The video and what I see in the comments make it seem as though Einstein was fooled by the question. And yet at 11:31 he correctly calculated that there was no time available for the trip down. I would say that he got it!
It means he was fully going calculation mode and wanted to subtract t trip-tascent. He was never tricked. He had the right method throughout@@juncheok8579
@@juncheok8579 yeah exactly, that's the human error everyone has when encountering a problem like 7 * 8 you have to calculate it in your head for a few moments
@@juncheok8579 But he will have been aware that 45 was not the answer and that some calculation was needed instead. (Of course in this case no calculation is needed to see that time is already up after travelling exactly half of the distance at half of the required speed.)
I like how asking someone very bad at math could actually have them solve it immediately! Me: "I want you to solve this problem: [...]" Friend (bad at math): "Impossible!" (as impossible for him) Me: "That's the correct answer!"
6:57 I've seen this one before. In order to average 30 mph for a 2 mile trip, the car has to make that trip in 4 minutes. But the car already took 4 minutes to travel the first mile, therefore the car would have to teleport the rest of the way! If you average a speed of x for the first half of any trip, it is impossible to average a speed of 2x for the full trip.
@@Sonny_McMacsson And then, there has been rumors that Einstein's spirit got cloned by magical equation sheets, so when he was proclaimed dead, his clone spirit quantum materialized a clone body and teleported underground into a base, where he would then change his name to "Doc Brown", and then he would then suffer a blow to the head which will cause him to use his theory of relativity to figure out the way to create a flux capacitor to make wormholes, as well as a hyperinfinipotent liquid known as "Time Juice" to power it, and then created a DeLorean Time Machine to the future, where he then drank a potion of immortality.
I've seen this before too, and my answer is that "average" is ambiguous for this - I'll stick with the arithmetic mean anyway, but if you average WRT distance instead of time you get that for the second mile the speed must have been 45mph. Remember - you can average anything WRT any variable you want. For what averaging speed WRT distance is (in the general case) that you integrate the (varying) speed WRT distance, then divide by total distance. When you're given averages for intervals that sum to the overall interval, that simplifies to a weighted average of those discrete intervals. When those smaller intervals are all equal, that simplifies to the sum-all-the-values-and-divide-by-the-number-of-values average - in this case (15+k)/2 = 30 which solves to k=45. Just because averaging WRT time is what you'd normally expect for problems involving speeds, doesn't mean it's the only possible option. People will say that speed is by definition an average given by distance per unit time, but they're wrong - speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously, and while it's natural to average WRT time when using a derivative WRT time (giving you a convenient integral-is-the-inverse-of-derivative simplification), you can average WRT any variable you like. ["speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously" is a slip-up, and unsurprisingly confusing. As mentioned in a much later comment, think in terms of a graph of speed WRT whatever - here, either time or distance, and averaging the height of the curve without caring what units are along the x-axis. The average is a speed irrespective - the units of the y-axis. The average WRT whatever based on integration is the integral of speed WRT whatever, divided by the total interval of whatever. The integral of speed WRT distance is in different units (distance squared per time) vs. the integral of speed WRT time (distance), but then you divide by the interval width along the x-axis in whatever units apply to that particular graph, and get back to speed either way. So WRT "but it's correct" below - well, no it wasn't (sigh) but that's an error in writing it out while half-asleep, not in the underlying principle.] It's not a popular answer, based on the last time I gave it, but it's correct - averages aren't even always arithmetic means (median, mode, geometric mean, harminic mean, many other kinds of mean) but, assuming the arithmetic mean, you can average over intervals of any kind you want, or you can even have averages of a list of numbers with no intervals specified (so assuming equal weight to each number) - again (15+45)/2 = 30. Personally, I'd consider my answer wrong normally (we generally know what is meant by "average speed" - ie. WRT time), but if someone asks an absurd question, I say exploit any ambiguity because ignoring conventional interpretation is the lesser absurdity. Averaging per distance travelled makes a lot more sense in reality when looking at fuel use - many road conditions vary with the interval of distance you're travelling on (others varying with time due to weather and traffic, others varying depending on the mood of the driver). Gallons per mile times miles = total gallons.
@@stevehorne5536 He explained why 45 is incorrect, you must go by total distance divided by total time to get average speed, even if the car drove 200 mph down the hill, the 4 minutes required for a 30 mph average speed are already "used up" so as additional seconds tick by, so the maximum possible average speed of the trip ticks down from that 30 mph target. After 8 minutes, it similarly becomes impossible to average 15 mph for the trip, because the required time threshold has already been passed. And as additional minutes pass from there, the maximum possible average speed gets lower and lower, no matter how fast the trip is completed.
There's a story about someone giving John von Neumann this puzzle: Two trains 180 miles apart are on the same track going opposite directions toward each other, one traveling 10 miles/hour and the other 20 miles/hour. A fly starts on the front of one train and flies toward the other at 60 miles/hour. As soon as it reaches the other it turns around and flies back at that same speed to the first train. When it reaches that it reverses, and so on, until the fly finally is crushed when the two train collide. Question: how far does the fly fly? von Neumann instantly gives the correct answer. The person then said "Oh, you saw the trick! Most people just try to sum an infinite series!". von Neumann than said "I did sum the infinite series! There's a trick?"
Just so we're all on the same page, it's 360 miles, right? Of course I solved the answer in my head and I also summed the series. But the series was the sum of the two speeds giving me the relative speed between the trains (30mph), giving me a time to collision (6hr), and the total distance traveled by the fly (60mph x 6hr = 360miles).
I did it by combining the speeds, 30mph, 180miles, 6hrs, then how far the fly can get at 60mph. After solving, I realized, I could have solve as follows. Combined the speed, observe that the fly is traveling at double the speed, and note that it would cover double the distance, and arrived at 360mi
6:46 ...I "invented" this problem a few years ago, trying to figure out why if I could ride my bike at 20 mph on the flat, why it was so hard to do so on a hilly course, if you get back all the energy you gained by going up the hill. I realized, that if the first half of the trip was up a hill you could only go up at 10 mph, you'd need to teleport down the hill to average 20 mph, and that allowed me to realize why the 20 mph assumption was wrong!
The hill problem is something that I've often thought about as both a bicyclist and a runner. The problem is intuitive if you're talking about differences in speed over fixed amounts of time, but not distance. This is basically why dealing with hills, even on a round trip is more of a pain than just a level path.
When I was young I used to run such calculations while staring at my head unit on my bike. These days I just ride without caring much about the numbers.
@@mrosskne That would imply that the speed increase of going downhill is greater than the speed decrease of going uphill. Do you assert that to be true? I can't say it's wrong for sure, bit it strikes me as counterintuitive.
@@plentyofpaper It doesn't matter. The point of the problem is that you travel one speed for the first leg and a different speed for the second. It could be talking about a submarine or a spaceship. The problem's flavor is irrelevant.
@@mrosskne The point of the problem is that if an equal distance is traveled at 2 different speeds, the slower speed will be closer to the average than the faster speed. Because the slower trip takes longer. I honestly have no idea what you're trying to get at.
The sum of dice property is used to great effect in table-top games such as D&D. For instance, when rolling a character's stats, you roll 3 6-sided dice and sum them. The probability creates a bell curve that makes the midpoint between 1 and 18 the most common, which works out well at making 10 and 11 the most common stat, with either extreme (3 or 18) the least likely. So just as the "geniuses" are least common in reality, the rolling of a genius stat is also the least likely.
