Ok

here is a little problem that will get the brain cells going.......

You are driving by car to a particular destination, and our only assumption is that you are free to drive at any speed you choose - no traffic jams or anything like that. For the first half of the journey (i.e.half the distance) you drive at 20 miles per hour. You then realise that this is all taking much too long, and that you are going to be late. You therefore decide that you will increase your speed so that your overall average speed for the whole journey will be 40 miles per hour. How fast do you have to drive for the remaining part of your journey in order for your average speed for the whole journey to be 40 miles per hour?


enjoy

Its not possible !

Even at 100,000mph for the second half you can only average 39.99mph. If rounding up is allowed then the answer is very bloody fast. If it has to be mathmatically correct then it is not possible as Snuggs has quickly concurred.

60

relative to whom? are we allowed to mention time dilation?!

technically, if you did the second half at 100% the speed of light, no time would pass for somebody (i forget who...) and to them, your average speed would be 40mph.

I agree with marcyboy 60.

(Av speed first half + av speed second half)/2 = 40
(20+x)/2=40
Rearrange for x x = 60

Therefore you could drive at any speed you like so long as your average speed for the second half of the journey was 60 ie for one sixth total journey distance you could travel at 40, another sixth 60 and the last sixth 80, that would give you and average speed of 60mph

20mph for 1hr = 20mls

20mls x 2 = 40 mls

To average 40mph you would have to cover the remaining 20mls in 0hrs and 0 minutes and 0 seconds.

Not possible.

[Edited on 2/8/05 by Snuggs]

who said it was for an hour?

20mph for 10hrs = 200mls

200mls x 2 = 400 mls

To average 40mph you would have to cover the remaining 200mls in 0hrs and 0 minutes and 0 seconds.

Still not possible.

OK here you go.....

Suppose you started out at 2 o'clock, and your destination is exactly 40 miles away. For your average speed to be 40 miles per hour, you need to arrive at 3 o'clock. But you drove the first half of the journey, 20 miles in this case, at 20 miles per hour. Which has taken you an hour. So it is now 3 o'clock, and you have only covered half the distance. There is no speed that you can drive for the second half of the journey that will make the average speed 40 miles per hour, since you need to be at your destination at this moment. And this applies no matter what the distance of the journey.



thought it would get a few people thinking - it caught me out

Thats two assumptions too many. Youve assumed that your journey is 40 miles and you only have 1 hour to do it in. I would agree that for a 40 mile journey in 1 hour to travel the first half ie 20miles at 20 mph would only get you half way but thats still assuming the journey is 40 miles and you only have 1 hour to do it in The question says the only assumption is that you can travel at any speed.


[Edited on 2/8/05 by jos]

OK..

so your journey is 120 miles,

you need to average 40 mph which gives a total journey time of 3 hours

you travel half the way at 20 mph which is 60 miles and has taken you 3 hours

therefore you need to travel the last 60 miles in no time at all to make your average speed for the trip 40mph........

it works for any distance because by the time you have travelled half way at half speed you are out of time.

Yeah but say the journey was 1 mile, its not impossible then.

Its defo not possible Jos - put the maths on the next post and try and prove us wrong.

i would assume the same theory goes if you do the first half of a journey at 60mph but want to average 120mph overall. Botton line is that you use your total allowed time travelling half the distance at half the speed. It always ends up that you have to travel th eremaining half of the journey in 0hrs, 0mins, 0secs etc.

Journey = 1 mile.
To average 40mph, 1 mile has to be covered in 1.5minutes.

Half a mile at 20mph = 1.5minutes.

Net result - final half a mile has to be completed in 1.5 - 1.5 = 0 minutes.

I'm probably digging myself into a deep hole but until I'm sensible enough to realise it I'm happy digging.

There was no mention of time in the first post.

20mph for 1 million years = 175200000000mls

175200000000mls x 2 = 350400000000mls

To drive 350400000000mls at an average of 40mph would take 8760000000hrs

8760000000hrs = 1 million years

As you have already been driving for 1 million years the the second half of the journey (175200000000mls) must be driven in 0 years 0 days 0 hours 0 minutes and 0 seconds

[Edited on 3/8/05 by Snuggs]

Ok. We have now tried 1 mile and 1,000,000 miles and a few in between. Would anyone like another permutation working through????

The first post doesn't give any distance or time for the journey so it's not possible to work it out from what you have. If you gave these details it could be worked out. Your working assumes there is a time limit on the journey of the time at the speed it takes to complete the first half.

OK...algebra time

time = distance / velocity

v=overall velocity
d=overall distance
v1=speed for first half (20mph)
v2=speed for second half (x mph)
d1=distance travelled for first half of journey
d2=distance travelled for second half of journey

t = d/v = (d1/v1) + (d2/v2)

t is a constant and irrelevant other than to be equal to both sides; we can safely disregard it

d1=d2=d/2, so we can substitute d1 and d2 for d/2

t=d/v=(d/(2v1))+(d/(2v2))

d is common on the right

d/v=d((1/(2v1))+(1/(2v2)))

divide both side by d

1/v=(1/(2v1))+(1/(2v2))

v=40
v1=20

=> 1/40=(1/(2x20))+(1/(2v2))

=> 1/40=(1/40)+(1/(2v2))

subtract 1/40 from each side

=> 0 = 1/(2v2)

multiple both sides by 2

=> 0 = 1/v2

Therefore impossible

PS Yes - I am bored!


