"Average Speed" math question.

[techs]

Me reaches for another pain pill as I realize the question is not so simple.

 

[AstroManLuca]

The two are not equal to eachother

avg speed = distance / time

scenario 1: (1000+100)/(1000/3.6+100/3.8)
scenario 2: (1000+50)/(1000/3.6+50/4.0)

Scenario 1: 3.617308 mile/hr
Scenario 2: 3.617225 mile/hr

Scenario 1 is a higher average speed.

*edit* well technically MikeyLSU is doing distanced averaged mph. I would do time average mile/hr so 50 miles @ 4.0 and 50 miles @ 3.6 mph would be an average speed of (50+50)/(50/4.0+50/3.6) = 3.789474 not 3.8. However, I don't see why anyone would actually want distanced averaged mph.

This. Just did the math myself.

1000 miles at 3.6 mph = 277.77... hours

100 miles at 3.8 mph = 26.315789 hours
50 miles at 4.0 mph = 12.5 hours

So with the first scenario, it's 1100 miles in (277.77... + 26.315789) hours, which works out to 3.617308 mph.

In the second scenario, it's 1050 miles in (277.77... + 12.5) hours, which is 3.617225 mph.

 

[DominionSeraph]

Huh? It's pretty easy that it'd be the first. You increased both by the same amount (0.2mph) when it's not a linear change with the 1000 miles dragging things down.

If you had 1 inch in which to do it, how fast would you have to go to increase the average to 100mph? Obviously a helluva lot when you have 1000 miles at 3.6mph dragging you down! The curve is extremely obvious at that point. The direction of the curve tells you that at 50 miles you'd need more than an average of 4mph to equal 100 miles at 3.8mph. The "more miles" has greater weight as it reduces the effect of the 1000 miles. The greater the mileage added, the closer you are to linearity. (Add a billion trillion miles at 500mph and you're pretty much at an average of 500mph. Those initial 1000 miles aren't doing much at that point)

 

[AstroManLuca]

Huh? It's pretty easy that it'd be the first. You increased both by the same amount (0.2mph) when it's not a linear change with the 1000 miles dragging things down.

If you had 1 inch in which to do it, how fast would you have to go to increase the average to 100mph? Obviously a helluva lot when you have 1000 miles at 3.6mph dragging you down! The curve is extremely obvious at that point. The direction of the curve tells you that at 50 miles you'd need more than an average of 4mph to equal 100 miles at 3.8mph. The "more miles" has greater weight as it reduces the effect of the 1000 miles. The greater the mileage added, the closer you are to linearity. (Add a billion trillion miles at 500mph and you're pretty much at an average of 500mph. Those initial 1000 miles aren't doing much at that point)

I understand your reasoning but if you look at the solution you see that the numbers are very, very close. The examples you listed are easy to work out logically without having to do the math, but the OP's problem is close enough you really have to write it all out to be sure.

Basically, you say that with only 50 miles (small compared to the initial 1000) you'd need to increase your speed to more than 4 mph to have much of an effect on the average, but by the same token, you could also say that 100 miles isn't much more (again, in comparison with the initial 1000) and that 3.8 mph is barely any faster than 3.6 mph.

 

[MotF Bane]

You and Zeze should hang out.

This.

 

[Macamus Prime]

I have traveled 1000 miles at an average speed of 3.6 miles per hour.

Which of the following added to the above will increase my average speed more(or are they equal)?
100 miles at 3.8 miles per hour
50 miles at 4.0 miles per hour

Neither, it's 450,000.

No?

Sorry - I'm bad at math.

 

[HumblePie]

But from what you posted before it sounded like you say the trips should be equal if the total trip is 1000miles(900+100 and 950+50)? or did I read what you wrote incorrectly?

You read it incorrectly.


He already traveled 1000 miles at 3.8mph.

The next part is when you ADD, and the original problem specifically states add to what is already done, to raise the average mph more.

1000/3.6 + 100/3.8

or

1000/3.6 + 50/4


Which is higher? It's simple when you think about it that way.

 

[CLite]

Huh? It's pretty easy that it'd be the first. You increased both by the same amount (0.2mph) when it's not a linear change with the 1000 miles dragging things down.

If you had 1 inch in which to do it, how fast would you have to go to increase the average to 100mph? Obviously a helluva lot when you have 1000 miles at 3.6mph dragging you down! The curve is extremely obvious at that point. The direction of the curve tells you that at 50 miles you'd need more than an average of 4mph to equal 100 miles at 3.8mph. The "more miles" has greater weight as it reduces the effect of the 1000 miles. The greater the mileage added, the closer you are to linearity. (Add a billion trillion miles at 500mph and you're pretty much at an average of 500mph. Those initial 1000 miles aren't doing much at that point)

First of all the curve is not extremely obvious 4.005 is not very different from 4.

