RIDDLE: Average Car Speed (Warning: Tricky!)

[DarloRich]

An even smarter-arse answer would be "wherever the passing loop is, hopefully".

I didn't want to get overly technical

My English teacher would have got it. My maths teachers was a car enthusiast. Humour generally was lost on him. Got me through my exams mind and the only person who recoginsed my trouble with maths ( Dyscalculia). He was a good guy.If dour.

 

[Strat-tastic]

If the car could travel at the speed of light then to an observer in the car it would appear to take zero time to complete the second lap so the total time for that observer to complete two laps would be 2 minutes.

But what would the stopping distance be?

 

[Howardh]

This isn't a puzzle, it's a genuine question; if a human was travelling at the speed of light - maybe towards the earth in a spacecraft - what would he see in front of him, and behind him? And if he shone a torch forward...and then backwards...? I would suggest, probably wrongly, that as he would be relatively still, the light from the torch would go forth at the SoL, therefore the light is travelling twice the speed of light which can't happen.....I need a beer.
If he's still then effectively everythings approaching him at the SoL so the torch light wouldn't emerge?
Another beer, please.

 

[DaleCooper]

The speed of light is the same for all observers regardless of their speed relative to some arbitrary frame of reference.

 

[DynamicSpirit]

If the car could travel at the speed of light then to an observer in the car it would appear to take zero time to complete the second lap so the total time for that observer to complete two laps would be 2 minutes.

Unfortunately, it would also appear to the observer in the car that the distance travelled over the 2nd lap was zero (space contraction and all that), which would kinda bugger up the average speed calculations

This isn't a puzzle, it's a genuine question; if a human was travelling at the speed of light - maybe towards the earth in a spacecraft - what would he see in front of him, and behind him? And if he shone a torch forward...and then backwards...? I would suggest, probably wrongly, that as he would be relatively still, the light from the torch would go forth at the SoL, therefore the light is travelling twice the speed of light which can't happen.....I need a beer.
If he's still then effectively everythings approaching him at the SoL so the torch light wouldn't emerge?
Another beer, please.

There is a slight matter that it's impossible for a human to travel at the speed of light (also a problem for my answer to DaleCooper above) because it would require infinite energy. You can only travel at the speed of light if you have zero rest mass. But, if you could find a zero-mass human being to travel at the speed of light, he wouldn't perceive anything, because time dilation and distance contraction would mean, that, no matter how far and for how long the beam of light appears to us to travel, it would all be instantaneous (and zero distance) to him.

The speed of light is the same for all observers regardless of their speed relative to some arbitrary frame of reference.

Agreed - for anything travelling below the speed of light relative to a (real) observer. I'm not sure whether you can apply that for something travelling at the speed of light itself though, since you'd probably end up with zero divided by zero in any actual calculation. I suspect the actual get-out is to say that something travelling at the speed of light cannot be an observer (which would appear to be 'obviously' true because no time would ever pass for something travelling at the speed of light, and you need a finite interval of time to be able to observe something!)

 

[Bevan Price]

that makes sense but i have no idea how you work that out from the information given. I could sit all week and never be able to understand it. My brain just doesnt work like that.

We had a question at school in a paper which was basically if train A leaves station A travelling at 100mph and train B leaves station B traveling towards train A at 60 mph where will they meet and after how long.

I put:
a) Northallerton.
b) depends on the timetable and any disruption.

Result =A telling off.

There is insufficient information to answer that question. You also need to know the distance between A & B.

 

[DarloRich]

There is insufficient information to answer that question. You also need to know the distance between A & B.

The typical response of a mathematician

 

[SS4]

Just thought we'd all have a bit of forum fun with this very tricky riddle. I love doing this one, so here it is below...

There is a racecourse loop that, if laid out flat, would stretch exactly one mile. So there is basically a mile of racecourse track.

A driver does a whole lap around the racecourse, averaging a speed of 30mph. How fast would the car have to go on the second lap in order to average a speed of 60mph over the course of BOTH laps?

Enjoy getting your head wrapped around this one

90mph (spoilers below)








Let x be the average speed of the second lap.

