AT A SPEED OF 60km/hr
Yes, it's because distance is proportional to time at a given speed. Speed in kilometers per hour (or whatever) refers to how far you would have gone at that speed if you had driven for one hour.

How can we say that it was going 60 km/hr? Because 20/20 = 60/60, or 20/(1/3) = 60/1. It's a ratio.
 
What accounts for this? The car traveled at the lower average speed for a LONGER time. The arithmetic mean ignores this fact. If, however, as you discovered, we take the arithmetic mean of average speeds for equal times, we get the right overall average speed.
I know that the second trip is one hour extra and The car traveled at the lower average speed for a LONGER time but still cant get the intuition how the arithmetic mean ignores this fact? This line i need to understand how it is working inside how it is ignoring . Its not hitting my instinct.
Please try a liitle bit more to make me understand sir. @JeffM
I have seen Dr p method but is there any other way.
 
I know that the second trip is one hour extra and The car traveled at the lower average speed for a LONGER time but still cant get the intuition how the arithmetic mean ignores this fact? This line i need to understand how it is working inside how it is ignoring . Its not hitting my instinct.
Please try a liitle bit more to make me understand sir. @JeffM
I have seen Dr p method but is there any other way.
You said two observations in your original post. Effectively you are giving both observations equal weight even if they do not cover equal periods.

We could use a weighted mean and derive the correct answer. What would our weights be? Why the length of the period each average speed lasted.

[MATH]\dfrac{2 \text { hours} * \dfrac{75 \text { miles}}{\text{hour}} + 3 \text { hours} * \dfrac{35 \text { miles}}{\text{hour}}}{2 \text { hours} + 3 \text { hours}} =\\ \dfrac{150 \text { miles} + 105 \text { miles}}{5 \text { hours}} = \dfrac{255 \text { miles}}{5 \text { hours}} = \dfrac{51 \text { miles}}{\text { hour}}[/MATH]
Does this help?
 
You said two observations in your original post. Effectively you are giving both observations equal weight even if they do not cover equal periods.

We could use a weighted mean and derive the correct answer. What would our weights be? Why the length of the period each average speed lasted.

[MATH]\dfrac{2 \text { hours} * \dfrac{75 \text { miles}}{\text{hour}} + 3 \text { hours} * \dfrac{35 \text { miles}}{\text{hour}}}{2 \text { hours} + 3 \text { hours}} =\\ \dfrac{150 \text { miles} + 105 \text { miles}}{5 \text { hours}} = \dfrac{255 \text { miles}}{5 \text { hours}} = \dfrac{51 \text { miles}}{\text { hour}}[/MATH]
Does this help?
This thing that u did was average speed=(d1+d2)/(t1+t2) this i know it . But i was not asking for it.

I was asking for the intuitive explanation of "how The arithmetic mean ignores the fact that The car traveled at the lower average speed for a LONGER time "

when we do this (75 + 35)/2 = 55. how the Am ignores the fact . ?
Perhaps i think there is no explanation other than doing a sum and then checking avg speed per trip not equals to AM
@Dr.Peterson @JeffM @Subhotosh Khan
 
When you take an unweighted mean, you are assuming that each item being averaged is of equal importance. So if you walk at 3 km/hr for 6 minutes to get to your car and then drive for 3 hours at 45 km/hr, does it seem at all intuitive that your average speed was 24 km/hr? The effect of the three hour's driving is much greater than 6 minutes walking. 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 kn/hr, you will travel 135.3 km.

It makes no sense to say that your speed for 1/10th of an hour has an equal effect as your different speed for 3 hours.
 
Suppose for a moment that you can change your speed instantaneously. Now you drive for 1 minute to the highway at 30mph and then for 24 hrs on the highway at 60 mph. Are you really averaging just 45 miles per hr?

This is just what JeffM said. I just felt you that you needed to hear it again. I assure you that the only way to understand this is with weighted averages. A short portion of a trip has less influence then the much longer portion of the trip.
 
