Hypothetically this should work for any distance, any value of D. D is constant and is not related to Velocity 1 or Velocity 2. Lets assume D is 120 miles in this case, a common factor of 60 and 40. T(1) for Way #1: 120 miles / 40 miles/hr = 3 hrs
T(2) for Way #2: 120 miles / 60 miles/ hr = 5 hrs
Our average velocity is equal to the total distance/total time.
One way to understand this intuitively is to use more extreme numbers. What about you drive one way at 1 mph and the other way at 99 mph. You can clearly see that 1 mph trip kills your chances of going 50 mph, even if that's what the naïve calculation still shows with these numbers.
If T1=T2, the arithmetic mean would work. Since T1=/=T2, a weighted mean is necessary. In this case, it's a harmonic (weighted) mean, similar to the formula for resistors in parallel in an electric circuit. (A weighted mean is the mean of two or more samples of different sizes.)
Well, I both agree and disagree with the 48. It could be 50, depending on perspective. You say "average" but average is pretty generic in and of itself. Average speed with respect to time is 48, but average speed with respect to distance is 50. When we ask to compute the average, why do you assume it needs to be done with respect to time? That is a leap in reasoning in itself.
A car goes round a one mile track at an average speed of 30mph. What speed must it do a second lap at to average 60mph for both laps? Answers as replies please.
This, too, is also extremely clever. I had to see it in my mind to get an answer.
Imagine the clock face is a track. 2 cars race, one at 30 m.p.h. and one at 60 m.p.h.
They both take off! When the 60m car goes around once, the 30m car is at the ^ O'Clock Position. When the 30m Car reaches a full lap, the 60m Car has done 2 laps.
Therefore it is mathematically too late to average a speed in physical time. In virtural time, the car must travel at instentaneous m.p.h. Very Clever.
@EighteenCharacters I think what you're saying is that since both displacements are 0 the method of averaging speeds is inadequate? well, that's not true. In that case you would either consider rotational velocity (in radians/second) or you would convert the "lap" into its linear translation before doing the calculation by working out the circumference. Either way, you will get a result consistent with the fact that one car is travelling 2x as fast as the other
@messakg123 No, not at all. The original question supposes that there is a racer traveling at 30 m.p.h. around a 1 mile track. He does a lap. Then the question is asked "How fast must he go in order to average 60 m.p.h." for both laps. En exploring that, the time needed to go 60 miles per hour (60 miles in 60 minutes) is 2 minutes for 2 laps. He's been running 30 miles per hour for 2 minutes already and mathematically cannot go fast enough to do a second lap to average 60 for both laps.
@rbwannasee Oh that's just mean. The answer is it's impossible for your average speed to be double or more the speed of the first lap. You will always finish the second lap at the higher speed before enough time elapsed to 'pull' the average that high.
Because when you hit college, you're no longer stupid and capable of rational, independant thought. I know this is an old comment, but I felt complelled to reply for any who read it.
It teaches young kids to study what is infront of them. There are many pits and traps in the world. It isn't a mathematical study in as much as it is a study in the deceptive nature of the world (at times).
It is a life lesson if anything else. Where else better than college (outside of the home)?
Actually, this isn't accurate. I know this is an old comment, but just so anyone reading can know, the problem said 40 miles per hour and 60 miles per hour.
Assume the distance is d in miles. Then we're dealing with D- distance, and 40 & 60- also Distance. Yet the professor handles the 40 & 60 as if they were time.
It's tricksy. Dmiles/40 miles vs Dmiles/60 Miles does not give you Miles per hour. Hour isn't in the equation. It is actually an average miles travled.
Imagine that you have to travel 240 from Cambridge to Oxford, if you travel in one direction 6 hours in 40 mph and 4 hours in 60 mph in the end you will get distance of 480 miles and your total time of travel will be 10 hours. Divide 480 with 10 and what you get is 48mph... If you don't believe me, use any other numbers for distance between Oxford and Cambridge :)
@elirox100 Speed is a ratio. 40 and 60 are distances. Hence 40 and 60 miles. This is thus divided by a time, hour, giving you 40 m.p.h and 60 m.p.h. 40 and 60 are distances. What you've said is akin to "Nails are houses, you idiot". They are part of a house, and make up the house, but are not the house.
This is a 400 day old comment. Do you have such a lack of understanding about fellow man, that you would flail about insults aimed at hurting them without regard?
Hypothetically this should work for any distance, any value of D. D is constant and is not related to Velocity 1 or Velocity 2. Lets assume D is 120 miles in this case, a common factor of 60 and 40. T(1) for Way #1: 120 miles / 40 miles/hr = 3 hrs
T(2) for Way #2: 120 miles / 60 miles/ hr = 5 hrs
Our average velocity is equal to the total distance/total time.
120(2)/5hrs = 240miles/5 hrs = 48mph
Ramoola1 2 months ago
Cambridge and Oxford are like right next to me.. Are those common town names or is he talking about Massachusettes?
sc81kdeterb 3 months ago
One way to understand this intuitively is to use more extreme numbers. What about you drive one way at 1 mph and the other way at 99 mph. You can clearly see that 1 mph trip kills your chances of going 50 mph, even if that's what the naïve calculation still shows with these numbers.
