what about accelleration from 0 to 60 then 60 to 120 and decelleration from 120 to 0
or am I just being pedantic :D
what about accelleration from 0 to 60 then 60 to 120 and decelleration from 120 to 0
or am I just being pedantic :D
The version of that mathematical exercise that I prefer, because I think it makes the point more dramatically is:
If you travel from A to B at 30km/h, how quickly will you have to drive back from B to A to average 60 km/h?
The answer is of course that even travelling at the speed of light will not give you a 60 km/h average because the first leg has already taken all the time needed to do the whole trip at 60 km/h.
For example if A to B was 30 km, it would take an hour to get there and that is how much time you need to average 60 km/h for the 60km return trip.
blitz wrote,...Hello Blitz,Quote:
what about accelleration from 0 to 60 then 60 to 120 and decelleration from 120 to 0
or am I just being pedantic
The question was only simple so the speed is assumed to be constant. Once you start to include acceleration and negative acceleration both linear and non linear, the complexity quickly appreciates.
I can if you like ask some questions where there is considerble complexity,..will keep the brain active and healthy for quite a while...:)
Ron.
To drive bak from B to A to average 60km/h you would only need to drive at 60km/h. :DQuote:
Originally Posted by vnx205
Oh you mean... :D
:DQuote:
Originally Posted by Slunnie
I probably became confused when the word "average" came up or maybe I just panicked about how some people might react to the word and forget to conclude the sentence with "60 km/h for the entire trip". :D:D:D
Quote:
Originally Posted by RoverP6B
I was just stirring Ron it's a good question as the answer is not the obvious.
I must be dum I dont get it.
If A to b is 30kph and b to a you drive 90 kph wont average kph be 60 ?
(Maths was never a strong point with me )
If we assume that the distance between each point is 100km, although you can substitute in any distance - the answer will still work out the same.Quote:
Originally Posted by VladTepes
A->B will take 3.33 hours (30km/h for 100km)
B->C will take 1.11 hours (90km/h for 100km)
The total trip of 200km therefore takes 4.44 hours, simplifying to a 100km trip taking 2.22 hours.
100km / 2.22hours will simplify by dividing the numerator and denominator by 2.22 to:
45km / 1 hour or 45km/h as you know it.
You would be correct however if you drove from A to B at 30km/h for x time and then B to C at 90km/h for x time.
However this question is that you travelled from A to B at 30km/h for x distance and then from B to C for x distance.
Look closely at my example in post #15.Quote:
Originally Posted by VladTepes
If you drive half the distance at half the speed you want to average, you have already used up all your time.
The key is to look at how much time each half of the trip takes, not how much distance each half of the trip is.
Or consider a more extreme example. You have 60 km to travel. If you drive the first half at 10 km/h, that will take you 3 hours to cover those 30 km, so there is no way you can average 60km/h by driving faster for the rest of the trip. You have already taken 2 hours more than the one hour you would need to do the trip if you wanted to average 60.
Driving at 110km/h will not get your average up to 60 even if (10+110)/2 = 60
Why do you keep asking maths questions ???????????????????
Are we doing your homework