I dont want to bore you with the very old classical riddles, like the Monty Hall Problem or Achilles and the turtle, so here is a rather unpopular riddle (for solving you'll also need to use some physics laws):
So, there is a dog wanting to run around the world along the equator. Slowing obstacles like water or mountains or stamina may be negligible. The dog has a tin attached to its tail which makes a loud noise whenever it hits the ground. Now, the dog starts running at the prime meridian of earth with 1 m/s. Everytime, he hears the sound of the tin hitting the ground he doubles his speed, so e.g. the first time he changes his speed to 2 m/s. The tin hits the ground about one time every 1 second.
How fast is the dog when he passes the end (again the prime meridian)?
This seems less like a riddle and more like just a math problem... Where's the riddle-y part?
I agree, it's just a math problem. Not even a very novel one, since the solution is found just by integrating a step function.
Pretty sure the end velocity is equal to watermelon. Hear me out. In Algebra I learned Susie had 5 apples in a basket and traded 2 for an orange and John gave her a pineapple for the fruit in her basket. Which is how she ended up trading a grapefruit for blue paint.

Jake has 25,000 apples because he owns an apple plantation, which is in foreclosure. He has decided to give away 2x Apples a day and close down his plantation forever when he gives away the last apple. Where X is the number from the prior day. Using a simple while loop, we discover he's given away about half his apples half way through the 14th day and the rest of his apples on the 16th. He moved a whopping 16,000 apples out of his plantation on the last full day of operation. How many oranges did he accept as payment?
Tari wrote:
I agree, it's just a math problem. Not even a very novel one, since the solution is found just by integrating a step function.

I am sorry, but you are not right. When you look closer, you will see that the way you want to solve it is not right.
Since you dont see the riddle in there, thus dont try to find the solution, I will just post it here:
Calculating the speed by using the length of the equator and determining the speed by integrating 2^x is wrong! (And yeah, you would be right then, it would not be a riddle then)
Solution: Look carefuly to the sentence:
Quote:
Everytime, he hears the sound of the tin hitting the ground he doubles his speed

The dog does not double speed, when the tin hits the ground, but when he hears it hitting the ground.
Now considering that the equator is long enough to reach maximum speed, the speed at the end is also the first speed faster than the speed that the noise travels with (since then the noise does not reach the dogs ear anymore). Mathematically, we have to find the first value, where 2^x>330 (330 m/s = sonic sound), with x ε integer.
Now you can try this out with 1,2,3... and will come to the result soon, but since we are programmers, we know that 512 is the first 2^x number , greater than 330. Therefore, the dogs final speed will be 512 m/s, because he wont hear the tin anymore after that.
I am sure that someone would have solved the riddle (maybe I formulated it so, that everyone thought you would have to do complicated mathematical stuff, then sry for that), but since already 2 people posted their thoughts (which would not have encouraged others to try solving), I had to post the solution.
Maybe, after this fail, we will see many other riddles from others?
Now, I will try out the one from comicIDIOT...
Edit: I guess the „The tin hits the ground about 1 time every second“ was the one sentence, that made things so confusing?
Well darn, that actually got me; however, this solution is only valid for Earth at STP. Generalized solutions, such as one for dogs on Kerm's home planet of Mars, are needed since the original problem did not specify on which planet this occurred (while it can be assumed "the world" is specifying Earth, it's a rather terracentric usage).
CVSoft wrote:
Well darn, that actually got me; however, this solution is only valid for Earth at STP. Generalized solutions, such as one for dogs on Kerm's home planet of Mars, are needed since the original problem did not specify on which planet this occurred (while it can be assumed "the world" is specifying Earth, it's a rather terracentric usage).