@JeffM @Dr.Peterson

When you take an unweighted mean, you are assuming that each item being averaged is of equal importance.
1: May be i am interpreting wrong or not understanding but suppose in my first eg :
if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes .

UNWEIGHTED Arithmetic mean of 60 and 20 = (60 km/hr + 20km/hr)/2 in this case also we do unweighted mean too. So what is the problemn with unweighted mean? NOTE: I dont get the feel or connection of why u refered the "unweighted mean" in the fisrt sentence ...I know that AM of speed means unweighted mean no weighted mean. "
I only have doubt regarding only the Arithmetic mean "how The arithmetic mean ignores the fact" not with the weighted average that u showed this one

{2hours *75mph + 3hours * 35miles/hour}/{2hours+ 3 hours} ---- THIS thing i know ;this is by the average speed formula ( (d1+d2)/(t1+t2)). I dont get the connection why u have shown this maybe u are trying to tell me something . is it this ---> that one hour is extra in the second case .


So if you walk at 3 km/hr for 6 minutes to get to your car and then drive for 3 hours at 45 km/hr, does it seem at all intuitive that your average speed was 24 km/hr? The effect of the three hour's driving is much greater than 6 minutes walking. 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 kn/hr, you will travel 135.3 km.

2: NOW this part i dont get it why u said that "does it seem at all intuitive that your average speed was 24 km/hr?" U added both the speeds per hour = (45 +3) and then divide by 2 which gives 24 . YEs thats the normal way of doing AM. Why u say "does it seem at all intuitive"?


3:
And 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 kn/hr, you will travel 135.3 km."


I am clear that with 24km/hr speed in 4 hours i can complete 96km and its also clear my actual journey has been about 135.3km.

BUT With this above line I know u are trying to tell me something that i am not able to catch . Are u trying to tell me that the distance that came from avg speed(i.e 96km) must/ has to be greater than the total distance travelled(135.5km) within a specific time or there is some disbalance of time .

Wht hint are u givng me?

I know every line is giving me some kind of intutive explanation but cant decipher it.
 
Suppose you want to average 10 test scores where 9 people got 90 and 1 person got a 10.
Now suppose that 1 person got a 90 and 9 people got a 10.
Shouldn't the first case yield a higher average?
It seems that you are saying that the average in the both cases is (10+90)/ = 50. This simply is not true!
The true average for the 1st case is (9*90 + 1*10)/10 = 820/10 =82. Nowhere near 50.
 
Suppose you want to average 10 test scores where 9 people got 90 and 1 person got a 10.
It seems that you are saying that the average in the both cases is (10+90)/ 2= 50
Why are u applying arithmetic mean here like this : (10+90)/ = 50 i dont get it .
9 people have got 90 and 1 has got 10 so it should be (9*90 + 1*10)/10 thats the arithemtic mean and also average of score per person
 
@JeffM @Dr.Peterson


1: May be i am interpreting wrong or not understanding but suppose in my first eg :
if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes .

UNWEIGHTED Arithmetic mean of 60 and 20 = (60 km/hr + 20km/hr)/2 in this case also we do unweighted mean too. So what is the problemn with unweighted mean? NOTE: I dont get the feel or connection of why u refered the "unweighted mean" in the fisrt sentence ...I know that AM of speed means unweighted mean no weighted mean. "
I only have doubt regarding only the Arithmetic mean "how The arithmetic mean ignores the fact" not with the weighted average that u showed this one

{2hours *75mph + 3hours * 35miles/hour}/{2hours+ 3 hours} ---- THIS thing i know ;this is by the average speed formula ( (d1+d2)/(t1+t2)). I dont get the connection why u have shown this maybe u are trying to tell me something . is it this ---> that one hour is extra in the second case .




2: NOW this part i dont get it why u said that "does it seem at all intuitive that your average speed was 24 km/hr?" U added both the speeds per hour = (45 +3) and then divide by 2 which gives 24 . YEs thats the normal way of doing AM. Why u say "does it seem at all intuitive"?


3:
And 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 kn/hr, you will travel 135.3 km."


I am clear that with 24km/hr speed in 4 hours i can complete 96km and its also clear my actual journey has been about 135.3km.

