Off-Topic

[NolanRyansSnowmonkey]

re: Off-Topic

My Dad suffered a heart attack yesterday morning and is once again back in the hospital. He is in pretty good spirits as they work to get him stabilized again. His heart is extremely weak and damaged from past incidents that there is nothing they can do for his heart other than treat with medicine to prolong life.

He currently has fluid around his heart that they are using dialysis to remove. Once they get that off he should be able to come home. He spent Father's Day in the PCU at the hospital and will now most likely spend his 65th birthday in there, which is Thursday.

Sorry for the Debbie Downer post...just needed to put something to paper.

Thoughts and prayers are with you, him, and your family, CB. And i'm not just saying that.

[pietasterp]

re: Off-Topic

why do you need to know that?




is it possible to go 15 mph over a one mile distance? yes.




is it possible to go 45 mph over a one mile distance? yes.




so the average is 30 mph

Distance = Rate x Time

In order to average 30 MPH over 2 miles, you must cover that 2 miles in 4 minutes.

Spoiler

2 Miles = 30 Miles/Hour x Time
Time = 1/15 Hour = 1/15 x (60 minutes/ 1 Hour) = 4 minutes

If you go 15 MPH over 1 mile, it took you 4 minutes of time (assuming constant rate of speed).

Spoiler

1 Mile = 15 Miles/Hour x Time
Time = 1/15 Hour = 1/15 x (60 minutes/ 1 Hour) = 4 minutes

Ergo....infinity?

I'm not saying my logic is correct. I'm saying that's how I approached the problem.

[pietasterp]

re: Off-Topic

My Dad suffered a heart attack yesterday morning and is once again back in the hospital. He is in pretty good spirits as they work to get him stabilized again. His heart is extremely weak and damaged from past incidents that there is nothing they can do for his heart other than treat with medicine to prolong life.

He currently has fluid around his heart that they are using dialysis to remove. Once they get that off he should be able to come home. He spent Father's Day in the PCU at the hospital and will now most likely spend his 65th birthday in there, which is Thursday.

Sorry for the Debbie Downer post...just needed to put something to paper.

Oh no! Really sorry to hear that. Hoping for a speedy recovery for your pops. Hang in there.

[l3ulvl]

re: Off-Topic

Blzer has to be laughing his *** off right now

[Caulfield]

re: Off-Topic

Seems like 45mph for the second mile would average out to 22.5 mph for the entire 2 mile stretch...

Look at it this way, lets just say that each mile is travelled by two different riders. Lance Armstrong rides his bike the first mile at a constant 15 mph. Travis Pastrana drives his bike at a constant 45 mph. why does that average 22.5 mph?

[NolanRyansSnowmonkey]

re: Off-Topic

Distance = Rate x Time

In order to average 30 MPH over 2 miles, you must cover that 2 miles in 4 minutes.

Spoiler

2 Miles = 30 Miles/Hour x Time
Time = 1/15 Hour = 1/15 x (60 minutes/ 1 Hour) = 4 minutes

If you go 15 MPH over 1 mile, it took you 4 minutes of time (assuming constant rate of speed).

Spoiler

1 Mile = 15 Miles/Hour x Time
Time = 1/15 Hour = 1/15 x (60 minutes/ 1 Hour) = 4 minutes

Ergo....infinity?

I'm not saying my logic is correct. I'm saying that's how I approached the problem.

We are approaching it the exact same way it seems

[pietasterp]

re: Off-Topic

Seems like 45mph for the second mile would average out to 22.5 mph for the entire 2 mile stretch...

Correct.

If you went 15 MPH for the first mile, it would take you 4 minutes to cover the mile.

If you then went 45 MPH for the second mile, it would take you ~1.3 minutes to cover the second mile.

5.3 minutes is about 0.0883 hours

So for the trip, it would be R = D/T = 2 miles/0.0883 hour = ~22.6 MPH

The issue is that there is a time component to this question, so arithmetic mean is not the right way to calculate the average.