[Edited on 3/8/05 by stevebubs]

quote:

---

Originally posted by Peteff
The first post doesn't give any distance or time for the journey so it's not possible to work it out from what you have. If you gave these details it could be worked out. Your working assumes there is a time limit on the journey of the time at the speed it takes to complete the first half.

---

No not half the time, only that you realise youre going to be late. Time is not the most important.

In order of importance is

Average speed, & percentage distance

S1=20
D1=50%=?
T1=?
S2=?
D2=50%=?
T2=?
SA = ((S1+S2)/2)=40
DT=?
Too many unknowns so have to make an additional assumption based on distance only.
Assume DT=20 therefore D1 = 10 & D2=10
S1=20, D1=10 therefore T1 = 0.5hr = 30mins
SA=((20+S2)/2)=40 therefore S2 = 60

Therefore you would have to travel the second half of the journey ie 50% total distance at 60mph to average 40mph over an assumed total journey distance of 20miles.

this is all over complicating a very simple answer....

you have a set journey to do and wish to average 40 mph

if you do half that that journey at 20mph you have taken the same amount of time as it would take to do the entire journey at 40mph

so therefore you are out of time, you need to cover the second half of the journey instantly to have an average speed of 40 mph

The flaw in my calcs has been that I'm using a numerical average over the total distance which is incorrect. As soon as you assuem a distance to substitute for 50% distance that needs to be applied to the total journey average speed to give you total time. Then your point is proven that your have used the allowed time in the first half of the journey and are therefore out of time, unless you travel the last have instantaneously as you say which is just a little slower than my car will travel once its finished so I'll get there but none of you lot will so ner ner ner ner ner



Runs and hides under a bush.


Twas a good laugh bending my mind around to try to prove my point and arguing myself around to show I was wrong.

Whats next weeks???? give me a fighting chance

quote:

---

Originally posted by Snuggs
Its not possible !

---

what about ......


how many locosters doe sit take to change a lightbulb ?

none - its done by the other half, fed up of waiting ..........

Already had that one a while back

quote:

---

How many forum members does it takes to change a light bulb?

1 to change the light bulb and to post that the light bulb has been changed
14 to share similar experiences of changing light bulbs and how the light
bulb could have been changed differently
7 to caution about the dangers of changing light bulbs
1 to move it to the Lighting section
2 to argue then move it to the Electricals section
7 to point out spelling/grammar errors in posts about changing light bulbs
5 to flame the spell checkers
3 to correct spelling/grammar flames
6 to argue over whether it's "lightbulb" or "light bulb" ... another 6 to
condemn those 6 as stupid
2 industry professionals to inform the group that the proper term is "lamp"
15 know-it-alls who claim they were in the industry, and that "light bulb"
is perfectly correct
19 to post that this forum is not about light bulbs and to please take this
discussion to a lightbulb forum
11 to defend the posting to this forum saying that we all use light bulbs
and therefore the posts are relevant to this forum
36 to debate which method of changing light bulbs is superior, where to buy
the best light bulbs, what brand of light bulbs work best for this technique
and what brands are faulty
7 to post URL's where one can see examples of different light bulbs
4 to post that the URL's were posted incorrectly and then post the corrected
URL's
3 to post about links they found from the URL's that are relevant to this
group which makes light bulbs relevant to this group
13 to link all posts to date, quote them in their entirety including all
headers and signatures, and add "Me too"
5 to post to the group that they will no longer post because they cannot
handle the light bulb controversy
4 to say "didn't we go through this already a short time ago?"
13 to say "do a Google search on light bulbs before posting questions about
light bulbs"
1 forum lurker to respond to the original post 6 months from now and start
it all over again.

---

top banana

can't wait for the next 6 months

quote:

---

Originally posted by stevebubs
OK...algebra time

time = distance / velocity

v=overall velocity
d=overall distance
v1=speed for first half (20mph)
v2=speed for second half (x mph)
d1=distance travelled for first half of journey
d2=distance travelled for second half of journey

t = d/v = (d1/v1) + (d2/v2)

t is a constant and irrelevant other than to be equal to both sides; we can safely disregard it

d1=d2=d/2, so we can substitute d1 and d2 for d/2

t=d/v=(d/(2v1))+(d/(2v2))

d is common on the right

d/v=d((1/(2v1))+(1/(2v2)))

divide both side by d

1/v=(1/(2v1))+(1/(2v2))

v=40
v1=20

=> 1/40=(1/(2x20))+(1/(2v2))

=> 1/40=(1/40)+(1/(2v2))

subtract 1/40 from each side

=> 0 = 1/(2v2)

multiple both sides by 2

=> 0 = 1/v2

Therefore impossible

PS Yes - I am bored!


[Edited on 3/8/05 by stevebubs]

---