OT: What simple rule of math am I missing here?

bgebrent

theyolksonyou said:

bgebrent said:

I started out as a math major. It became clear to me very early on that a chemistry major would get hotter chicks. It's all about plastics. And math majors sucked at physical chemistry. 

Worked out for you. 

Indeed it did. Chemistry is sexy. 

HeavyG

maso said:

maso said:

JohnInCarolina said:

Velocity is distance over time. So time is distance over velocity.

The original time is one hour.  If you want that to remain the same using two different velocities that just happen to average to your original 5mph, then you will have to walk *different* distances for each of them.  Let's see if we can show this.

The original velocity is 5 miles per hour.  So we have:

1 = (5mi)/(5 mph) 

or

T = D/V 

We want the same T for two different Ds and two different Vs, or

T = D1/V1 + D2/V2, where D1+D2 = D.  

In your case, the V1 (6mph) is 1.2 V, and the V2 (4mph) is 0.8V.  So we have

T = D1/ 1.2V + D2/ 0.8V

or

T = D1/1.2V + 1.5 D2/ 1.2V = (D1 + 1.5D2) / 1.2V

So this will be the same as the above if

(D1 + 1.5D2)/1.2 = D

and we also know that:

(D1 + D2) = D.  

I'm assuming you all know how to solve a simultaneous system of linear equations for two unknowns, so I'll leave it there.   But suffice it to say that it is easy to show that D1 = D/2 and D2 = D/2 is not a solution.  

I mean that's wrong but ok. 

IF you want an formula for average speed, with each speed over a same distance, then:

Speed = 2ab/(a + b), where a and b are the speeds. so in this situation, a = 4, b = 6 so

2*4*6/(4+6) = 48/10 or 4.8 mph (or 5 miles in 62.5min)

theyolksonyou

HeavyG said:

maso said:

maso said:

JohnInCarolina said:

Velocity is distance over time. So time is distance over velocity.

The original time is one hour.  If you want that to remain the same using two different velocities that just happen to average to your original 5mph, then you will have to walk *different* distances for each of them.  Let's see if we can show this.

The original velocity is 5 miles per hour.  So we have:

1 = (5mi)/(5 mph) 

or

T = D/V 

We want the same T for two different Ds and two different Vs, or

T = D1/V1 + D2/V2, where D1+D2 = D.  

In your case, the V1 (6mph) is 1.2 V, and the V2 (4mph) is 0.8V.  So we have

T = D1/ 1.2V + D2/ 0.8V

or

T = D1/1.2V + 1.5 D2/ 1.2V = (D1 + 1.5D2) / 1.2V

So this will be the same as the above if

(D1 + 1.5D2)/1.2 = D

and we also know that:

(D1 + D2) = D.  

I'm assuming you all know how to solve a simultaneous system of linear equations for two unknowns, so I'll leave it there.   But suffice it to say that it is easy to show that D1 = D/2 and D2 = D/2 is not a solution.  

I mean that's wrong but ok. 

IF you want an formula for average speed, with each speed over a same distance, then:

Speed = 2ab/(a + b), where a and b are the speeds. so in this situation, a = 4, b = 6 so

2*4*6/(4+6) = 48/10 or 4.8 mph (or 5 miles in 62.5min)

Oh boy....MATH FIGHT!!!

One of you young dudes, break out common core!

bgebrent

theyolksonyou said:

HeavyG said:

maso said:

maso said:

JohnInCarolina said:

Velocity is distance over time. So time is distance over velocity.

The original time is one hour.  If you want that to remain the same using two different velocities that just happen to average to your original 5mph, then you will have to walk *different* distances for each of them.  Let's see if we can show this.

The original velocity is 5 miles per hour.  So we have:

1 = (5mi)/(5 mph) 

or

T = D/V 

We want the same T for two different Ds and two different Vs, or

T = D1/V1 + D2/V2, where D1+D2 = D.  

In your case, the V1 (6mph) is 1.2 V, and the V2 (4mph) is 0.8V.  So we have

T = D1/ 1.2V + D2/ 0.8V

or

T = D1/1.2V + 1.5 D2/ 1.2V = (D1 + 1.5D2) / 1.2V

So this will be the same as the above if

(D1 + 1.5D2)/1.2 = D

and we also know that:

(D1 + D2) = D.  

I'm assuming you all know how to solve a simultaneous system of linear equations for two unknowns, so I'll leave it there.   But suffice it to say that it is easy to show that D1 = D/2 and D2 = D/2 is not a solution.  

I mean that's wrong but ok. 

IF you want an formula for average speed, with each speed over a same distance, then:

Speed = 2ab/(a + b), where a and b are the speeds. so in this situation, a = 4, b = 6 so

2*4*6/(4+6) = 48/10 or 4.8 mph (or 5 miles in 62.5min)

Oh boy....MATH FIGHT!!!

One of you young dudes, break out common core!

Common core will insure their both right. 

maso

Moreover, @theyolksonyou, if you want a formula for average speed for any distance and any speed:

avg speed = total distance/total time

D = D1 + D2
T = T1 + T2

T1 = D1/V1
T2 = D2/V2

T = D1/V1 + D2/V2

S = (D1 + D2) / (D1/V1 + D2/V2)

There's no system of equations to solve. You have to know the distances of each run and their speeds or you can't solve it.