[NolanRyansSnowmonkey]

re: Off-Topic

Look at it this way, lets just say that each mile is travelled by two different riders. Lance Armstrong rides his bike the first mile at a constant 15 mph. Travis Pastrana drives his bike at a constant 45 mph. why does that average 22.5 mph?

Spoiler

Because youve ridden 4 minutes on the first mile. If you ride 45 mph on the second mile it will take 1.33 minutes. Thats a total of 5.33 minutes for 2 miles.

5.33/2 = 2.665 minutes per mile.

60mph/2.665 minutes = 22.51 mph...

[NolanRyansSnowmonkey]

re: Off-Topic

Correct.

If you went 15 MPH for the first mile, it would take you 4 minutes to cover the mile.

If you then went 45 MPH for the second mile, it would take you ~1.3 minutes to cover the second mile.

5.3 minutes is about 0.0883 hours

So for the trip, it would be R = D/T = 2 miles/0.0883 hour = ~22.6 MPH

The issue is that there is a time component to this question, so arithmetic mean is not the right way to calculate the average.

Yes there has to be more to it

[Caulfield]

re: Off-Topic

In order to average 30 MPH over 2 miles, you must cover that 2 miles in 4 minutes.

this is where I get disconnected. I dont get why the trip must take exactly 4 minutes

[NolanRyansSnowmonkey]

re: Off-Topic

Hey stop typing faster than me

[NolanRyansSnowmonkey]

re: Off-Topic

this is where I get disconnected. I dont get why the trip must take exactly 4 minutes

If you average 15mph wouldnt 60minutes/15 miles = 4 minutes?

[pietasterp]

re: Off-Topic

Look at it this way, lets just say that each mile is travelled by two different riders. Lance Armstrong rides his bike the first mile at a constant 15 mph. Travis Pastrana drives his bike at a constant 45 mph. why does that average 22.5 mph?

So this is a good example of the difference in the problems. The problem comes from the term "average", which can be a bit imprecise (or inadequate) in some situations. The average speed of the 2 riders at any given point in time during their rides is indeed 30 MPH (assuming constant speed). However, that's not the question being posed.

I think the question might be better phrased as, over a 2-mile span, if the first mile is covered at a rate of 15 MPH, how fast does the second mile have to be covered in order for the average over the entire 2 mile span to equal 30 MPH?

I still can't wrap my mind around the result of having to cover the second mile in 0 minutes, but I can't see the flaw in my logic or math right now.

Blzer has to be laughing his *** off right now

I had the same thought....diabolical bastard. I expect him to drop some serious knowledge on us about this at some point. How long is he going to let us make fools of ourselves?

BTW, I'm deleting all my posts about this topic when the right answer is revealed....

[pietasterp]

re: Off-Topic

My Dad suffered a heart attack yesterday morning and is once again back in the hospital. He is in pretty good spirits as they work to get him stabilized again. His heart is extremely weak and damaged from past incidents that there is nothing they can do for his heart other than treat with medicine to prolong life.

He currently has fluid around his heart that they are using dialysis to remove. Once they get that off he should be able to come home. He spent Father's Day in the PCU at the hospital and will now most likely spend his 65th birthday in there, which is Thursday.

Sorry for the Debbie Downer post...just needed to put something to paper.

BTW country, I don't want your post to get lost in all the blather about trivial math problems. This this hits us all in the feels for sure, I know we're all thinking about you guys this evening.

[NolanRyansSnowmonkey]

re: Off-Topic

So this is a good example of the difference in the problems. The problem comes from the term "average", which can be a bit imprecise (or inadequate) in some situations. The average speed of the 2 riders at any given point in time during their rides is indeed 30 MPH (assuming constant speed). However, that's not the question being posed.

So the time frame would be the same though, since it doesn't matter if you go 10mph part of the way, and 20 part of the way, if you average 15mph over a mile, the time should always be the same. 4 minutes. If not, then 15mph wouldnt be the average.

Edit: i misread your statement, sorry. But my point remains.