Well done Dozey.
I think that is the only number with that property - at least in English.
Actually I think it requires steady acceleration. Once steady speed is reached, everything returns to its original equilibrium.
Thanks for the explanation, I get the apple one now.
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The railway distance from London to Glasgow is 400 miles.
A train leaves Glasgow for London at 11:00 and travels at an average speed of 80mph.
At 10:30 another train leaves London for Glasgow and travels at an average speed of 70mph
Which train is nearer London when they pass ?
The train departing London travels 35 miles from 10:30am before the train departing Glasgow leaves 0.5 hours later at 11:00am.
The midpoint between London and Glasgow is 200 miles. The train leaving London will met the midpoint in (200 - 35)/70 = 2.36 hours from 11:00am.
The train leaving Glasgow at 11:00am will met the midpoint in 200/80 = 2.50 hours from 11:00 am.
So both trains are closer to Glasgow when they pass, and are both the same distance from London when they pass 
I’ll never get that 5 minutes back……
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Well done Mike. (and sorry about the lost 5 minutes 
Yes, no need for any calculation to get the required answer !! - they are both the same distance (205.33 miles) from London when they pass.
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One could muse as to whether the trains have actually passed each until the last carriages have passed each other, in which case the train leaving Glasgow will be a little bit closer to London 
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Neither. They will both be cancelled due to “staff shortages”.
Well that’s been my recent experience!
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As we allow heavier lorries on our roads, we need to strengthen our bridges, at some considerable cost.
An engineer (let’s call him Mike !) has suggested that we transfer much of our transport of goods to our extensive canal system. Apart from being more environmentally friendly (debateable), our engineer has stated that even if we introduce barges that can carry much heavier loads than at present, we won’t need to strengthen the bridges that carry the canals over roads and railways etc.
Is Mike correct ?
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OK, I’ll bite!
Assuming the canals are deep enough to take them, the large barges would displace large amounts of water along the canal network (ie Archimedes Principle), so the bridges wouldn’t be supporting both the barge and the original volume of water. So maybe Mike is correct.
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I believe so, as the additional weight of the barge loads is compensated for by the corresponding weight of water being displaced in the canal by the barge along both approaches of the bridge.
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I’m sure Mike the engineer is invariably correct.
It took me a few minutes to work out what happens to the displaced water. I can’t add anything to the succinct explanations from Steve and, er, Mike.
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Steve, Mike and Ravvie … well done.
Transport canals are generally level from one lock to the next. The maximum level of water is determined by one or two weirs that discharge any over-fill back into the adjacent river or drainage system.
As barges are loaded/unloaded, so the water level in the canal will rise/fall within the limits set by the weirs. It’s irrelevant whether a barge is on an aqueduct or not. The aqueduct is designed to carry the maximum water depth and that’s all that it basically needs to do (plus any dynamic effects and safety factors etc). As you have all indicated, Archimedes sorted out the physics some while back.
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Anybody remember the “Five Star TriStars” that used to be operated by Gulf Air ?
I thought these three-engined aircraft were long-gone … but …
I noticed that a new start-up airline (I think they are called Fifty-Fifty) has a fleet of “refurbished” TriStars.
I say “refurbished” because each engine has a 20% chance of failure on a given flight.
a) What is the probability that a given flight will be made without an engine failure ?
b) What is the probability of losing only one engine ?
c) And the probability that all three will fail ?
No need to answer all three parts, just try a, b or c or any combination that suits
c) Let’s be thankful that most flights are being cancelled
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Ravvie has no doubt done the arithmetic (correctly) that for three events to occur, each with a probability of 0.2 (20%) you need to find 0.2 x 0.2 x 0.2 which of course is 0.008 or 0.8% which is slightly less than 1%.
Who would be keen to fly with a outfit where 1 in 100 flights terminate with complete engine failure ?
You might now recall how to solve Part (a) and even Part (b)
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a) 24.8 %
b) 4.0 %
c) 0.8 %
I’ll use my bike at those odds.
Hmm! I get different answers to Mike.
a) 51.2%
b) 38.4%
c) 0.8%
and the remaining probability that 2 engines fail is 9.6%, with all four possible outcomes adding to 100%.
I expect I’m wrong 
Well done Steve, spot-on.
Hi Mike, your expectation is spot-on as well
I’ll run through the arithmetic later, unless Ravvie or Steve wish to outline the maths instead ?