Haha, trying to exploit the solution?
I am not sure, but I think there is no prime meridian defined on Mars.
Therefore, Earth is the only possibility
Given the heavy mapping we've done of the Martian surface, there's a defined coordinate system. Other bodies that have been mapped also have coordinate systems, it's how we locate features.
MasterChief56 wrote:

Now considering that the equator is long enough to reach maximum speed, the speed at the end is also the first speed faster than the speed that the noise travels with (since then the noise does not reach the dogs ear anymore). Mathematically, we have to find the first value, where 2^x>330 (330 m/s = sonic sound), with x ε integer.
Now you can try this out with 1,2,3... and will come to the result soon, but since we are programmers, we know that 512 is the first 2^x number , greater than 330. Therefore, the dogs final speed will be 512 m/s, because he wont hear the tin anymore after that.
I dispute this conclusion. You assume a stationary airstream relative to the course of travel, which is extremely unlikely to be true. I'm no fluid dynamics expert, but the time to audible effect definitely does not have a strictly linear relationship to speed as measured externally.

If you really want to get pedantic (which you've shown you do), you need to account for reflected sound and perhaps waves that propagate backwards relative to motion which are met head-on (depending on the sort of minimum amplitude you want to accept). Is a blueshifted oncoming signal even 'scary' enough to spur additional acceleration?

Stil sounds like a math problem to me, but with a "haha, gotcha!" element to seem clever. Not very riddle-like.
MasterChief56 wrote:
Now, I will try out the one from comicIDIOT...

I'll be frank and admit I was answering your riddle (albeit, now it appears I'm wrong) with another question in case this turned out to be a math problem for school or something, since your account was a day old and had few posts. I rounded a few numbers and changed their corresponding values to be safe.

Number of Apples was the distance at the equator, which is about 24,800 miles. The days corresponded to seconds and the apples given away was speed. You're more than welcome to figure out the problem though.
Tari wrote:
MasterChief56 wrote:

Now considering that the equator is long enough to reach maximum speed, the speed at the end is also the first speed faster than the speed that the noise travels with (since then the noise does not reach the dogs ear anymore). Mathematically, we have to find the first value, where 2^x>330 (330 m/s = sonic sound), with x ε integer.
Now you can try this out with 1,2,3... and will come to the result soon, but since we are programmers, we know that 512 is the first 2^x number , greater than 330. Therefore, the dogs final speed will be 512 m/s, because he wont hear the tin anymore after that.
I dispute this conclusion. You assume a stationary airstream relative to the course of travel, which is extremely unlikely to be true. I'm no fluid dynamics expert, but the time to audible effect definitely does not have a strictly linear relationship to speed as measured externally.

If you really want to get pedantic (which you've shown you do), you need to account for reflected sound and perhaps waves that propagate backwards relative to motion which are met head-on (depending on the sort of minimum amplitude you want to accept). Is a blueshifted oncoming signal even 'scary' enough to spur additional acceleration?

Stil sounds like a math problem to me, but with a "haha, gotcha!" element to seem clever. Not very riddle-like.
I think another consideration is if the dog can potentially hear the sound through the vibrations in his collar to his neck to his ear--kind of like how crickets hear through their feet or how a jet pilot "hears" the jet engine even at mach speeds.
Ok, ok.... I definetely shouldnt have posted that riddle...
Alex wrote:
Pretty sure the end velocity is equal to watermelon.

Like the astrophysicist said, No comet...

(I was laughing so hard...)
It really depends which way he was running.

he's got the lucky, or unlucky way
At 512 m/s, it takes the dog 78,125 seconds, or roughly 21.7 hours, to circle the equator (40,075.017 km according to Wikipedia, I simplified it to 40,000 km). In 21.7 hours, the sound waves from the tin hitting the ground would have travelled more than halfway around the earth, meaning the dog encounters them significantly earlier than he reaches his destinations. If we assume that the tin indeed made A LOT of noise, the dog would hear them and accelerate even more. Here's the true solution to the riddle.

Bonus points for whomever is willing to calculate that.
Interesting. Though...

Wikipedia told me the earth spins equatorial speed of 465.1 m/s. (of significance)

If his speed is in function of the earths center, and the surface from x=0 circumference travelled, I believe that if he runs one side...

He will go faster, or not fast enough.

It's like, his has no air speed no water speed so land speed and acceleration, to say time to complete the task is significantly relative to the earth's movement.

We slowpokes don't see this type of effect of course.
Should we also consider the reduced G acting upon the dog due to centrifugal forces? If so, the faster the dog, the less often the tin must be hitting the ground.
EDIT: That is, if the dog is running from east to west.

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