JohnInCarolina

@maso you are answering a different question from the one I have addressed.  

I turned it around to make the general point as to why halving the distance for each of the speeds wouldn't work, and then showed how to determine the proper distances for each speed such that you walk the same total distance in the same amount of time.  

bgebrent

maso said:

Moreover, @theyolksonyou, if you want a formula for average speed for any distance and any speed:

avg speed = total distance/total time

D = D1 + D2
T = T1 + T2

T1 = D1/V1
T2 = D2/V2

T = D1/V1 + D2/V2

S = (D1 + D2) / (D1/V1 + D2/V2)

There's no system of equations to solve. You have to know the distances of each run and their speeds or you can't solve it.

You must declare your school of education. It will enrich the conversation. 

maso

JohnInCarolina said:

@maso you are correct, but you are answering a different question from the one I have addressed.  

I turned it around to make the general point as to why halving the distance for each of the speeds wouldn't work, and then showed how to determine the proper distances for each speed such that you walk the same total distance in the same amount of time.  

Well I was answer the question he asked; but anywho, you still had wrong conversions, for what it's worth.

Wolfie51sb

bgebrent said:

Common core will insure their both right. 

Now see, this is in my wheelhouse:

Common Core will ensure they're both right.

nolaegghead

nolaegghead

nolaegghead

bgebrent

nolaegghead said:

E still equals mc x2 right?

nolaegghead

nolaegghead

</troll>

HeavyG

Wolfie51sb said:

bgebrent said:

Common core will insure their both right. 

Now see, this is in my wheelhouse:

Common Core will ensure they're both right.

Oh SNAP!

bgebrent

Still open to pinks. 

Thatgrimguy

Holy ****. This thread reminded me why I miss this place so much. I got to get more internets in

maso

bgebrent said:

You must declare your school of education. It will enrich the conversation. 

Fellow SEC'er. We do occasionally stop facking our cousins to learn atleast some of the three R's. 

bgebrent

maso said:

bgebrent said:

You must declare your school of education. It will enrich the conversation. 

Fellow SEC'er. We do occasionally stop facking our cousins to learn atleast some of the three R's. 

Hate the 'backs, you're the exception brother. 

maso

bgebrent said:

maso said:

bgebrent said:

You must declare your school of education. It will enrich the conversation. 

Fellow SEC'er. We do occasionally stop facking our cousins to learn atleast some of the three R's. 

Hate the 'backs, you're the exception brother. 

It's ok, most of the time I hate them too! Lucky y'all have basketball (even if you did steal Monk!)!

bgebrent

maso said:

bgebrent said:

maso said:

bgebrent said:

You must declare your school of education. It will enrich the conversation. 

Fellow SEC'er. We do occasionally stop facking our cousins to learn atleast some of the three R's. 

Hate the 'backs, you're the exception brother. 

It's ok, most of the time I hate them too! Lucky y'all have basketball (even if you did steal Monk!)!

Monk was an amazing steal. When it comes to round ball, we're better off not talking. 

Wolfie51sb

HeavyG said:

Wolfie51sb said:

bgebrent said:

Common core will insure their both right. 

Now see, this is in my wheelhouse:

Common Core will ensure they're both right.

Oh SNAP!

Yeah, everyone hates copy editors.

JohnInCarolina

maso said:

JohnInCarolina said:

@maso you are correct, but you are answering a different question from the one I have addressed.  

I turned it around to make the general point as to why halving the distance for each of the speeds wouldn't work, and then showed how to determine the proper distances for each speed such that you walk the same total distance in the same amount of time.  

Well I was answer the question he asked; but anywho, you still had wrong conversions, for what it's worth.

I understand the confusion now.  You're focusing on the use of average speed.  I was just using the OP's notion - meaning that (4+6)/2 = 5, even though that doesn't necessarily turn out to be the average velocity you'll have over the interval.  

If you solve my system, it will give you the right distances to walk at a velocity of 4mph and 6mph, such that you walk a total of five miles in one hour.  The OP had asked why, if he alternates quarter miles (in other words, equal distances), it ends up taking longer.  All I did was to demonstrate why that's the case.  

DoubleEgger

If a westbound train leaves the station at 8am traveling 58mph...

nolaegghead

Wuddya, some kinda "math professor" or something John?

bgebrent

nolaegghead said:

Wuddya, some kinda "math professor" or something John?

He hasn't solved that impossible train equation yet.  Could take years. 

JohnInCarolina

nolaegghead said:

Wuddya, some kinda "math professor" or something John?

Nah I just stayed at the Holiday Inn Express last night.   

bgebrent

JohnInCarolina said:

nolaegghead said:

Wuddya, some kinda "math professor" or something John?

Nah I just stayed at the Holiday Inn Express last night.   

Need to see the receipt, otherwise you're a crackpot. 

JohnInCarolina

The solution is pretty simple by the way.  Instead of trading speeds every quarter mile, you trade them in a ratio of 3 to 2 (or 6 to 4).  So you walk the first two at 4mph, then you walk the next three at 6mph.  Then the next two at 4mph, and so on.  

You end up covering a total of 3 miles at a speed of 6mph and 2 at a speed of 4mph.  You can certainly figure this out without any of the algebra I used.  But that's not quite as fun.