Just a correction there - it is not a bell curve. With 2 dice it's actually just an upside down V with no curves at all, just 2 straight lines. With 3 dice it's a bit more complicated, but still isn't a normal distribution. Also, the mid point with 3 dice is half way between 3 and 18, not 1 and 18.
@@asdfqwerty14587 He was referring to 3 dice, not 2, and he didn't say it was a normal distribution. He said it was a bell curve, which it is. This is an impressive amount of being wrong and simultaneously not responding to what the comment you're replying to actually said.
@@mrosskne A bell curve is the normal distribution. That is the dictionary definition of it. Google the definition of it and link on the front page will say that it refers to a normal distribution, including the one coming from a dictionary.
@@mrosskne Astonishing how bad your reading comprehension is and how wrong you are. Bell curve means normal distribution, not a "curve shaped like a bell". And the comment replies to what OP said. Corrects the mistake of it not being a bell curve and corrects the range to be from 3 to 18 not 1 to 18.
"I was going 70 miles an hour and got stopped by a cop who said, "Do you know the speed limit is 55 miles per hour?" "Yes, officer, I know but I wasn't going to be out that long..." -- Stephen Wright
Hah! That Einstein problem... I was getting annoyed with myself because I was getting an infinite speed/zero time for the descent and thought I was calculating it wrong. And then I thought, nah, it's a trick question 🙂
I was sure there was an answer, and so after confusing myself with the fact both require the same time, I watched the video, to find I was not actually mistaken.
They use this paradoxical question in 'Speed awareness courses' in the UK. I was allowed to leave when I answered "Easy just reduce your mass to a negative value and travel at 4C as you will arrive before you set off" it hurt the guy's head so much he knew he couldn't handle 4 hours with me 😂😂😂😂
7:59 i remember one video where you argued "half of two plus two" is 1/2 * 2 + 2 and not 1/2 * (2 + 2), because 1/2 (2 + 2) would be said as "half of *the sum* of two and two" or "half of *the quantity* two plus two" but look where we are now
Words aren't precise, that's why we don't use them for math. You can avoid ambiguity by saying "one half, multiply, open bracket, two, add, two, close bracket". But then you might as well just use the symbols.
@@mrosskne i just remember watching the "half of two plus two" video and they sounded very serious about it not being 1/2 * (2 + 2) and spent like half the video arguing so
@@mrosskne I believe the video being questioned was the use of the divide sign, AND juxtaposition of a(b+c), and whether people treat that as a*(b+c) verses (a*(b+c)), and was a lot of controversy because people learn to treat those differently, thus pointing out the problem with using such symbols. So to the point OP is making, half of 15 plus x, hasn't been explicitly stated as the sum of 15 and x, and should, by strict adherence to BEDMAS (or whatever you learned), be treated at half of 15, plus x.
Translation: "If I have to be at work by 8am, and I'm only halfway there when the clock strikes 8am, how fast do I have to drive to get to work on time?" YOU CAN'T. You will be late to work no matter how fast you go.
Technically with time dilation it is possible and you wouldn't need to hit the speed of light because there is also very very very slight time dilation on the ascent as well. So it would be like (4 -(10^-32)) mins for the accent and 1O^-32 mins for the descent.
@@ericpaul4575 so I think you save 1.44 e-13 seconds on the ascent so I think you'd have to travel the equivalent of 6.944e12 miles per sec (without time dilation) I'm not sure how to convert that back to relative light speed (with time dilation) or I guess unwilling to figure it out.
The third one does something to our brains, it seems obvious yet that's because we haven't accounted for time in the t/d/s triangle and it has already been spent. Thanks for explaining.
So the first question, the answer is independent of a and c, which is not immediately obvious in the solution of the quadratic. I guess there is some interplay between a, b and c that would make the a and c terms cancel
For the first puzzle, given the 2 results you can get a third result for r in terms of a and c only which comes to r = (a+c) + sqrt(2ac) or r = (a+c) - sqrt(2ac) (both values work as a,c > 0, so a+c > sqrt(2ac))
However, for the second solution the side lengths of the rectangle are a - sqrt(2ac) and c - sqrt(2ac) and at least one of them is guaranteed to be negative, which we probably don't want to accept.
The Feynman question has a problem in it. a, b, c are not independent of each other. That's why you get such different answers. Construct only a segment of length = a. From there if you draw a line straight up to the intersection with the circle, and from there strictly left, you can see that b and c are uniquely determined because of just fixing the a value.
@annanay007 if you fix only a, and "let b and c be free", and try to recreate the same picture from ONLY fixed a value, you will find out, that there would be anyway the only possible value for b and for c (because in the construction we have two lines with fixed right angles)
So Feynman's solution actually presents a restriction on viable values for a,b,c. If you instead use b for r in his equation and then solve for b you get that b=a+c-sqrt(2ac) so for any other value of b given a,c then the problem is unsolvable because the combination is invalid. For example if a=2,c=9 then we get b=17. If instead we are given instead b=19 then there is no possible rectangle with diagonal 19 such that to given quarter circle has the extra lengths a,c as given.
You got the wrong sign there. It should be PLUS sqrt not minus. There’s no way 2+9 minus a positive number equals 17 so you used it correctly in your first test.
Actually, if the 30mph is rounded to the nearest integer, this allows you to travel at an average of 29.5mph and still qualify. This works out at around 885mph for the downhill portion, which is less than 20% over the current land speed record. And so, of course, far less than infinity
@@kobeballer if you think specifying the level of precision changes the problem, please never consider becoming a scientist, or at the very least an engineer
@@mickdavies5647 lol so if my engineering manager asks me to design a part or process to 0.001 precision, then I can just present something to 0.01 and tell them "you shouldn't be in engineering if you think changing the precision changes the problem"? 😂😂
@kobeballer well, since you seem to think you can just consider the precision to be irrelevant, or ignore the fact it hasn't been provided, I pity both your manager and your clients
The first question felt so obvious that I was SURE I was doing it wrong, especially when you went into your explanation. Nope, turns out it was just obvious. ;)
At that point you need to ask what clock you are going by. Because by the clocks of the rest of the universe it still takes you some time to travel that last bit, even at the speed of light.
There are a few problems with this. 1) If you're moving at/very close to the speed of light and you're talking about the frame of reference of the car, then the distance will not be 1 mile anymore. Relativity causes the lengths to change too, not just time. Even if you were talking about a wacky frame of reference the math would still work out the same and it would still be impossible. The only way you could come up with different answers is if you're talking about things like trying to calculate the speed of an object by taking the distance from 1 frame of reference and the time from a different frame of reference, which is a nonsense measurement which doesn't actually calculate anything useful (by that logic you could come up with literally any number, you could even say the car is travelling backwards if you wanted to). 2) If you're talking about the frame of reference of the object itself, then "the car isn't moving at all" - there's no such thing as measuring the speed of an object from its own frame of reference, because obviously the car is always at the same position relative to itself.
I must be better at math than I thought. I had the b = r almost immediately. I've played craps in Vegas after reading a book on how to do it and already know that 11 is twice as likely as 12. I thought that the last one was a trick right away, too. It just didn't register exactly how it was tricky until the math was done. Another great video Talwalker!
