attention ded. reckoning navigators....

[james ozzie]

Here's one for all you experienced navigators out there. I got it from my daughters Year 11 maths homework. I also got it wrong...

If you travel at a steady 40 km/hr from point A to point B, how fast must you travel on the reverse leg B to A to average 80km/hr for the whole journey?

[Solid Rust Twotter]

Still conditions?

Errmmmm, 120 km/h?

[Sultan Ismail]

At light speed, in a nanosecond, or without further delay

or something like that

[sky waiter]

I agree with solid rust, 120km/h in still wind or is there something hidden? Wind velocity would help:p

[Farmer 1]

If the distance from A to B is 40km, it takes 1 hour at 40kph.

The distance A - B - A is 80km. At an average speed of 80kph, that will take 1 hour, which has already expired reaching point B.

I believe the Sultan is 1 nanosecond late.

When you see the word "average" in a question like this, always smell a rat. Bit cruel for a class of 11-year olds, though, or are our standards of education improving?

[Cap Loko]

Nice trick,

After some puzzling figured out this: Not Possible.

Only if you extend the return leg B-A-C while doing 120 kph, you will get 80 kph average.

Like farmer said. Upon arriving at B, your time is already expired and given the distance=speed x time formula you already have 2 variables (distance and speed) fixed and therefore the time determined.

The confusion is that the question suggests a number as an answer.

James, a question for the Maths teacher to return (it's a classic):

A plane is flying from A to B and back to A with a constant IAS. Now there's some wind which will be constant in speed and direction at all altitudes at all times; the pilot will have to deal with exactly the same wind on the way to B and then back to A.

Will this flight take: shorter, the same, or longer in time to fly A-B-A then the same A-B-A flight with no wind at all?

[Recuperator]

Correct me if I'm wrong, but there is not many 11 year olds in year 11 and secondly, it is not a trick question because it uses the word "average". The question is not a time question, it is a question to determine average speed. The question states:

how fast must you travel on the reverse leg B to A to average 80km/hr for the whole journey?not what distance must you cover in 1 hour.

As my teacher always said: RTFQ!! - (Read the f:mad:ing question):hmm:

So look at the post again or quietly get back to the business of farming, as reading might not and doesn't seem to be your strongest point.;) :p

Fire away...:)

[tired]

160 km/h

If you travel the outbound leg at 1/2 the speed req, you need to travel at twice the speed req on the return.

[Farmer 1]

Dear Recuperator.

I agree - I missed the point of the phrase "year 11". Mine is a different culture from yours, and I failed to pick up on the relevance. I apologise. However, this obviously has no bearing on the question, or the answer.


So, let me begin again. I will type this very slowly, specially for you:

The known facts: two points, A and B; and you travel between them at a steady speed of 40 kph. These are all we are given - nothing more. We have no need to know where points A and B are, how far apart they are, what the wind speed is, the magnetic variation, how much fuel you have or what you had for breakfast. This is not an aviation question, it is a question for year 11 students. It is not even a trick question, and I never suggested it was.

When it says a steady 40 km/hr, it means that this is the AVERAGE speed for the journey A to B.

Are you still with me, or would you like me to start again? For the moment, I will continue.

Cap Loko mentioned the distance=speed x time formula, which is what this is all about. But we only have one of the afore-mentioned - the speed. So, we need to invent one of the others in order to work out the third.

This is why I suggested a distance of 40 km. Because it is easy. But, if you prefer to make life difficult for yourself, then you may choose any distance you wish - the world is your lobster. I am typing this at 1133 local time. I know - why not choose that number as the distance from A to B? (I will stick with km, but if you prefer, you can have anything from nanometers to light years, in any unit you wish. As I said, the world is...).

1133 km at 40 km/hr = 28.325 hours.

If the distance A - B is 1133 km, the distance A - B - A = 2 x 1133 km. Which is 2266 km.

To travel 2266 km at an AVERAGE speed of 80 km/hr will take you 2266 / 80 hr.
Which is - wait for it - wait for it - 28.325 hours.

I am forced to repeat myself from my initial post - "... which has already expired reaching point B."

Recuperator - did you ever listen to your teacher? Is it necessary to be so crude and offensive to someone you've never met? Why not take a day off once in a while. And did I say I was actually a farmer, or do you have something against them as well?

James ozzie - great question, but did your daughter get it right? You have this knack of raising emotions in the lower orders. By the way, could you please tell me how old a year-11 student is?