BUT With this above line I know u are trying to tell me something that i am not able to catch . Are u trying to tell me that the distance that came from avg speed(i.e 96km) must/ has to be greater than the total distance travelled(135.5km) within a specific time or there is some disbalance of time .

Wht hint are u givng me?

I know every line is giving me some kind of intutive explanation but cant decipher it.
@JeffM @Dr.Peterson is my doubts not making any sense . Please read it once sir . I think jeffm is trying to give me something but cant decipher . Please go through my post once
 
I can't answer this, because it is what JeffM has said, which is not how I would approach the problem -- especially as talk about weighted and unweighted averages doesn't seem to help you.

Please restate your concerns as if we were starting over, and without reference to specific things that have been said.
 
@Dr.Peterson
NOTE:I AM CONCERNED ABOUT THE ARITHMETIC MEAN ONLY. I KNOW HOW TO FIND AVERAGE SPEED THE FORMULA WAY.((d1 + d2)/(t + t)

U SAID " When the times are equal, that means equal denominators, and you can add the fractions directly. That's why the average speed, (d1 + d2)/(t + t), is the same as the average (arithmetic mean) of the speeds, (d1/t + d2/t)/2."

I HAVE UNDERSTOOD UR METHOD but is there any other way to understand this . Thats why i was quoting jeffm answer because he was trying to say something important in post #9 Whcich i could not understand may be u can help me .


MY Doubt remains the same
if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes .
D1=20km d2=20/3km
Avg speed per trip=40 km/hr

ALSO , Arithmetic mean of 60 and 20 = avg of 2 speeds = (60 km/hr + 20km/hr)/2= sum of observation / no of observation=40km/hr

So avg speed=Arithmetic mean (only when t1=t2)
DOUBT : Is there any intuitive explanation why avg speed = AM ? other than doing a sum and finding out Avg speed per trip and then AM and then checking at last whether they are equal or not . I know that another way to identify is if the time is same but i am asking for other intutive or logical method to know that oh in this problemn AM will not give average speed ( JeffM in post9 was giving me some hint i failed to understand this line "does it seem at all intuitive that your average speed was 24 km/hr? " )
 
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@Dr.Peterson
NOTE:I AM CONCERNED ABOUT THE ARITHMETIC MEAN ONLY. I KNOW HOW TO FIND AVERAGE SPEED THE FORMULA WAY.((d1 + d2)/(t + t)

U SAID " When the times are equal, that means equal denominators, and you can add the fractions directly. That's why the average speed, (d1 + d2)/(t + t), is the same as the average (arithmetic mean) of the speeds, (d1/t + d2/t)/2."

I HAVE UNDERSTOOD UR METHOD but is there any other way to understand this . Thats why i was quoting jeffm answer because he was trying to say something important in post #9 Whcich i could not understand may be u can help me .


MY Doubt remains the same
if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes .
D1=20km d2=20/3km
Avg speed per trip=40 km/hr

ALSO , Arithmetic mean of 60 and 20 = avg of 2 speeds = (60 km/hr + 20km/hr)/2= sum of observation / no of observation=40km/hr

So avg speed=Arithmetic mean (only when t1=t2)
DOUBT : Is there any intuitive explanation why avg speed = AM ? other than doing a sum and finding out Avg speed per trip and then AM and then checking at last whether they are equal or not . I know that another way to identify is if the time is same but i am asking for other intutive or logical method to know that oh in this problemn AM will not give average speed ( JeffM in post9 was giving me some hint i failed to understand this line "does it seem at all intuitive that your average speed was 24 km/hr? " )
What I stated is the MEANING of "average speed", and I believe you know that. Any other method you devise has to agree with that.

In particular, I said that if you take two parts of a trip in the same TIME, then the average speed is the average (arithmetic mean) of the two speeds. I demonstrated why that is true. So that is the "other way" you ask for.

If the two parts are not in the same time, then you can't just take the arithmetic mean. You don't need more of a rule than that.