There are many variations of 'rate problems' that fool many people, this is a classic. Similar ones go something like, "If Bob can do a job in 2 hours, and Bill can do the same job in 3 hours..." Many folks will incorrectly combine the times vs rates.
looking at the car problem, to get a 30mph average, and you can only do 15mph for the 1st half of the trip you would have 0 time to do the 2nd mile, it takes the same amount of time to travel 2*D at 2*T as it does to travel 1*D at 1*T. this problem is deceptive in that it is really a time problem, not a speed problem.
Averaging speeds over the same distance requires taking a harmonic mean. 2/v_avg = 1/v1 + 1/v2 The reason is that you've defined the problem as saying the two legs of the trip cover the same _distance._ So to calculate it, the _distance_ needs to be in the denominator (since they're the same distance), just like when you add two fractions they need to have the same denominator. Since miles/hr has distance in the numerator, you need to flip it to put distance in the denominator before calculating the average. If the problem had been the car spends 1 hour driving 15 mph, and 1 hour driving v2 mph, then the time is equal for both legs, so time needs to go in the denominator. Which it already is for mph, so you can solve _that_ as v_avg = (v1 + v2)/2. And v2 = 45 mph would be correct. Same issue crops up with fuel efficiency in the U.S. (which uses miles per gallon). If you've got a car which gets 20 mpg and another car which gets 40 mpg and you drive both cars the same distance over the year, then the average mpg is the harmonic mean. Resulting in an average mileage of 26.7 mpg. OTOH, if you put 15 gallons in the 20 mpg car, and 15 gallons in the 40 mpg car (same gallons, instead of same distance), and drive both until their tank is empty, then the average mileage will be 30 mpg. Since the vast majority of driving you do is based on distance, not how much fuel you have, the rest of the world uses "liters per 100 km" to avoid this inversion issue (puts distance in the denominator).
None of this is necessary. If you want an average speed of 30 mph (~45 fps) over 2 miles (~10600 ft), the trip must take ~236 seconds. You traveled 5300 ft at 22.5 fps, so it took you 236 seconds. You have zero seconds to travel the remaining distance. If you want to reach an average speed of 30 mph, it will take you more than one mile of additional travel to reach it, no matter how fast you accelerate.
I agree with you...time has nothing to do with this problem. It's all about speed and distance. You go 15 for half and you have to go 45 the second half to average the SPEED. Who cares that one half takes 1/3 the time to complete. Speed is all that was asked and all that matters. Stick to the same unit of measure...don't bring in something that doesn't matter and use it to "prove" the correct solution wrong. If you are driving down the highway at 50 for a mile...and then for another mile you travel at 60...your average speed is 55. How long it took isn't the question.
the Einstein one is easier to visualize with different numbers: suppose you have an average 100 km/h to go 100km. What speed do you need to go through the next 100km to average 200km/h in the total trip? It's more immediately evident that a 100km trip at 100km/h would take 1h, and that a 200km trip at 200km/h would have to take 1h as well.
Question in the thumbnail... I went directly to : How much time it take for the first part (4 minutes) and how much time do I have for the full trip (also 4 minutes) ?
I saw the average speed problem in a book many years ago, and I remember when I was 14 presenting this problem to my friends. They refused to believe that reaching the target average speed was impossible. Play confidently asserted the time had nothing to do with speed. One friend even claimed that his truck could accomplish it. I think they know better now. At least, I hope so.
You know, I actually thought about it using the second method you were talking about, with the idea of 4 minutes. I am not the best at math in my head for equations like this, but I did notice that it seemed like there wasn't enough time remaining to go that fast although I wasn't sure exactly why.
That's made me feel great about my hitherto averagish brain. I got that Einstein-stumper straight away, just by looking at the thumbnail. Go brain! Go brain!
When driving long distances over expressways where the speed limit is sometimes 70 mph or more, I TRY to have an average speed of about 65 or so. But it's really hard when you have to stop every 3 hours or so, even for just a few minutes (for gas, food, and the call-of-nature). Even brief stops quickly drops the 'average speed'.
Now im curious, if you add relativity to the car puzzle, would it be possible that going up the hill time is little less then 4 minutes and then it goes to light speed with whatever time is left? ;)
well because the other diagonal would also be b, it would be a straight line going from the center to the edge of the quarter circle, which means b is the radius of the circle. as for a and c, idk lol maybe somehow would be able to compute it in terms of r and b using trig functions.
Actually, at the speed of light, the last mile would take 5.3µs, and the whole thing feels pretty 1s.f. so I don't think anyone would say you hadn't made it.
No3 is a great example of something that happens all the time in industry and the media. I call it the average of averages and so many decisions are based on this flawed logic! This is a great way of pointing out that flaw.
Who is measuring the time spent on the decent? If you're travelling at the sped of light, time doesn't exist. As such, if you're the one timing the decent, the answer is the speed of light. If it's an external measurement, then yes - it's not possible.
Haha! This was a dead giveaway to me but out of interest I subjected my family to this riddle and for sure my wife responded 45mph. Next my kid who is in second grade middle school gave the exact same answer, so I asked him if he could calculate what the resulting times for both distances would be for the given average speeds. For sure he did calculate it was 4 minutes for both but due to the nature of the question he then concluded that the additional mile had to be run in less than a minute and ended up answering 60mph. So next I asked him if he could use seconds instead of minutes and the big question mark started rising above his head - ehm 3600mph? Granted I don't really know if he already learned the concept of infinity slash division by zero, I think back then I already had it in first grade, but I believe he really did get it when I gave him the answer.
My first thought was, "speed of light" The reason is simple, 15mph and to travel 1 mile, you have to spend the entire time as you would do with average for 2 miles. That is, there isn't time to spare!
I actually came a variation of the Einstein problem decades ago (as a kid), so I already knew the answer to it. But it’s also a good teaching tool-the “average” isn’t always the arithmetic mean. Especially when dealing with complex units (a complex unit is one that combines 2 or more basic units).
When looking at the radius problem, I didn't notice the r=b shortcut at first. But when I looked at the quadratic solution, I immediately saw the 2ac - a² - c², and in my head converted it to -(a-c)², and now I had 2b²-(a-c)² and thus b² + b²-(a-c)² = b² + (a+b-c)(a+b+c), and slowly, it dawned me, that I missed something.
I got the b=r one instantly but that's probably because Feynman was primed for it to be a maths puzzle and tunnel visioned himself (I assume, I don't know anything about him outside that)
So for the first problem, I got both solutions on my own, but how are they the same? What is missing to simplify the complicated expression from the quadratic formula to just come out to be b?
I love your videos. They are beautifully designed and presented in an absolutely clear way. I'd just point out two mistakes: 1. The symbol for hour is just (upright or roman) h 2. Since time depends on the observer's perspective, going at light speed does solve the problem, assuming the one person interested in completing the trip on such average speed is the driver. From his perspective, going at light speed does take him to the 2-mile mark instantly, whereas an observer (e.g., someone who had an appointment with him) would be pissed due to him arrving a fraction if a second late.
@tezzerii that seems likely. Also, technically, a wheeled vehicle can only reach half light speed. This is because the highest speeds on a vehicle are those of the upper part of the wheels. If those where at light speed, their center would be at light speed divided by two.