Farmer

[james ozzie]

Yes, the solution is that it is not possible to achieve that average speed. Mr Sulu could get close to it with his warp drive engines on the star ship Enterprise but could never reach it.

Most people (including me) initially offer 120 km/h as the answer.

There were some good explanations posted here - thanks everyone for taking an interest and having a shot at it. I thought it a good example of how a problem can be disguised to appear simple.

No. she did not get it right. Year 11 here in Queensland is mainly 16 year olds.

But I am sorry to read some of the sprightly reparti appearing here - please folks, it is just a fun riddle.....

[Farmer 1]

But let's not stop here - what about Cap Loko's question?

"A plane is flying from A to B and back to A with a constant IAS. Now there's some wind which will be constant in speed and direction at all altitudes at all times; the pilot will have to deal with exactly the same wind on the way to B and then back to A.

Will this flight take: shorter, the same, or longer in time to fly A-B-A then the same A-B-A flight with no wind at all?"


Once again - let's keep things simple:

The distance A to B is (shall we say?) 100 nm, and you fly at a constant airspeed of (shall we say?) 100 kts. Choose other figures as you will - refer to authors above, no names, no packdrill. With no wind, the return trip will take two hours.

If you now have a headwind component of (shall we say?) 50 kts, this reduces your groundspeed to 50 kts, and it will take you two hours to do the 100 nm. Once again, your time is expired on the outbound leg.

What matters here is the strength of the wind, not the direction. No matter what anybody tries to tell you, if you have any wind AT ALL on a round trip, it will take you longer than if there is no wind.

I have a suspicion that the direction can make a difference, but I believe only slight. I wait to be corrected, preferably by someone with more knowledge on the subject than I.

[Recuperator]

Farmer 1, james ozzie,

No offence meant mates, a bit of bantering that was construed as a personal attack, which it wasn't. Apologies if it came across like that and by the way some of my best friends are farmers and Aussies, and some really don't read that well!;) Just kidding!!I must admit, for a non-farmer you have quite a bite with the sarcasm...:p

I still maintain that you are looking at it and giving me an answer in time, when the question asks for a speed.

Albeit so, even if humanly not possible, if your steady speed is 40km/hr even over 1133 km's or what ever A-B distance, takes 28.325 hours, as you rightly said, what speed would you have to increase to on the B-A leg, to average 80 km's/hr for the journey?

Surely any average speed over two legs will be speed of each leg added to each other and divided by 2, 4 legs divided by 4 and so on? How do they get an average lap time on the Formula 1 circuit? If I do one lap at 40 km/hr and the next lap at 120km/hr, surely my average speed is 80 km/hr?:confused:

Please oblige me, I stand humble in your presence and knowledge, a mere mortal and uninformed soul!

And yes you were right Farmer 1, I don't know if the educational systems we live with are better these days, but the educational standards does seem higher, but one thing is sure, our children are much wiser than what we were when we were their age, doing more and more advanced stuff. E.G. sometimes when the computer gets the better of me, I call my son and he sorts the bl**dy thing out. Times are changing!

Take care!:ok:

[forssi]

It's 120km/h

2 different speeds ( 40 km/h A - B) ( ? km/h B- A)

average is 80km'h x 2 = 160

160- 40 = 120

simple!

[Sultan Ismail]

So, I got it right then :D

[Milt]

Navigators and members of that illustrious Red Indian Tribe Wherethehellarewe

When and by what edict did "Dead Reckoning" change to "Ded Reckoning"?

[Sultan Ismail]

Milt
Did you not notice the full stop after ded?

Signifies an abbreviation, thus: ded. = deduced

[Solid Rust Twotter]

To simplify things:

Inbound speed would have to be the speed required (80 km/h) plus the difference between speed required and actual outbound speed, ie 40 km/h, thus 40 + 80 = 120.

If you did 60 km/h outbound, you'd only have to do 100 km/h inbound to average 80 km/h.

Or: 80x2(both legs) = 160
160 - 40 = 120

Why are we waffling on about this again? :confused:

[Recuperator]

Gives us something to do other than work, following good advice to take a day off once in a while. ;)

[WideGlide]

Suppose half the distance is x. Time = dist/speed = x/40

Then for the second half of the journey, Time = x/v (v = speed on
second half)

Total time = x/40 + x/v Total distance = 2x

Average speed for whole journey = Total dist/Total time
= 2x/(x/40 + x/v)
= 2/(1/40 + 1/v)

and this must equal 80. So 2/(1/40 + 1/v) = 80
1/40 = 1/40 + 1/v
0 = 1/v so v = infinity

This shows that you would require an infinite speed on the second half
of the journey to get an average speed of 80 mph.