The Average speed of both laps is given by (u1 + u2)/2 where u1 and u2 are the average speeds of the first and second laps respectively and must be equal to 60 as given in the question.

We know that:
u2 = x (by my definition of x)
u1 = 30 (given)

and so (30+x)/2 = 60

Solving for x gives x = 90mph



-------------------------


More generally we can say that for speeds u1 and u2 then u2 is given by u2 = 2µ-u1 where µ is the average speed.

Suppose though that we have n laps with speeds u1, u2, u3, ..., un

µ = (u1+u2+u3+...+un)/n

un = µn - (u1+u2+u3+...+u[n-1])

--------------------------

It would be easier if we were asked to work out the average velocity which would be 0

 

[DaleCooper]

90mph (spoilers below)








Let x be the average speed of the second lap.

The Average speed of both laps is given by (u1 + u2)/2 where u1 and u2 are the average speeds of the first and second laps respectively and must be equal to 60 as given in the question.

We know that:
u2 = x (by my definition of x)
u1 = 30 (given)

and so (30+x)/2 = 60

Solving for x gives x = 90mph



-------------------------


More generally we can say that for speeds u1 and u2 then u2 is given by u2 = 2µ-u1 where µ is the average speed.

Suppose though that we have n laps with speeds u1, u2, u3, ..., un

µ = (u1+u2+u3+...+un)/n

un = µn - (u1+u2+u3+...+u[n-1])

--------------------------

It would be easier if we were asked to work out the average velocity which would be 0

Lap 1 at 30mph takes 2 minutes
Lap 2 at 90mph takes 0.667 minutes
Total time for 2 miles = 2.667 minutes, average speed = 45mph

 

[Up_Tilt_390]

Thanks to DaleCooper for reminding people why 90mph is wrong

Everyone ought to know that you can spend all day trying to come up with every possible answer, but because it took 2 minutes to do one mile at 30mph, you will never average 60mph over the course of the two laps. Not to mention too, we haven't even considered the force of acceleration when the first and second lap starts...

 

[DaleCooper]

I think this thread should really be in "Quizzes & Games" however here's another one:

I take a 100g potato of a certain (fictitious) variety which is 99% water and dry it in an oven until the water content is reduced to 98%. Assuming only water has been lost how much does the potato now weigh?

 

[Jonny]

I think this thread should really be in "Quizzes & Games" however here's another one:

I take a 100g potato of a certain (fictitious) variety which is 99% water and dry it in an oven until the water content is reduced to 98%. Assuming only water has been lost how much does the potato now weigh?

Dry weight is 1g (100g less 99%).
Remaining dry weight, 1g, now makes up 2% (of 98% water)
1g of dry mass represents 2% of 50 grams (being the final answer)

 

[JoeGJ1984]

I think this thread should really be in "Quizzes & Games" however here's another one:

I take a 100g potato of a certain (fictitious) variety which is 99% water and dry it in an oven until the water content is reduced to 98%. Assuming only water has been lost how much does the potato now weigh?

50g (is this the correct answer?)

 

[DaleCooper]

Dry weight is 1g (100g less 99%).
Remaining dry weight, 1g, now makes up 2% (of 98% water)
1g of dry mass represents 2% of 50 grams (being the final answer)

50g (is this the correct answer?)

Both of you are correct, I wouldn't recommend that variety of potato.

 

[Bevan Price]

Bevan Price said: ↑
There is insufficient information to answer that question. You also need to know the distance between A & B.

The typical response of a mathematician

An alternative answer is that they never met, because if station A was Wick, one train "fell off" trying to traverse the Inverness avoiding line at excessive speed.

 

[DynamicSpirit]

90mph (spoilers below)

Let x be the average speed of the second lap.

The Average speed of both laps is given by (u1 + u2)/2 where u1 and u2 are the average speeds of the first and second laps respectively and must be equal to 60 as given in the question.

That formula is not correct. It would only work if the time spent travelling at the two speeds was equal, which is not the case here.

If you want to reason in the algebraic way that you've attempted, then the correct approach is to say, average speed (call it u_av) = total distance/total time.