Also if the problem read.

I have traveled 889 miles at an average speed of 3.6 miles per hour.

Which of the following added to the above will increase my average speed more(or are they equal)?
100 miles at 3.8 miles per hour
50 miles at 4.0 miles per hour.

Then the 50miles@4mph would be a higher average.

You are telling me you can just logically tell the difference between 889 miles and 1000 miles without doing at least some basic multiplication and addition? I highly highly doubt this. It is most definitely an analytical solution and not a logical solution.

 

[TecHNooB]

You read it incorrectly.


He already traveled 1000 miles at 3.8mph.

The next part is when you ADD, and the original problem specifically states add to what is already done, to raise the average mph more.

1000/3.6 + 100/3.8

or

1000/3.6 + 50/4


Which is higher? It's simple when you think about it that way.

it's not that simple lol

 

[AstroManLuca]

You read it incorrectly.


He already traveled 1000 miles at 3.8mph.

The next part is when you ADD, and the original problem specifically states add to what is already done, to raise the average mph more.

1000/3.6 + 100/3.8

or

1000/3.6 + 50/4


Which is higher? It's simple when you think about it that way.

It's actually:

(1000+100)/[(1000/3.6)+(100/3.8)]

vs.

(1000+50)/[(1000/3.6)+(50/4)]

Plenty of parentheses added for clarity.

 

[silverpig]

A better question is why the fuck are you walking 1000 miles?

 

[AstroManLuca]

A better question is why the fuck are you walking 1000 miles?

just to be the man who walked a thousand miles to fall down at your door

 

[HumblePie]

It's actually:

(1000+100)/[(1000/3.6)+(100/3.8)]

vs.

(1000+50)/[(1000/3.6)+(50/4)]

Plenty of parentheses added for clarity.

well yah, I didn't put the full formula doh.

which if I did would actually be longer.

(1000 miles + 100 miles) / [(1000 miles /3.6 miles /1 hour) + (100 miles /3.8 miles /1 hour)]

vs

(1000 miles + 50 miles) / [(1000 miles /3.6 miles /1 hour) + (50 miles /4 miles /1 hour)]


That would be the full long and correct formula.

 

[DominionSeraph]

You are telling me you can just logically tell the difference between 889 miles and 1000 miles without doing at least some basic multiplication and addition? I highly highly doubt this.

I never said I wasn't using math. It's just really informal.

I can tell that the point of intersection is below 1000 miles at 3.6mph. But if you gave me 500 miles at 3.6mph and asked me whether 50 miles at 4.0 or 100 miles at 3.8 would give the higher result, I'd have a hard time eyeballing it, even though now the difference is greater in favor of the 3.8 than it was before with the 4.0. The mileage added is just proportionally too great -- a small mistake in estimation there results in a large mistake on the other side. I can't hold them to a small enough range to shave against 4.0 and 3.8. They overlap in my margin for error, so all I get is the fuzz of noise.

 

[CLite]

I never said I wasn't using math. It's just really informal.

I can tell that the point of intersection is below 1000 miles at 3.6mph. But if you gave me 500 miles at 3.6mph and asked me whether 50 miles at 4.0 or 100 miles at 3.8 would give the higher result, I'd have a hard time eyeballing it, even though now the difference is greater in favor of the 3.8 than it was before with the 4.0. The mileage added is just proportionally too great -- a small mistake in estimation there results in a large mistake on the other side. I can't hold them to a small enough range to shave against 4.0 and 3.8. They overlap in my margin for error, so all I get is the fuzz of noise.

From your earlier post, what do you mean both increase by 0.2mph but the 100 miles weighs it more.

What is the exact curve you are referring to? If you can produce one the inflection point is actually at a base of 900 miles, I don't see how 500 miles is less obvious for you when it is 400 miles from the inflection point as compared to 1,000 miles only being 100 miles away from the inflection point.

 

[ShawnD1]

And yes, I know I should know the answer. What can I say? I'm medicated today.

Sounds more like you're off meds today :hmm:

He already traveled 1000 miles at 3.8mph.
The next part is when you ADD, and the original problem specifically states add to what is already done, to raise the average mph more.
1000/3.6 + 100/3.8
or
1000/3.6 + 50/4
Which is higher? It's simple when you think about it that way.

Let's try it based on time. Original distance of 1000 at 3.6 would take 277.8 hours.
50 at 4.0 would take 12.5 hours
100 at 3.8 would take 26.3 hours

scenario A:
(1000 + 50) miles at (277.8 + 12.5) hours = 3.616 miles per hour

scenario B:
(1000 + 100) miles at (277.8 + 26.3) hours = 3.617 miles per hour


So basically I don't know. They could be the same or I could have rounding errors because I have rounding errors.