I don't know what is "intuitive" to you. Some people have broken intuition, and have to learn not to follow it; that's why we have proofs. If the proof doesn't seem intuitive to you, then abandon your intuition (on this point). Or rather, make this your new intuition, by getting used to it.

Here is the piece of post #9 you quoted from:
When you take an unweighted mean, you are assuming that each item being averaged is of equal importance. So if you walk at 3 km/hr for 6 minutes to get to your car and then drive for 3 hours at 45 km/hr, does it seem at all intuitive that your average speed was 24 km/hr? The effect of the three hour's driving is much greater than 6 minutes walking. 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.
What is unclear to you about that? He was asking if that (wrong) answer seems intuitive to you, expecting you to say "no". Perhaps you misunderstood that because your English is imperfect. He was not telling you that what he said before the question was true.

If you walk for 6 minutes at 3 kph, then drive for 3 hours at 45 kph, do you think that the average speed would be (3 + 45)/2 = 24 kph? That's what he is asking. If you do, then you intuition is broken, and needs to be fixed. If you see immediately that this is not equivalent to driving for 3:06 at 24 kph, then you do have the intuition you need.
 
Why are u applying arithmetic mean here like this : (10+90)/ = 50 i dont get it .
9 people have got 90 and 1 has got 10 so it should be (9*90 + 1*10)/10 thats the arithemtic mean and also average of score per person
Why are u applying arithmetic mean here like this : (10+90)/ = 50 i dont get it .
Why are you applying AM to: when you drive 1 min at 30mph and then 24 hrs at 60 mph.

Now that you said that my average is wrong (and you are correct) maybe I can help you.

Driving for a length of time is a what is called continuous while getting some scores on an exam is not--it is called discrete.

Lets make driving discrete by looking at your speed every 20 second. Say you drive your car at 30mph for 1 minute and then 60 mph for two minutes.
Now the average would be (30 + 30 + 30 + 60 + 60 + 60 + 60 + 60 + 60)/(3+6) = (3*30 + 6*60)/9. I think that you will agree with this.
Now suppose you record your speed every second. Then every tenth of a second. Then every 100th of a second. Do you see that this is weighted mean?
What do you think would happen if you recorded your speed instantaneously?
 
If you see immediately that this is not equivalent to driving for 3:06 at 24 kph, then you do have the intuition you need.
@Dr.Peterson @JeffM PLEASE READ IT ONCE

the actual average speed taking the info given in post no9 ----> AVG SPEED PER TRIP=43.6451612903 km/hr

NOW if i travel 3/10 km with this avg speed time t1 =0.000687hr & 135 km with this avg speed time t2=3.0931307 hr

Total time taken =3.09 + 0.000687 hr =3.1 hr = 3 hr 6 minutes

CONCLUSION: So it means AVG SPEED should be such that if object travels the same total distance(135.3 km) with avg speed then it should take the exact total time (3hr 6min) that was taken in the actual journey with different speeds .

NOTE:
I think thats why jeffm said in post 9 "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" ------> because suppose if 24 km/hr was the avg speed for real ; then it should take the same time to cover 135.3 km i.e 3hrs 6minutes but it is clearly visible that the time has been 4 hours still THE object with 24km/hr avg speed has covered 96 km short of 39.3 km (135.3-96) and on the top of that it has taken 54 minutes extra (4 hr -3hr 6min).
That why 24km/hr can't be avg speed in this case found through AM method!

I HOPE MY CONCLUSION IS RIGHT?


JeffM said:
What accounts for this? The car traveled at the lower average speed for a LONGER time. The arithmetic mean ignores this fact.

Yes from the beginning my doubt was " i wanted to know how the arithmetic mean ignores the fact" . But i think my above paragraph(IN THE NOTE SEC) gives me the answer how does the AM ignores the fact (only when t1 not equal t2)

AND if i ask myself why AM= avg speed (when t1=t2) , @Dr.Peterson demonstrated why that is true (the formula part) in post #18. I need to accept the proof and there is no intutive explan..

I THINK MY INTERPRETATION IS RIGHT . please go through once sir and this post will be completed then.
 
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