I didn't overthink the first one, it seemed obvious using trig that r=b. The second one is deeply obvious unless you think 6+5 and 5+6 are different. The last one takes a moment to realise the average speed is already sucked up by the ascent time. Sometimes overthinking or expectation stops the otherwise obvious from being apparent.
Strictly the Leibniz problem is ambiguous: if you don't assume commutivity then you can interpret "a sum of" to be "a specific sum of" so that 5+6 is not the same sum as 6+5.
There are several lines of thought to immediately get the answer to the Einstein puzzle. One of them is: If you you have a certain speed for a given distance and you want double the speed on double the distance then of course the time has to be exactly the same. This is true for any non-relativistic speed and any non-relativistic distance. If it's easier for you to calculate with certain numbers. Just imagine the first part of the hill to be exactly 15 retard units. Then the ascent takes 1 hour. The total distance would be 30 retard units and hence you also have 1 hour for the full distance to get the desired average speed. Which is of course impossible unless you are a massless particle...
I refer to it as "a 5th-gear brain working on a 2nd-gear problem." If you think of the brain as a car engine, you can think of working on higher-end problems as being in a higher gear. Solving a simple problem with your brain in low gear is like trying to drive at a slow speed in a low gear; the engine handles it just fine. Trying to drive slow in a high gear, however, causes so few RPMs that the engine stalls and dies out, so it can't do it. ...A bit of a weird analogy, I admit, but it seems to get the point across.
Just to reword the question before watching: If a car travels one mile in four minutes, what speed does it need to go to cover a second mile in zero additional minutes? In Star Trek, they call it warp ten.
I got the last puzzle right within seconds without calculations. I ride a bicycle a LOT, and I calculate my ride times by my average speeds. If I'm X miles from home, how fast do I have to go to get there in Y minutes? Yes, I call ahead and tell people I'm going to be late often. Why do you ask?
einsteins riddle is only tricky because time is involved. since time doesnt exist in real life, its only a mathematical measurement, you can actually just average the total distance and take an eternity to cross it.
I have a weirdo way of thinking about the Einstein trick since I've seen it before, and it kinda works out by making the answer also a trick. So in some versions of the problem, it is vague about what average is being taken over and asks simply "what speed do you have to go for the total trip to average 30mph". This is generally assumed to be the average speed over the entire distance of the trip (2 miles). As the video shows, there's no way to get an average speed of 30 mph over the distance without traveling instantaneously. However, if you consider the average over time instead, then you end up back with 45 mph IFF you drive for another 4 minutes. Because it takes 4 minutes to go up at 15 mph, then another 4 minutes at 45 mph, you end up with an average speed of 30 mph. The distance you have traveled will end up being much further (4 miles here vs 2 in the original problem). If you reduce the time component out of this entirely (since it is constant), then it ends up being just the simple average of 15 and 45 like shown in the video, but it's important to clearly state the same amount of time as a constant.
Correction: the notion that the speed of light is the same in all directions is unprovable in principle (at least, not without FTL capability). The speed of light is considered to be the same in all directions only by convention. Therefore, it is entirely possible for the speed of light to be infinite in one direction and 1/2c in the other. Since the car only needs to travel at infinite speed in one direction, it can, provided that it becomes massless on the descent. What’s funny is that Einstein should have noticed such a loophole given that he noted in his own paper that the one-way speed of light is unknowable and he set it equal to c by convention.
The video and what I see in the comments make it seem as though Einstein was fooled by the question. And yet at 11:31 he correctly calculated that there was no time available for the trip down. I would say that he got it!
"Not until calculating did i notice..." implies he got tricked, even if it was for a few moments
It means he was fully going calculation mode and wanted to subtract t trip-tascent. He was never tricked. He had the right method throughout@@juncheok8579
@@juncheok8579 yeah exactly, that's the human error everyone has when encountering a problem like 7 * 8 you have to calculate it in your head for a few moments
@@juncheok8579 But he will have been aware that 45 was not the answer and that some calculation was needed instead. (Of course in this case no calculation is needed to see that time is already up after travelling exactly half of the distance at half of the required speed.)
I agree. I've heard it suggested that this problem stumped Einstein or that he couldn't solve it. But that's clearly not true.
I kept trying to do Einstein's question, getting to the divide by 0 part, and assuming I did something wrong and starting over for like an hour
Therefore, Officer, it was impossible for me to be speeding.
I like how asking someone very bad at math could actually have them solve it immediately!
Me: "I want you to solve this problem: [...]"
Friend (bad at math): "Impossible!" (as impossible for him)
Me: "That's the correct answer!"
I drank six glasses of 5% ABV beer, you honour. But I only drank 95% of each glass.
Not impossible, simply highly improbable…
@@kjamison5951 No. The Improbability Drive is not enough, you need the Impossibility Drive (ref: Hitchhiker's Guide to the Galaxy).
6:57 I've seen this one before. In order to average 30 mph for a 2 mile trip, the car has to make that trip in 4 minutes. But the car already took 4 minutes to travel the first mile, therefore the car would have to teleport the rest of the way!
If you average a speed of x for the first half of any trip, it is impossible to average a speed of 2x for the full trip.
Einstein was figuring out HOW to do it, considering he's gonna need a wormhole.
@@Sonny_McMacsson And then, there has been rumors that Einstein's spirit got cloned by magical equation sheets, so when he was proclaimed dead, his clone spirit quantum materialized a clone body and teleported underground into a base, where he would then change his name to "Doc Brown", and then he would then suffer a blow to the head which will cause him to use his theory of relativity to figure out the way to create a flux capacitor to make wormholes, as well as a hyperinfinipotent liquid known as "Time Juice" to power it, and then created a DeLorean Time Machine to the future, where he then drank a potion of immortality.
Oh that’s why I was so confused. I kept using d=rt while looking at he thumbnail
I've seen this before too, and my answer is that "average" is ambiguous for this - I'll stick with the arithmetic mean anyway, but if you average WRT distance instead of time you get that for the second mile the speed must have been 45mph. Remember - you can average anything WRT any variable you want. For what averaging speed WRT distance is (in the general case) that you integrate the (varying) speed WRT distance, then divide by total distance. When you're given averages for intervals that sum to the overall interval, that simplifies to a weighted average of those discrete intervals. When those smaller intervals are all equal, that simplifies to the sum-all-the-values-and-divide-by-the-number-of-values average - in this case (15+k)/2 = 30 which solves to k=45.
Just because averaging WRT time is what you'd normally expect for problems involving speeds, doesn't mean it's the only possible option. People will say that speed is by definition an average given by distance per unit time, but they're wrong - speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously, and while it's natural to average WRT time when using a derivative WRT time (giving you a convenient integral-is-the-inverse-of-derivative simplification), you can average WRT any variable you like.
["speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously" is a slip-up, and unsurprisingly confusing. As mentioned in a much later comment, think in terms of a graph of speed WRT whatever - here, either time or distance, and averaging the height of the curve without caring what units are along the x-axis. The average is a speed irrespective - the units of the y-axis. The average WRT whatever based on integration is the integral of speed WRT whatever, divided by the total interval of whatever. The integral of speed WRT distance is in different units (distance squared per time) vs. the integral of speed WRT time (distance), but then you divide by the interval width along the x-axis in whatever units apply to that particular graph, and get back to speed either way. So WRT "but it's correct" below - well, no it wasn't (sigh) but that's an error in writing it out while half-asleep, not in the underlying principle.]