Gameboygenius 5 months ago 4
@Gameboygenius lol ya
TheDoncruz11 5 months ago
If T1=T2, the arithmetic mean would work. Since T1=/=T2, a weighted mean is necessary. In this case, it's a harmonic (weighted) mean, similar to the formula for resistors in parallel in an electric circuit. (A weighted mean is the mean of two or more samples of different sizes.)
ktd66 5 months ago
pause at 0:13
he looks like a ken doll xP
likeRAWRnshit 7 months ago
It's not surprising when you realise that you spend longer going 40mph because you're going slower.
messakg123 7 months ago
Well, I both agree and disagree with the 48. It could be 50, depending on perspective. You say "average" but average is pretty generic in and of itself. Average speed with respect to time is 48, but average speed with respect to distance is 50. When we ask to compute the average, why do you assume it needs to be done with respect to time? That is a leap in reasoning in itself.
CogitoErgoCogitoSum 1 year ago
I understand why this might be surprising, but I don't think this would be considered a paradox
stikpet 2 years ago
A car goes round a one mile track at an average speed of 30mph. What speed must it do a second lap at to average 60mph for both laps? Answers as replies please.
rbwannasee 2 years ago
faster than light !
SuperRaptorz 2 years ago
This, too, is also extremely clever. I had to see it in my mind to get an answer.
Imagine the clock face is a track. 2 cars race, one at 30 m.p.h. and one at 60 m.p.h.
They both take off! When the 60m car goes around once, the 30m car is at the ^ O'Clock Position. When the 30m Car reaches a full lap, the 60m Car has done 2 laps.
Therefore it is mathematically too late to average a speed in physical time. In virtural time, the car must travel at instentaneous m.p.h. Very Clever.
EighteenCharacters 1 year ago
@EighteenCharacters I think what you're saying is that since both displacements are 0 the method of averaging speeds is inadequate? well, that's not true. In that case you would either consider rotational velocity (in radians/second) or you would convert the "lap" into its linear translation before doing the calculation by working out the circumference. Either way, you will get a result consistent with the fact that one car is travelling 2x as fast as the other
messakg123 7 months ago
@messakg123 No, not at all. The original question supposes that there is a racer traveling at 30 m.p.h. around a 1 mile track. He does a lap. Then the question is asked "How fast must he go in order to average 60 m.p.h." for both laps. En exploring that, the time needed to go 60 miles per hour (60 miles in 60 minutes) is 2 minutes for 2 laps. He's been running 30 miles per hour for 2 minutes already and mathematically cannot go fast enough to do a second lap to average 60 for both laps.
EighteenCharacters 7 months ago
@EighteenCharacters sincerest apologies, i didnt see the comment you were replying to.
messakg123 7 months ago
Keep them coming by the way! I love these!
EighteenCharacters 1 year ago
@rbwannasee Oh that's just mean. The answer is it's impossible for your average speed to be double or more the speed of the first lap. You will always finish the second lap at the higher speed before enough time elapsed to 'pull' the average that high.
fatsomamacheese 1 year ago
why is this being tought in college. the answer is extremely obvious
thejohnofsteel 2 years ago
Because when you hit college, you're no longer stupid and capable of rational, independant thought. I know this is an old comment, but I felt complelled to reply for any who read it.
It teaches young kids to study what is infront of them. There are many pits and traps in the world. It isn't a mathematical study in as much as it is a study in the deceptive nature of the world (at times).
It is a life lesson if anything else. Where else better than college (outside of the home)?
EighteenCharacters 1 year ago
it's because it takes longer to travel when you go at 40 mph. more of your time is spent at 40 miles per hour.
trump3t 3 years ago 21
Yeah, isn't this more technically the expectation value of speed?
chaztikov 3 years ago
That was...A little obvious....
Xjkjk3264X 2 years ago
Actually, this isn't accurate. I know this is an old comment, but just so anyone reading can know, the problem said 40 miles per hour and 60 miles per hour.
Assume the distance is d in miles. Then we're dealing with D- distance, and 40 & 60- also Distance. Yet the professor handles the 40 & 60 as if they were time.
It's tricksy. Dmiles/40 miles vs Dmiles/60 Miles does not give you Miles per hour. Hour isn't in the equation. It is actually an average miles travled.
They are tricksy
EighteenCharacters 1 year ago
Imagine that you have to travel 240 from Cambridge to Oxford, if you travel in one direction 6 hours in 40 mph and 4 hours in 60 mph in the end you will get distance of 480 miles and your total time of travel will be 10 hours. Divide 480 with 10 and what you get is 48mph... If you don't believe me, use any other numbers for distance between Oxford and Cambridge :)
MakingYouSad 1 year ago
@EighteenCharacters
This has to be the stupidest thing I have ever read on the internet.
hamsterpoop 11 months ago
@hamsterpoop Smart enough for you to reply to it.
What did you expect to achieve by telling me this? Because I smell irony an internet mile away.
EighteenCharacters 11 months ago
@EighteenCharacters 40 and 60 are speed not distance you idiot
elirox100 10 months ago
@elirox100 Speed is a ratio. 40 and 60 are distances. Hence 40 and 60 miles. This is thus divided by a time, hour, giving you 40 m.p.h and 60 m.p.h. 40 and 60 are distances. What you've said is akin to "Nails are houses, you idiot". They are part of a house, and make up the house, but are not the house.
This is a 400 day old comment. Do you have such a lack of understanding about fellow man, that you would flail about insults aimed at hurting them without regard?
EighteenCharacters 10 months ago
@EighteenCharacters ]
Just admit you're wrong, you moron.
The 40 and 60 are NOT distances, you stubborn simpleton.
frazzzer8888 10 months ago
@frazzzer8888 40 miles and 60 miles are not distances. I admit you're wrong you moron.
There. Find a new hobby.
EighteenCharacters 10 months ago