 

[Ricemarine]

I thought it could be done easier than that...

1000 miles * 3.6 miles/hr = 3600 miles^2/hr

100 miles * 3.8 miles/hr + 3600 miles^2/hr= 3980 miles^2/hr
3980 miles^2/hr / (1000 miles + 100 miles) = 3.61818 mph

50 miles * 4.0 miles/hr + 3600 miles^2/hr= 3800 miles^2/hr
3800 miles^2/hr / (1000 miles + 50 miles) = 3.61905 mph

Thus, 50 miles at 4.0 mph is faster. But, it seems to be going against the other method so... something went wrong here lol. It's probably because I didn't treat the initial info and the options separately properly.

 

[AstroManLuca]

I can tell that the point of intersection is below 1000 miles at 3.6mph.

How?

The difference between the two possible answers in the OP is less than 0.0001 mph. Don't see how you could possibly "eyeball it." I understand there are situations where you COULD figure it out without doing all of the math, but this isn't one of them.

Either you have really amazing mental math skills or you simply got lucky in your guess and think that you have a better grasp of the situation than you actually do.

 

[ShawnD1]

weighted average maybe?

scenario A:
((1000 * 3.6) + (50 * 4.0)) / 1050
= (3600 + 200) / 1050
= 3.619 mph average

scenario B:
((1000 + 3.6) + (100 + 3.8)) / 1100
= (3600 + 380) / 1100
= 3.618 mph average


I don't know. Rounding errors maybe?

 

[AstroManLuca]

You're doing it wrong. Keep the units in mind.

1000 miles at 3.6 miles per hour. Think about it. What piece of information is missing? The number of hours. To get that, you DIVIDE 1000 by 3.6.

1000 * 3.6 is what you'd do if you wanted to know how many miles you'd go if you traveled at 3.6 miles per hour for 1000 hours.

This is why they taught unit canceling in school.

1000 miles * (1 hour/3.6 miles)
miles cancel out
277.777... hours

The way you're doing it, you end up with a nonsense unit (square miles per hour).

 

[TecHNooB]

weighted average maybe?

scenario A:
((1000 * 3.6) + (50 * 4.0)) / 1050
= (3600 + 200) / 1050
= 3.619 mph average

scenario B:
((1000 + 3.6) + (100 + 3.8)) / 1100
= (3600 + 380) / 1100
= 3.618 mph average


I don't know. Rounding errors maybe?

you have to weight by time and the solution probably decomposes into something already posted cuz its essentially the same.

 

[ShawnD1]

The way you're doing it, you end up with a nonsense unit (square miles per hour).

???

miles * miles/h / miles = miles/hour

I have no idea how you ended up with a squared unit.

 

[DominionSeraph]

From your earlier post, what do you mean both increase by 0.2mph but the 100 miles weighs it more.

Out that far, a large increase in distance would be preferred over a small increase in speed. Far easier to have a significant change in the average over 1000 miles than 1 inch. 50 miles just doesn't weigh the 4.0 heavily enough to overcome 100 miles at 3.8. It's a greater percentage difference in distance to 1000 miles than speed to each other.

What is the exact curve you are referring to? If you can produce one the inflection point is actually at a base of 900 miles, I don't see how 500 miles is less obvious for you when it is 400 miles from the inflection point as compared to 1,000 miles only being 100 miles away from the inflection point.

The percentage of 100/50 miles to 1000 miles is close enough to that of 1100/1050 miles to use that with just a nudge for correction. 100/50 to 500 is not. That adds four calculations with all of the following calculations based off those, and none of those are easy fractions. So the margin for error goes through the roof.
I'm not gonna do the inverse of 11/12ths of 9% vs 16% in my head. (or something like that. I really don't know how I work these things.)

 

[Ricemarine]

weighted average maybe?

scenario A:
((1000 * 3.6) + (50 * 4.0)) / 1050
= (3600 + 200) / 1050
= 3.619 mph average

scenario B:
((1000 + 3.6) + (100 + 3.8)) / 1100
= (3600 + 380) / 1100
= 3.618 mph average


I don't know. Rounding errors maybe?

The scenarios are swapped. Therefore, you have the same answer as mine.

 

[HumblePie]

???

miles * miles/h / miles = miles/hour

I have no idea how you ended up with a squared unit.

He was doing the units wrong.

(1000 miles / 3.6 miles) * hour. The miles cancels out and the only unit left is hours. Then you divide total distance by hours driven to get miles per hour.