"A car travels the first half of its journey at 60km/h.

>>8154872
Trick question.

Before the car can reach the end of the journey it has to travel half the distance, before it reaches half the distance it has to travel half of half the distance and before it reaches that it has to travel half that distance again and so on an infinite number of times.

Hope this helps :^)

>>8154872
back to high school kiddo. And yes it is solveable

>>8154872
x/70 = x/120 + x/2y
or
1/70 = 1/120 + 1/2y
-1/2y = 1/120 - 1/70
y= 168 km/h

>>8154972
Good one.

>>8154872
It is 80km/h unless they impose finite acceleration in which case the time of acceleration can vary giving a continuum of possible solutions.

>>8154989
It's true though.

>>8154995
They didn't say half the time. They meant half the distance.

>>8155006
Gotcha. I think that's bad wording, because I think the average person is more concerned with the time elapsed through a trip than the distance covered. I know when I say "I'm halfway there" I'm talking about time, not distance.

>>8155008
Almost all textbooks have a problem just like this.
As far as time, it's not determined until speed is set.

>>8154872
Can't believe I"m doing this.

Suppose the car travels distance d in the first half of the journey, and averages speed v in the second half.

It takes d/60 to complete the first half and d/v to complete the second half, so d/60+d/v to complete the entire journey.

Therefore, the proportion of time spent completing the first half is (d/60)/(d/60+d/v)=v/(v+60) and the proportion of time spent completing the second half is (d/v)/(d/60+d/v)=60/(v+60).

So the time-average speed is

60*v/(v+60)+v*60/(v+60)=120v/(v+60).

I'll admit that I first thought the answer was 80 km/h. Ultimately it depends on how you interpret "average speed", and more precisely whether you average over distance or over time. If you average over time, then you get the answer above, but I think it's also reasonable to average over distance, in which case 80 km/h is correct.

>>8154972
Isn't it supposed to be
2y=168
y=84

>>8154872
Its 80

>>8155048
yeah. oops.

>>8154872
Harmonic mean
70 = (2*60*x) / (60+x)
4200 + 70x = 120x
4200 = 50x
x = 84 km/h

Let's say the whole journey is J km. He travels J/2 km at 60 km/h, so it takes him J/120 hours for the first part. Let's say X is he second speed in km/h, so it will take him J/(2 X) hours for the second part. His average speed V is total distance divided by total time, so V = J /
(J/120 + J/(2 X) ) =(120 X)/(X+60). Note that there is no J in there! We want V to equal 70, and solving (120 X)/(X+60) = 70 for X gives X=84 km/h.

84 km/h.