Another brain stumper (this one involves cars) ----->

DETAILKING · 11-29-05, 07:15

A car is to go from Point A to Point B and return. If the car maintains 30 mph on the trip from Point A to Point B.......How fast must the car go on the return to average 60 mph for the whole round trip??

rjstaaf · 11-29-05, 07:24

You would have to go faster than light to do that. Let's say it is 30 miles from point A to point B. It would take an hour averaging 60mph to do the whole 60 mile round trip. Going 30mph on the one leg has already used up that time.

Neothin · 11-29-05, 12:30

but there is no time limit, you just have to average a speed over any amount of time. you would have to travel 90 mph on the return trip from Point B to Point A to average 60 mph for the round trip.

rjstaaf · 11-29-05, 12:40

Well, that is what I initially thought too. That just seems to obvious, I guess maybe I started over-analyzing the question

JBM · 11-29-05, 12:42

Point A to B is one mile-say, 30 mph takes 90 seconds, 60 mph takes 60 seonds and 90 mph takes 45 seconds, 120 is 30 seconds.

On the first trip it takes 90 seconds, means you owe 30 seconds.

Return trip needs to be done in 30 seconds to have both at 60/60 seconds.

Answer is 120 .

PhilS · 11-29-05, 02:04

The return journey would have to be completed instantaneously (i.e. you would travel from point B back to point A in zero time) to average 60mph for the entire journey - this, of course, is impossible.

I'll explain it empirically rather than algebraically:

Average speed is the total distance travelled divided by the total time taken i.e S = D / T. So, T = D / S. Outward journey represents half the distance travelled. Calculations are for the time taken to travel outward journey at average of 30mph, followed by whole journey at average of 60mph.

1mile @ 30mph => 2mins. 2miles @ 60mph => 2mins.
10miles @ 30mph => 20mins. 20miles @ 60mph => 20mins.
100miles @ 30mph => 3hrs20mins. 200miles @ 60mph => 3hrs20mins
etc

Actually, if you hopped into a jet fighter for your return journey and flew back at an average speed of mach 4 (just over 3000mph) then you'd get fairly close with an average speed of 59.4mph - any conveyer belt would have no effect.

JBM · 11-29-05, 03:42

Ahh yes, i was wrong, to go a mile at 30mph takes 120 seconds.

I am confused about the answer though.

If you go half as slow to get there, you have to go twice as fast to get back.

1 mile@30 mph= 2 min.

1 mile@ 60 mph =1 min.

....never mind, the whole trip(based on a mile) is only going to take 2 minutes, it would be impossible to return, you are stuck in oblivion hahahahhaa.

DETAILKING · 11-30-05, 09:06

This goes to show that when you get slowed down in traffic......well.....no matter how fast you go, you can never really make up the time!

JBM · 11-30-05, 01:10

Lost time and lost money, when its gone its gone.

az57chevy · 11-30-05, 07:36

I hate to be a math geek but it can’t be done unless you believe in instantaneous speed !

For the first half of the trip the driver went the distance AB at an AVERAGE rate of 30 miles/hour

- or -

1 hour/30 miles x AB miles = AB/30 hours

Now for the total trip the subject wants to average 60miles/hour
The round trip – total distance – is 2AB (twice the distance)

60miles/hours = 2ABmiles/(second time hours +AB/30 hours)

Multiply both sides of equation by (second time + AB/30) hours

- and –

60 miles (second time +AB/30) hours = 2AB miles hours

- simplify-

60 (second time)hours +2AB hours= 2AB hours

And subtract 2AB hours from each side leaves your with

60 (second time) hours = 0 hours
- divide by 60 –

Second time (hours) = 0 hours

So the second trip back has to occur in zero time

Time for coffee