It's not a popular answer, based on the last time I gave it, but it's correct - averages aren't even always arithmetic means (median, mode, geometric mean, harminic mean, many other kinds of mean) but, assuming the arithmetic mean, you can average over intervals of any kind you want, or you can even have averages of a list of numbers with no intervals specified (so assuming equal weight to each number) - again (15+45)/2 = 30. Personally, I'd consider my answer wrong normally (we generally know what is meant by "average speed" - ie. WRT time), but if someone asks an absurd question, I say exploit any ambiguity because ignoring conventional interpretation is the lesser absurdity.
Averaging per distance travelled makes a lot more sense in reality when looking at fuel use - many road conditions vary with the interval of distance you're travelling on (others varying with time due to weather and traffic, others varying depending on the mood of the driver). Gallons per mile times miles = total gallons.
@@stevehorne5536 He explained why 45 is incorrect, you must go by total distance divided by total time to get average speed, even if the car drove 200 mph down the hill, the 4 minutes required for a 30 mph average speed are already "used up" so as additional seconds tick by, so the maximum possible average speed of the trip ticks down from that 30 mph target. After 8 minutes, it similarly becomes impossible to average 15 mph for the trip, because the required time threshold has already been passed. And as additional minutes pass from there, the maximum possible average speed gets lower and lower, no matter how fast the trip is completed.
There's a story about someone giving John von Neumann this puzzle: Two trains 180 miles apart are on the same track going opposite directions toward each other, one traveling 10 miles/hour and the other 20 miles/hour. A fly starts on the front of one train and flies toward the other at 60 miles/hour. As soon as it reaches the other it turns around and flies back at that same speed to the first train. When it reaches that it reverses, and so on, until the fly finally is crushed when the two train collide. Question: how far does the fly fly?
von Neumann instantly gives the correct answer. The person then said "Oh, you saw the trick! Most people just try to sum an infinite series!".
von Neumann than said "I did sum the infinite series! There's a trick?"
Just so we're all on the same page, it's 360 miles, right?
Of course I solved the answer in my head and I also summed the series. But the series was the sum of the two speeds giving me the relative speed between the trains (30mph), giving me a time to collision (6hr), and the total distance traveled by the fly (60mph x 6hr = 360miles).
I did it by combining the speeds, 30mph, 180miles, 6hrs, then how far the fly can get at 60mph.
After solving, I realized, I could have solve as follows.
Combined the speed, observe that the fly is traveling at double the speed, and note that it would cover double the distance, and arrived at 360mi
Well, 180 miles covered at a combined speed of 30 mph takes 6 hours. Fly always travels at 60 mph during that time = 360 miles. Easy peasy.
Harder question would be, how many times does the fly turns?
@@Nobody-tu5wt As the 2 trains get closer and closer, the number of turns approach infinity if I'm not mistaken.
I once thought that I made a mistake ... but I was wrong.
As a middle aged white man on the internet, I’ve never been wrong.
The correct expression is:
"I'm always right. I thought I was wrong once, but I was mistaken"
And that makes you right. Right?
6:46 ...I "invented" this problem a few years ago, trying to figure out why if I could ride my bike at 20 mph on the flat, why it was so hard to do so on a hilly course, if you get back all the energy you gained by going up the hill. I realized, that if the first half of the trip was up a hill you could only go up at 10 mph, you'd need to teleport down the hill to average 20 mph, and that allowed me to realize why the 20 mph assumption was wrong!
The hill problem is something that I've often thought about as both a bicyclist and a runner.
The problem is intuitive if you're talking about differences in speed over fixed amounts of time, but not distance.
This is basically why dealing with hills, even on a round trip is more of a pain than just a level path.
When I was young I used to run such calculations while staring at my head unit on my bike. These days I just ride without caring much about the numbers.
The trip problem has nothing at all to do with hills, that's basically a red herring. The result is the same if you're on flat ground.
@@mrosskne That would imply that the speed increase of going downhill is greater than the speed decrease of going uphill.
Do you assert that to be true? I can't say it's wrong for sure, bit it strikes me as counterintuitive.
@@plentyofpaper It doesn't matter. The point of the problem is that you travel one speed for the first leg and a different speed for the second. It could be talking about a submarine or a spaceship. The problem's flavor is irrelevant.
@@mrosskne The point of the problem is that if an equal distance is traveled at 2 different speeds, the slower speed will be closer to the average than the faster speed. Because the slower trip takes longer.
I honestly have no idea what you're trying to get at.
The sum of dice property is used to great effect in table-top games such as D&D. For instance, when rolling a character's stats, you roll 3 6-sided dice and sum them. The probability creates a bell curve that makes the midpoint between 1 and 18 the most common, which works out well at making 10 and 11 the most common stat, with either extreme (3 or 18) the least likely. So just as the "geniuses" are least common in reality, the rolling of a genius stat is also the least likely.
Just a correction there - it is not a bell curve. With 2 dice it's actually just an upside down V with no curves at all, just 2 straight lines. With 3 dice it's a bit more complicated, but still isn't a normal distribution.
Also, the mid point with 3 dice is half way between 3 and 18, not 1 and 18.
@@asdfqwerty14587 He was referring to 3 dice, not 2, and he didn't say it was a normal distribution. He said it was a bell curve, which it is. This is an impressive amount of being wrong and simultaneously not responding to what the comment you're replying to actually said.
@@mrosskne A bell curve is the normal distribution. That is the dictionary definition of it. Google the definition of it and link on the front page will say that it refers to a normal distribution, including the one coming from a dictionary.
No, it's any curve shaped like a bell, sorry.
@@mrosskne Astonishing how bad your reading comprehension is and how wrong you are.
Bell curve means normal distribution, not a "curve shaped like a bell".
And the comment replies to what OP said. Corrects the mistake of it not being a bell curve and corrects the range to be from 3 to 18 not 1 to 18.
If you average over distance rather than time, 45mph is correct. There's just the small problem that we never average speed over distance
"I was going 70 miles an hour and got stopped by a cop who said, "Do you know the speed limit is 55 miles per hour?" "Yes, officer, I know but I wasn't going to be out that long..."
-- Stephen Wright
I think that would work in court. 😂😂
Also Stephen Wright:
Cop: Why were you going so fast?
Stephen: Because I pressed the gas pedal all the way down to the floor.
Hah! That Einstein problem... I was getting annoyed with myself because I was getting an infinite speed/zero time for the descent and thought I was calculating it wrong. And then I thought, nah, it's a trick question 🙂
Like reverse psychology, it's a reverse trick question. Ie: Not the trick you're thinking of!! 🤣
I was sure there was an answer, and so after confusing myself with the fact both require the same time, I watched the video, to find I was not actually mistaken.
They use this paradoxical question in 'Speed awareness courses' in the UK. I was allowed to leave when I answered "Easy just reduce your mass to a negative value and travel at 4C as you will arrive before you set off" it hurt the guy's head so much he knew he couldn't handle 4 hours with me 😂😂😂😂
I feel smart when i instantly got the dice problem.
7:59 i remember one video where you argued "half of two plus two" is 1/2 * 2 + 2 and not 1/2 * (2 + 2), because 1/2 (2 + 2) would be said as "half of *the sum* of two and two" or "half of *the quantity* two plus two" but look where we are now
Words aren't precise, that's why we don't use them for math.