The distance is 2 miles. The time for each lap is distance/speed, so the total time is (1/u1) + (1/u2). Hence u_av = 2 / [(1/u1) + (1/u2)]. A bit of algebra turns that into u_av = (2 * u1 * u2) / (u1 + u2). If you use that formula, you'll find that, with u1=30mph, the only way to get u_av to be 60mph is to set u2 to be infinite - which is of course the correct answer (if you ignore relativistic effects ).

 

[PaulLothian]

Basic rule is that you cannot take a simple average of figures that are themselves averages. Same applies to taking an average of a series of percentages. In each case, you must go back to the original data. A rule drummed into me at school half a century ago of which I have frequently to remind young colleagues looking for shortcuts!

 

[Jonny]

As a slight aside, the driver would then have to go at 90mph for two whole laps thereafter to bring his average up to 60mph.

 

[Up_Tilt_390]

As a slight aside, the driver would then have to go at 90mph for two whole laps thereafter to bring his average up to 60mph.

1 lap at 30mph = 2 minutes
1 lap at 90mph = 40 seconds
1 lap at 90mph = 40 seconds

Total time is 3 minutes ten seconds. Since three laps have been done, this leaves three miles covered. Using the formula for speed calculation, the average speed works out at 57mph. Close, but not 60.

 

[DaleCooper]

As a slight aside, the driver would then have to go at 90mph for two whole laps thereafter to bring his average up to 60mph.

No matter how many laps are completed at 90mph the average speed will asymptotically approach 60mph but will never reach that figure.

 

[DynamicSpirit]

Lol! I love how this apparently simple problem is confusing so many people.

1 lap at 30mph = 2 minutes
1 lap at 90mph = 40 seconds
1 lap at 90mph = 40 seconds

Total time is 3 minutes ten seconds. Since three laps have been done, this leaves three miles covered. Using the formula for speed calculation, the average speed works out at 57mph. Close, but not 60.

I think you'll find it's 3 minutes 20 seconds! Doing 3 laps at 90mph after the 30mph lap will get the average speed to exactly 60mph.

No matter how many laps are completed at 90mph the average speed will asymptotically approach 60mph but will never reach that figure.

Are you sure of that?

Sufficiently many laps at 90mph will cause the average speed to asymptotically approach 90mph, not 60mph!

 

[Up_Tilt_390]

I think you'll find it's 3 minutes 20 seconds! Doing 3 laps at 90mph after the 30mph lap will get the average speed to exactly 60mph.

I was wrong about the time I admit. But the 3 miles in 3 minutes and 20 seconds only gives an average speed of 54mph.

As for doing three laps at 90 after the 30. Well, that adds another 40 seconds, so it takes 4 minutes to do 4 miles. That in itself does average it out to 60, but the riddle only specified 2 laps.

Then confusion comes from the fact that people are tying to average 60mph over the course of two laps, which is impossible if the first lap is done in two minutes at 30mph.

 

[DynamicSpirit]

As for doing three laps at 90 after the 30. Well, that adds another 40 seconds, so it takes 4 minutes to do 4 miles. That in itself does average it out to 60, but the riddle only specified 2 laps.

Then confusion comes from the fact that people are tying to average 60mph over the course of two laps, which is impossible if the first lap is done in two minutes at 30mph.

Agreed. I wasn't attempting to answer the original riddle, merely addressing the most recent posts that were commenting about how many laps you'd need at 90mph. Several people have already correctly pointed out that the answer to the original riddle is that you must complete the 2nd lap at infinite speed to get an average speed of 60mph in two laps.

 

[DaleCooper]

Sufficiently many laps at 90mph will cause the average speed to asymptotically approach 90mph, not 60mph!

Quite right, sorry.

 

[ComUtoR]

We had a question at school in a paper which was basically if train A leaves station A travelling at 100mph and train B leaves station B traveling towards train A at 60 mph where will they meet and after how long.

I always thought of this as a logic puzzle.

Ans : They meet in the same place.

 

[DynamicSpirit]

I always thought of this as a logic puzzle.

Ans : They meet in the same place.

But... The same place as what?

 

[ComUtoR]

Each other..