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"Average Speed" math question.

techs

techs

Lifer
Ok, so math isn't my best subject.

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.

If you could post the formula it would also be helpful.

And yes, I know I should know the answer. What can I say? I'm medicated today.
 
techs said:
Ok, so math isn't my best subject.

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.

If you could post the formula it would also be helpful.
Click to expand...
d = t * v

Distance = time of travel * average velocity.

That is about all you need. Use it several times and you'll get your answer. Answer these in this order and you'll answer your question:
1) What was the original distance?
2) What was the original velocity?
3) What was the original time (use the formula)?
4) What is the new total distance in the first option?
5) What is the new total time in the first option?
6) What is the new average velocity in the first option?
7) What is the new total distance in the second option?
8) What is the new total time in the second option?
9) What is the new average velocity in the second option?
10) Is the answer to #6 or #9 bigger?
 
you can just look at it and figure the answer to this one.

you go 50 miles at 4.0 mph, and 50 miles at 3.6 mph.
or
you go 100 miles at 3.8

Look at it that way(since when you are not changing speeds, you are going 3.6 mph), an the 2 are obviously equal to each other.
 
MikeyLSU said:
you can just look at it and figure the answer to this one.

you go 50 miles at 4.0 mph, and 50 miles at 3.6 mph.
or
you go 100 miles at 3.8

Look at it that way(since when you are not changing speeds, you are going 3.6 mph), an the 2 are obviously equal to each other.
Click to expand...

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.
 
Last edited:
CLite said:
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.
Click to expand...

average speed is a time thing anyway so ..
 
TecHNooB said:
average speed is a time thing anyway so ..
Click to expand...

yep, people in this thread who suggest it's rudamentary logic (ala cyclowizard, mikeyLSU) are incorrect. Average speed clearly is a time-averaged calculation.

If the speed was 4.005 instead of 4.0 then Scenario 2 would have been faster. It is an extremely marginal difference that would have been impossible to solve without going through the calculations.
 
CLite said:
yep, people in this thread who suggest it's rudamentally logic (ala cyclowizard, mikeyLSU) are incorrect. Average speed clearly is a time-averaged calculation.

If the speed was 4.005 instead of 4.0 then Scenario 2 would have been faster. It is an extremely marginal difference that would have been impossible to solve without going through the calculations.
Click to expand...

huh?

(100/3.8)<(50/4+50/3.6)

edited for writing the wrong formula
 
Last edited:
sdifox said:
that is why I changed the formula.
Click to expand...

That formula still has no relevence to the answer of this problem.

*edit* for instance if you use 4.005mph , your formula still has the same result, but the answer to the original problem changes.
 
CLite said:
yep, people in this thread who suggest it's rudamentary logic (ala cyclowizard, mikeyLSU) are incorrect. Average speed clearly is a time-averaged calculation.
Click to expand...
Apparently what I consider rudimentary logic is the limit of higher brain activity for someone like you. My mistake.
 
CycloWizard said:
Apparently what I consider rudimentary logic is the limit of higher brain activity for someone like you. My mistake.
Click to expand...

Get off your high horse, rudimentary logic implies a logical solution without having to do an analytical analysis. This problem requires doing very simple math which is an analytical based solution, it is not a logic based solution.

Assuming you have a college education you know the difference between a logic based solution and an analytical solution.
 
CLite said:
That formula still has no relevence to the answer of this problem.

*edit* for instance if you use 4.005mph , your formula still has the same result, but the answer to the original problem changes.
Click to expand...

(100/3.8)<(50/4+50/3.6)

I don't see why you think there is no relevance. 900km of the journey is identical, so you just reduce them out of the picture. The question becomes what yields shorter time to travel the 100km. And that is what my formula shows.
 
sdifox said:
(100/3.8)<(50/4+50/3.6)

I don't see why you think there is no relevance. 900km of the journey is identical, so you just reduce them out of the picture. The question becomes what yields shorter time to travel the 100km. And that is what my formula shows.
Click to expand...

They aren't traveling the same total distance, one travels 1,100 the other travels 1,050.
 
CLite said:
They aren't traveling the same total distance, one travels 1,100 the other travels 1,050.
Click to expand...


Fuck me, I read the question wrong...I thought it was within the 1000km, not in addition.

<=== still pissed at election result, not thinking straight.
 
sdifox said:
Fuck me, I read the question wrong...I thought it was within the 1000km, not in addition.

<=== still pissed at election result, not thinking straight.
Click to expand...

yeah, I took the questino to mean both were still part of the original 1000 miles so each scenerio still had 1000 miles.

Oh well.
 
MikeyLSU said:
yeah, I took the questino to mean both were still part of the original 1000 miles so each scenerio still had 1000 miles.

Oh well.
Click to expand...


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?
 
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