You can avoid ambiguity by saying "one half, multiply, open bracket, two, add, two, close bracket". But then you might as well just use the symbols.
@@mrosskne i just remember watching the "half of two plus two" video and they sounded very serious about it not being 1/2 * (2 + 2) and spent like half the video arguing so
ok. do you have an argument to contribute?
@@mrosskne search "half of two plus two" on their channel
@@mrosskne I believe the video being questioned was the use of the divide sign, AND juxtaposition of a(b+c), and whether people treat that as a*(b+c) verses (a*(b+c)), and was a lot of controversy because people learn to treat those differently, thus pointing out the problem with using such symbols. So to the point OP is making, half of 15 plus x, hasn't been explicitly stated as the sum of 15 and x, and should, by strict adherence to BEDMAS (or whatever you learned), be treated at half of 15, plus x.
Translation: "If I have to be at work by 8am, and I'm only halfway there when the clock strikes 8am, how fast do I have to drive to get to work on time?"
YOU CAN'T. You will be late to work no matter how fast you go.
If u go at infinite speed u can reach or just light speed
or it could still be 8am when you arrive.
What if the work place is in a timezone west of you? (could be as simple as a short drive crossing a border)
You are not thinking with portals
@@Hunni125 No, it could not.
That last puzzle reminds me of trying to bump up my average speed during a bike ride. Always harder than you think!
I wish my Garmin calculated my mode average instead of my mean average. I'd still be slow, but my Strava activities would look better.
Technically with time dilation it is possible and you wouldn't need to hit the speed of light because there is also very very very slight time dilation on the ascent as well. So it would be like (4 -(10^-32)) mins for the accent and 1O^-32 mins for the descent.
So what speed is that on decent?
@ashtonaughtband Except that you're in an old car that can only manage 15mph up the hill. =o)
@@ericpaul4575 so I think you save 1.44 e-13 seconds on the ascent so I think you'd have to travel the equivalent of 6.944e12 miles per sec (without time dilation) I'm not sure how to convert that back to relative light speed (with time dilation) or I guess unwilling to figure it out.
@@ashtonaughtbandyou just need to play with the Lorentz factor. I'll calculate it when I get up from my nap, if I remember
Yes, that's why the simple arithmetic problem didn't mention time dilation. Because you were supposed to consider time dilation. Christ.
The third one does something to our brains, it seems obvious yet that's because we haven't accounted for time in the t/d/s triangle and it has already been spent. Thanks for explaining.
So the first question, the answer is independent of a and c, which is not immediately obvious in the solution of the quadratic. I guess there is some interplay between a, b and c that would make the a and c terms cancel
For the first puzzle, given the 2 results you can get a third result for r in terms of a and c only which comes to r = (a+c) + sqrt(2ac) or r = (a+c) - sqrt(2ac) (both values work as a,c > 0, so a+c > sqrt(2ac))
However, for the second solution the side lengths of the rectangle are a - sqrt(2ac) and c - sqrt(2ac) and at least one of them is guaranteed to be negative, which we probably don't want to accept.
The Feynman question has a problem in it. a, b, c are not independent of each other. That's why you get such different answers.
Construct only a segment of length = a. From there if you draw a line straight up to the intersection with the circle, and from there strictly left, you can see that b and c are uniquely determined because of just fixing the a value.
Did not understand. Can you please elaborate
@annanay007 if you fix only a, and "let b and c be free", and try to recreate the same picture from ONLY fixed a value, you will find out, that there would be anyway the only possible value for b and for c (because in the construction we have two lines with fixed right angles)
@@Serg_144 In other words, with that additional information Feynman's answer can be reduced to "b".
@@jmodified yep, he just didn't used it. They are functions b=b(a), c=c(a)
@@Serg_144 now finally understood. Thanks brother
So Feynman's solution actually presents a restriction on viable values for a,b,c. If you instead use b for r in his equation and then solve for b you get that
b=a+c-sqrt(2ac)
so for any other value of b given a,c then the problem is unsolvable because the combination is invalid.
For example if a=2,c=9 then we get b=17. If instead we are given instead b=19 then there is no possible rectangle with diagonal 19 such that to given quarter circle has the extra lengths a,c as given.
You got the wrong sign there. It should be PLUS sqrt not minus. There’s no way 2+9 minus a positive number equals 17 so you used it correctly in your first test.
Actually, if the 30mph is rounded to the nearest integer, this allows you to travel at an average of 29.5mph and still qualify. This works out at around 885mph for the downhill portion, which is less than 20% over the current land speed record. And so, of course, far less than infinity
Sure, if you change the problem, then you get a different answer...
@@kobeballer if you think specifying the level of precision changes the problem, please never consider becoming a scientist, or at the very least an engineer
@@mickdavies5647 lol so if my engineering manager asks me to design a part or process to 0.001 precision, then I can just present something to 0.01 and tell them "you shouldn't be in engineering if you think changing the precision changes the problem"? 😂😂
@kobeballer well, since you seem to think you can just consider the precision to be irrelevant, or ignore the fact it hasn't been provided, I pity both your manager and your clients
"Never try! Trying is the 1st step to failure!" -- Grandmaster Ben Finegold
That means, Ben already knew the results before the match with Mittens
The first question felt so obvious that I was SURE I was doing it wrong, especially when you went into your explanation. Nope, turns out it was just obvious. ;)
It’s a trick question. The diagram is not an accurate representation of a quarter circle so there can be no radius
@bobross7473 diagrams in math questions are rarely to scale. That's not a trick, that's a convention
I'm pleased that I realised the car question was impossible before I saw the answer.
5:15 "Imagine we roll one *dice."* Nope, I can't. But I can imagine rolling one *die.* 🎲
“Dice” is now totally accepted as a singular term.
@@ollyrukesapparently it's not accepted by everyone
@@jerkisoni like to eat one hamburgers every wednesday
Dr. Evil's son: This is what I'm talking about... YOU ALWAYS DO THAT!
@@xaigamer3129 Dice doesn’t have an s at the end of an existing word.
It isn't necessary to travel at an infinite speed for the trip down. Just travel at the speed of light and time stands still.
At that point you need to ask what clock you are going by. Because by the clocks of the rest of the universe it still takes you some time to travel that last bit, even at the speed of light.
Love pedantry. ;)
Still, either way, the given problem has no practical solution..
The answer is C. Even though it isn't a multiple choice question.
In an old car which can only manage 15mph up the hill.
There are a few problems with this.
1) If you're moving at/very close to the speed of light and you're talking about the frame of reference of the car, then the distance will not be 1 mile anymore. Relativity causes the lengths to change too, not just time. Even if you were talking about a wacky frame of reference the math would still work out the same and it would still be impossible. The only way you could come up with different answers is if you're talking about things like trying to calculate the speed of an object by taking the distance from 1 frame of reference and the time from a different frame of reference, which is a nonsense measurement which doesn't actually calculate anything useful (by that logic you could come up with literally any number, you could even say the car is travelling backwards if you wanted to).
2) If you're talking about the frame of reference of the object itself, then "the car isn't moving at all" - there's no such thing as measuring the speed of an object from its own frame of reference, because obviously the car is always at the same position relative to itself.
Thank you, I needed that little confidence boost!
