Brain Teaser

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
Here is a little problem that you might like to have a crack at.

At time t = 0 (stopwatch set to zero), a single engine jet aircraft is flying due East at a speed of 12km per minute. At the same altitude and 208km directly ahead is a twin engine commercial aircraft, still at time t = 0. This second aircraft is flying due North at a speed of 8km per minute. Start the stop watch,....so t > 0. At what time are the two planes closest to each other? What is the minimal distance between them at that time?

Ron.
 
I'll be first to look stupid, it's probably a trick question, but, I'll have a stab at 17min 20secs , 138.67km away, but that doesn't take into account the curvature of the earth etc... :LOL:
 
I would say that after 10.4 minutes - 10 minutes and 24 seconds, the small jet would have travelled 124.8km and be 83.2 km from the starting point of the big jet.

The big jet would have travelled 83.2km from its starting point.

So a right angeled triangle of 83.2 x 83.2 has a hypotyneuse of 117.66km

Gotta be right 8)

Hmm... but a quick calculation shows that after 12 minutes, the distance would be 115.377 km :? From this time, the distance starts to increase again
 
Sorry Bri - just added a bit

Hmm... but a quick calculation shows that after 12 minutes, the distance would be 115.377 km From this time, the distance starts to increase again
Click to expand...
 
Well we have a correct answer and Richard (quattro) is the first to do so. Well done Richard... :D

I used Calculus,...related rates and implicit differentiation to solve the problem. Time 12 mins and distance 32 root 13 km which approximates to 115.38km. Using Pythagorean triangles repeatedly to narrow down the distance will also yield the same outcome.

Hmmm....I'll have to find a tougher teaser next time.

Ron,
 
Hi Ron

I never actually did calculus so had to rely on some older methods and the asumption that a right angled isosceles triangle would give the shortest distance, however I failed to realise that the northern route was advancing at 2/3 of the rate that the eastern was diminishing.

Quick check on a spread sheet, with a graph, showed the mistake and that a right angled traingle with the sides of 96 and 64 was indeed a bit shorter.

A few years ago my Brother, a maths teacher, got a problem stuck in his head and it started to interupt his work. He gave me this: -

If a ladder measuring 18' 6" was placed on a wall with a 5' diameter pipe sitting at its base, how far up the wall would the ladder reach if it touched the pipe, floor and wall. There are obviously two answers.

Have fun with that Ron 8)
 
We did something simliar to this at Uni today. All I found myself doing was drawing pictures of my p6 in my logbook and continuing the ongoing thought process beside (breathing, eating and women) on how am I going to turbo my 4 block without tearing the sideplates off?

I may look over this blog as revision.
 
Hello Richard,

Yes doing a problem and applying some logic can be a lot of fun,...not to mention frustrating too on occassions!

With the problem that you posted,...just to clarify,..the pipe has a diameter of 5 feet which sits on the floor on its side at the base of a wall, a ladder of 18 feet 6 inches in length is placed in such a way that it makes contact with the wall, pipe and floor simultaneously.

I'll have a go and see what I find. Glad you enjoyed my little puzzle too.. :D

I'll post another later today,...all good fun.

Ron.
 
The maximum distance that the ladder would extend up the wall while satisfying all three conditions,..namely being in contact with the wall, floor and pipe simultaneously would be the length of the ladder,..18' 6". The ladder, a tangent to the pipe, makes an angle of 90 degrees to the floor, being in contact with the wall along its entire length.

Ron.
 
The ladder AB is 18.5 and the pipe is 5.

The question is, what is the length of AC?
 

Attachments

  • ladder.png
    ladder.png
    5.5 KB · Views: 248
Oh that is different then.....the ladder making just one point of contact with the wall, with the pipe between it and the wall.

Looks fun.... :D

Ron.
 
Hello Richard,

Hmm, well I have an answer (which may well be wrong,...never a fan of geometry)....y being the distance that it reaches up the wall...( ( 171/2 - ( 5/2tan alpha + 5/2)^2 ) )^1/2. Substituting in for alpha yields....17.3 units.

Ron.
 
SydneyRoverP6B said:
Hello Richard,

Hmm, well I have an answer (which may well be wrong,...never a fan of geometry)....y being the distance that it reaches up the wall...( ( 171/2 - ( 5/2tan alpha + 5/2)^2 ) )^1/2. Substituting in for alpha yields....17.3 units.

Ron.
Click to expand...

Hmmm.... I don't understand any of that :shock:

When I get home, I am going to have to try and remember how I did it.

Clue - may not be similar, but certainly congruent
 
You must log in or register to reply here.
Back
Top