Dinner we've had
Meal we made of it!

[Sultan Ismail]

To make it even simpler, the question asks for average speed for the WHOLE journey.

Thus A to B and B to A

If AB = unit of distance

To = Elapsed time outbound

Tw = Elapsed time (outbound + Inbound)

Now:

To = AB/40

Tw = 2AB/80 which in turn = AB/40

Thus To = Tw

In other words you would have to arrive back at A at the same time as you arrive at B


And why is an Australian asking this on the Africa Forum?

[Farmer 1]

Recuperator,

You are right. I did construe your bantering as a personal attack. However, I accept your apology, so let’s be friends and say no more about that. Did I say I was actually not a farmer, by the way?

But you are still wrong about the rest. Remember my warning – if the word “Average” is used, always smell a rat. Cap Loko saw the nail and hit it right on the head – the question suggests a number as an answer. If the question had been, “If you travel at a steady 40 km/hr from point A to point B, IS IT POSSIBLE TO TRAVEL AT SUCH A SPEED ON THE RETURN JOURNEY TO AVERAGE 80 KM/HR”, that might have made more people scratch their heads and say, “Now just hang on a minute, let’s think about this one.” The obvious answer is “Yes, of course. What a stupid question.” But then you might think, “Why should an intelligent person ask such a stupid question? Is it, after all, really such a stupid question?” And maybe a tiny seed of doubt is planted.

Recuperator, you fatal error is adding two average speeds and assuming the average of those speeds is actually the average overall speed. You cannot do this. It does not work. To find the average speed, you add up the total distance, and you divide it by the time it takes to cover that distance. Distance = speed x time.

As regards Formula 1 (and I missed yesterday’s race), if you are simply recording lap times, then yes, you time each lap, add up the times, and divide by the number of laps. This gives you the average lap time, but that is all. You cannot use this to work out the average speed, it just does not work that way, my friend.

To use your example in motor racing, one lap at 40 km/hr and one lap at 120 km/hr does not give you an average speed of 80 km/hr. Please believe me.

Let us say a lap is 120 km long. I know this is ridiculous, and even the old Nurburgring didn’t come close, but it is an easy number to use, and once again you can use any distance you wish:

Lap one: 120 km / 40 km/hr will take you 3 hours.
Lap two: 120 km / 120 km/hr will take you 1 hour.
Total distance: 240 km.
Total time: 4 hours.
AVERAGE speed: 240 km / 4 hours gives you 60 km/hr.

Another question for you: You have two groups of people; the average age of one group is 40 years, and the average age of the other group is 60 years. What is the average age of both groups? Let me warn you again – I have used the word “Average” here, three times. Smell a rat. I might even admit that this is a trick question

Another point: I suspect (and hope) that you have more than the average number of fingers and toes, I certainly do.

James ozzie, like I said before, a great question, which is still confusing a lot of people. Thank you for enlightening my on year 11, that’s something I’ve learnt.

Farmer.

[james ozzie]

As the initiator of this debate, I think it time we laid it to rest. Thanks for the many well reasoned replies. Can I offer a simple illustration, using the numbers?

Assume A to B 40 km & outbound leg always 40 km/hr (therefore always 60 minutes)

If you return at 120 km/hr, return time is 20 minutes, total time is 80 minutes over 80 km so average speed is 60km/hr

If you return at 1,200 km/hr return time is 2 minutes, total time is 62 minutes over 80km so average speed is 77.43km/hr

If you retrun at 12,000 km/hr, return time is 0.2 minutes, total time is 60.2 minutes so average speed is 79.73 km/hr

If you return at 120,000 km/hr, return time is 0.02 minutes, total time is 60.02 minutes so average speed is 79.97 km/hr

You get the drift...

Asymtotic on 80 km/hr (lovely buzz word)

I will post another elegant problem sometime.

Well done to those who saw the problem very early (Sultan, Farmer, Loka and others)

[Recuperator]

As said, I stand humble in your presence and knowledge, and no, I can't go that fast, even if I tried, thus still a mere mortal soul...:\

[AfricanQueen]

Asymptotic? James Ozzie is a learned man. I had to go to dictionary.com to figure that out.

"A line whose distance to a given curve tends to zero. An asymptote may or may not intersect its associated curve". Eesh