I must be better at math than I thought. I had the b = r almost immediately. I've played craps in Vegas after reading a book on how to do it and already know that 11 is twice as likely as 12. I thought that the last one was a trick right away, too. It just didn't register exactly how it was tricky until the math was done. Another great video Talwalker!
There are many variations of 'rate problems' that fool many people, this is a classic. Similar ones go something like, "If Bob can do a job in 2 hours, and Bill can do the same job in 3 hours..." Many folks will incorrectly combine the times vs rates.
looking at the car problem, to get a 30mph average, and you can only do 15mph for the 1st half of the trip you would have 0 time to do the 2nd mile, it takes the same amount of time to travel 2*D at 2*T as it does to travel 1*D at 1*T.
this problem is deceptive in that it is really a time problem, not a speed problem.
A speed problem is a time problem.
@@mrosskneit’s relative
The "famous right triangle theorem" made me LOL 😂😂
Einstein got it right. Thats what the video said. Also, he had time left to make bagels!
Averaging speeds over the same distance requires taking a harmonic mean.
2/v_avg = 1/v1 + 1/v2
The reason is that you've defined the problem as saying the two legs of the trip cover the same _distance._ So to calculate it, the _distance_ needs to be in the denominator (since they're the same distance), just like when you add two fractions they need to have the same denominator. Since miles/hr has distance in the numerator, you need to flip it to put distance in the denominator before calculating the average.
If the problem had been the car spends 1 hour driving 15 mph, and 1 hour driving v2 mph, then the time is equal for both legs, so time needs to go in the denominator. Which it already is for mph, so you can solve _that_ as v_avg = (v1 + v2)/2. And v2 = 45 mph would be correct.
Same issue crops up with fuel efficiency in the U.S. (which uses miles per gallon). If you've got a car which gets 20 mpg and another car which gets 40 mpg and you drive both cars the same distance over the year, then the average mpg is the harmonic mean. Resulting in an average mileage of 26.7 mpg.
OTOH, if you put 15 gallons in the 20 mpg car, and 15 gallons in the 40 mpg car (same gallons, instead of same distance), and drive both until their tank is empty, then the average mileage will be 30 mpg.
Since the vast majority of driving you do is based on distance, not how much fuel you have, the rest of the world uses "liters per 100 km" to avoid this inversion issue (puts distance in the denominator).
...and because they use metric instead of (Hamburger*Eagle)/Shotgun^Footballfields.
None of this is necessary.
If you want an average speed of 30 mph (~45 fps) over 2 miles (~10600 ft), the trip must take ~236 seconds.
You traveled 5300 ft at 22.5 fps, so it took you 236 seconds. You have zero seconds to travel the remaining distance. If you want to reach an average speed of 30 mph, it will take you more than one mile of additional travel to reach it, no matter how fast you accelerate.
I agree with you...time has nothing to do with this problem. It's all about speed and distance. You go 15 for half and you have to go 45 the second half to average the SPEED. Who cares that one half takes 1/3 the time to complete. Speed is all that was asked and all that matters. Stick to the same unit of measure...don't bring in something that doesn't matter and use it to "prove" the correct solution wrong.
If you are driving down the highway at 50 for a mile...and then for another mile you travel at 60...your average speed is 55. How long it took isn't the question.
On the 1st one, where does a^2 + c^2 - b^2 = 2 come from?
the Einstein one is easier to visualize with different numbers: suppose you have an average 100 km/h to go 100km. What speed do you need to go through the next 100km to average 200km/h in the total trip?
It's more immediately evident that a 100km trip at 100km/h would take 1h, and that a 200km trip at 200km/h would have to take 1h as well.
Question in the thumbnail... I went directly to : How much time it take for the first part (4 minutes) and how much time do I have for the full trip (also 4 minutes) ?
the first puzzle actually made me do quadratic. After I went through the video I found out that I was just overthinking it LOL
Stop reuploading your old videos, this is a reupload of a 5 year old video, smh.
Watch full video smh
I ain't going back to watch 5 year old videos so it's new to me
I dont mind
I don't mind also.
No-one asked. Don't like it? Don't watch it then. Easy.
I saw the average speed problem in a book many years ago, and I remember when I was 14 presenting this problem to my friends. They refused to believe that reaching the target average speed was impossible. Play confidently asserted the time had nothing to do with speed. One friend even claimed that his truck could accomplish it. I think they know better now. At least, I hope so.
I love your videos
so: (a+c+b^2)/2 == b ??? I solved the 1st by saying a+(b^2)+c covers 2r, and then just divided.. am I wrong? (besides being overly complicated?)
I got r= b in under 3 seconds. Where is my Nobel prize ?
I came across the radius problem on an IQ test, many years ago. I solved it instantly. Take that Feynman!
Every one of these is trivially easy, IF you look at them a certain way.
and this is why traffic engineering differentiates between time average and distance average when it comes to average speed
Is there a copy of the original letter that Werthelmer sent to Einstein? Did Werthelmer really use miles as the unit of measurement?
Damn; you had me with the last one.
You know, I actually thought about it using the second method you were talking about, with the idea of 4 minutes. I am not the best at math in my head for equations like this, but I did notice that it seemed like there wasn't enough time remaining to go that fast although I wasn't sure exactly why.
That's made me feel great about my hitherto averagish brain. I got that Einstein-stumper straight away, just by looking at the thumbnail.
Go brain! Go brain!
When driving long distances over expressways where the speed limit is sometimes 70 mph or more, I TRY to have an average speed of about 65 or so. But it's really hard when you have to stop every 3 hours or so, even for just a few minutes (for gas, food, and the call-of-nature). Even brief stops quickly drops the 'average speed'.
i will literally confuse my friends with this question before our math exam saying that this is in our syllabus xdxd
Now im curious, if you add relativity to the car puzzle, would it be possible that going up the hill time is little less then 4 minutes and then it goes to light speed with whatever time is left? ;)
If I do the second half of the trip at the speed of light, then time would stop for me.
well because the other diagonal would also be b, it would be a straight line going from the center to the edge of the quarter circle, which means b is the radius of the circle. as for a and c, idk lol maybe somehow would be able to compute it in terms of r and b using trig functions.
Actually, at the speed of light, the last mile would take 5.3µs, and the whole thing feels pretty 1s.f. so I don't think anyone would say you hadn't made it.
No3 is a great example of something that happens all the time in industry and the media. I call it the average of averages and so many decisions are based on this flawed logic! This is a great way of pointing out that flaw.
Who is measuring the time spent on the decent?
If you're travelling at the sped of light, time doesn't exist. As such, if you're the one timing the decent, the answer is the speed of light.
If it's an external measurement, then yes - it's not possible.
Haha! This was a dead giveaway to me but out of interest I subjected my family to this riddle and for sure my wife responded 45mph. Next my kid who is in second grade middle school gave the exact same answer, so I asked him if he could calculate what the resulting times for both distances would be for the given average speeds. For sure he did calculate it was 4 minutes for both but due to the nature of the question he then concluded that the additional mile had to be run in less than a minute and ended up answering 60mph. So next I asked him if he could use seconds instead of minutes and the big question mark started rising above his head - ehm 3600mph? Granted I don't really know if he already learned the concept of infinity slash division by zero, I think back then I already had it in first grade, but I believe he really did get it when I gave him the answer.
Now I'm afraid to drive 30MPH.
My first thought was, "speed of light" The reason is simple, 15mph and to travel 1 mile, you have to spend the entire time as you would do with average for 2 miles. That is, there isn't time to spare!
I have a riddle how many times minute hand and hour hand makes 90deg between noon and mid night?
I actually came a variation of the Einstein problem decades ago (as a kid), so I already knew the answer to it. But it’s also a good teaching tool-the “average” isn’t always the arithmetic mean. Especially when dealing with complex units (a complex unit is one that combines 2 or more basic units).
When looking at the radius problem, I didn't notice the r=b shortcut at first. But when I looked at the quadratic solution, I immediately saw the 2ac - a² - c², and in my head converted it to -(a-c)², and now I had 2b²-(a-c)² and thus b² + b²-(a-c)² = b² + (a+b-c)(a+b+c), and slowly, it dawned me, that I missed something.
I got the b=r one instantly but that's probably because Feynman was primed for it to be a maths puzzle and tunnel visioned himself (I assume, I don't know anything about him outside that)
My father showed us the first one, a half-century back.
So for the first problem, I got both solutions on my own, but how are they the same? What is missing to simplify the complicated expression from the quadratic formula to just come out to be b?
6:35
mark antony disagrees:
"the evil men do lives after them; the good is oft interred with their bones"
I love your videos. They are beautifully designed and presented in an absolutely clear way. I'd just point out two mistakes:
1. The symbol for hour is just (upright or roman) h
2. Since time depends on the observer's perspective, going at light speed does solve the problem, assuming the one person interested in completing the trip on such average speed is the driver. From his perspective, going at light speed does take him to the 2-mile mark instantly, whereas an observer (e.g., someone who had an appointment with him) would be pissed due to him arrving a fraction if a second late.
Here's the thing though - it's an old car and can't manage more than 15mph up the hill ! So I reckon light speed is literally out of the question. =o)
@tezzerii that seems likely. Also, technically, a wheeled vehicle can only reach half light speed. This is because the highest speeds on a vehicle are those of the upper part of the wheels. If those where at light speed, their center would be at light speed divided by two.
It's an arithmetic problem, not a relativity problem. Get a clue.
It is a physics (kinematics) problem, @mrosskne. Make sure you are at least right before correcting others.
Oh thank goodness. I clued in on r=b immediately, but thought I must have missed something.
2 miles at 30 mph takes 4 minutes. So does 1 mile at 15 mph. The second mile would have to be instantaneous.
"Hooray! We found another way not to do it." - from "Meet The Robinsons"
You want to drive 1mile in 4mins and 2 miles in 4mins, so the 2nd mile has inf speed
You forgot about time dilation. If you travel at c on the way down, no time will pass for the car, so total trip time is still 4 minutes.
I didn't overthink the first one, it seemed obvious using trig that r=b.
The second one is deeply obvious unless you think 6+5 and 5+6 are different.
The last one takes a moment to realise the average speed is already sucked up by the ascent time.
Sometimes overthinking or expectation stops the otherwise obvious from being apparent.
Strictly the Leibniz problem is ambiguous: if you don't assume commutivity then you can interpret "a sum of" to be "a specific sum of" so that 5+6 is not the same sum as 6+5.
How is it ambigous? Rolling 11 is more likely and that can easily be tested.
I thought it was an interesting paradox, but then realized it’s just worded in a way that tricks you, which is uninteresting.
I remember the car riddle from an old Encyclopedia Brown book.
There are several lines of thought to immediately get the answer to the Einstein puzzle. One of them is: If you you have a certain speed for a given distance and you want double the speed on double the distance then of course the time has to be exactly the same. This is true for any non-relativistic speed and any non-relativistic distance.
If it's easier for you to calculate with certain numbers. Just imagine the first part of the hill to be exactly 15 retard units. Then the ascent takes 1 hour. The total distance would be 30 retard units and hence you also have 1 hour for the full distance to get the desired average speed. Which is of course impossible unless you are a massless particle...
I refer to it as "a 5th-gear brain working on a 2nd-gear problem." If you think of the brain as a car engine, you can think of working on higher-end problems as being in a higher gear. Solving a simple problem with your brain in low gear is like trying to drive at a slow speed in a low gear; the engine handles it just fine. Trying to drive slow in a high gear, however, causes so few RPMs that the engine stalls and dies out, so it can't do it. ...A bit of a weird analogy, I admit, but it seems to get the point across.
Just to reword the question before watching:
If a car travels one mile in four minutes, what speed does it need to go to cover a second mile in zero additional minutes?
In Star Trek, they call it warp ten.
I'll restart it... a car averages 15.1MPH for the first mile...
Leibniz definitely did not play enough Settlers of Catan, smh
No wonder Einstein got it wrong. :P But if you were in the car, I bet it would feel like only 4 minutes.
I got the last puzzle right within seconds without calculations. I ride a bicycle a LOT, and I calculate my ride times by my average speeds. If I'm X miles from home, how fast do I have to go to get there in Y minutes? Yes, I call ahead and tell people I'm going to be late often. Why do you ask?
It's hard to imagine Newton making such an elementary error of probability...
einsteins riddle is only tricky because time is involved. since time doesnt exist in real life, its only a mathematical measurement, you can actually just average the total distance and take an eternity to cross it.
Worked it out. Watched the video for confirmation. It’s good being intelligent 😊
I have a weirdo way of thinking about the Einstein trick since I've seen it before, and it kinda works out by making the answer also a trick.
So in some versions of the problem, it is vague about what average is being taken over and asks simply "what speed do you have to go for the total trip to average 30mph". This is generally assumed to be the average speed over the entire distance of the trip (2 miles). As the video shows, there's no way to get an average speed of 30 mph over the distance without traveling instantaneously.
However, if you consider the average over time instead, then you end up back with 45 mph IFF you drive for another 4 minutes. Because it takes 4 minutes to go up at 15 mph, then another 4 minutes at 45 mph, you end up with an average speed of 30 mph. The distance you have traveled will end up being much further (4 miles here vs 2 in the original problem). If you reduce the time component out of this entirely (since it is constant), then it ends up being just the simple average of 15 and 45 like shown in the video, but it's important to clearly state the same amount of time as a constant.
I got Leibniz's question right, and then I changed my answer to them being equal, and now I'm mad
Wow , so amazing technique of puzzle #1.
one way to look at the last one is if you double the speed required for average to travel you would double the distance in the same amount of time.
"Tricked"... in other words deceived intentionally, or unintentionally, so the failure belongs to the person stating the question.
@3:40 - end of the video. Thank you. I do not watch anything past this point.
Correction: the notion that the speed of light is the same in all directions is unprovable in principle (at least, not without FTL capability). The speed of light is considered to be the same in all directions only by convention. Therefore, it is entirely possible for the speed of light to be infinite in one direction and 1/2c in the other. Since the car only needs to travel at infinite speed in one direction, it can, provided that it becomes massless on the descent.
What’s funny is that Einstein should have noticed such a loophole given that he noted in his own paper that the one-way speed of light is unknowable and he set it equal to c by convention.
I barely drew the diagram when I realized that r=b.
Where's my Nobel Prize?
Ah thank God. I was thinking all the while “how is the radius not simply b?” Good thing it is. Was beginning to doubt myself.
Should you not be able to simplify the expression to get r=b?