/sci/ - Science & Math » Thread #10855110

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Anonymous Thu 01 Aug 17:31:55 2019 No.10855110 View Reply Original Report

Quoted By: >>10855119 >>10855180 >>10855191 >>10855200

A cyclist is riding his bike from point A to point B.
He rides at an average speed of 8km/h for half of the distance and then at 10km/h for the remaining half distance. What is his overall average speed?
Hint: it's not 9km/h

Anonymous

Anonymous Thu 01 Aug 2019 17:34:55 No.10855114 Report

Quoted By: >>10855120

Not doing your homework.

Anonymous

Anonymous Thu 01 Aug 2019 17:40:24 No.10855119 Report

Quoted By:

>>10855110
80/9 km/h ~ 8.9km/h
your homework solving speed: 0km/h

Anonymous

Anonymous Thu 01 Aug 2019 17:41:04 No.10855120 Report

Quoted By: >>10855180

Let's say you ride a 160 km long track.
The first 80km take you ten hours, the other 80km only take you eight. So in total it takes you 18 hours to ride 160km, 160/18=8.88...
>>10855114
It's holiday.

Anonymous

Anonymous Thu 01 Aug 2019 18:06:24 No.10855180 Report

Quoted By: >>10855185 >>10855217 >>10855224

>>10855110
>>10855120
Just take the harmonic mean, it accounts for the difference in speed.
$\frac{2}{\frac{1}{8} + \frac{1}{10}} = 8.888\ldots$

Anonymous

Anonymous Thu 01 Aug 2019 18:08:23 No.10855185 Report

Quoted By:

>>10855180
in general you want to use the harmonic mean for finding averages of rates

Anonymous

Anonymous Thu 01 Aug 2019 18:12:30 No.10855191 Report

Quoted By: >>10855196 >>10855199

>>10855110
$V_{ave}^{0 \to 1/2} = \frac{2}{L} \int_{0}^{L/2} v(x)dx = 8 km/s$ .

$V_{ave}^{1/2 \to 1} = \frac{2}{L} \int_{L/2}^{L} v(x) dx = 10 km/s$ .

$V_{ave}^{0 \to 1} = \frac{1}{L} [ \int_{0}^{L/2} v(x) dx + \int_{L/2}^{L} v(x) dx ] = \frac{1}{L} [(8)(\frac{L}{2}) + (10) (\frac{L}{2} ) ] = 9 km/s$ .

What am I doing wrong here?

Anonymous

Anonymous Thu 01 Aug 2019 18:14:58 No.10855196 Report

Quoted By:

>>10855191
If you're getting an arithmetic mean then somewhere you're probably assuming each leg of the trip takes the same amount of time. Despite involving distances, this is more of a discrete problem so integrals are unnecessary.

Anonymous

Anonymous Thu 01 Aug 2019 18:15:36 No.10855199 Report

Quoted By: >>10855211

>>10855191
Oh, fuck. I see. Speed is absolute value of velocity. Knowing someone's speed is not enough information to solve the problem.

Anonymous

Anonymous Thu 01 Aug 2019 18:15:44 No.10855200 Report

Quoted By:

>>10855110
well, if he rides d km, then he rides for (d/2)/8 + (d/2)/10 = 9d/80 hours. so his average speed is d/(9d/80) = 80/9 km/h

Anonymous

Anonymous Thu 01 Aug 2019 18:21:21 No.10855211 Report

Quoted By: >>10855222

>>10855199
no you moron, you're taking the average speed over distance when the average speed is defined as the total distance over the total time, which is the average taken over time.
think about it this way, i have a function f(0) = 0, f(x) = 1 for all nonzero x. of course the average value of the function is 1. but if you looked at function outputs instead, taking the "average" over values in the range, you'd guess that the average is 1/2 since there is a 0 and a 1. in your case, instead of that, we have a function L(t) which is the position at time t, and you are averaging over values in the range of L, not over the domain t.

Anonymous

Anonymous Thu 01 Aug 2019 18:25:51 No.10855217 Report

Quoted By: >>10855224

>>10855180
It's the exactly same thing, I just wanted to explain why it works. You wouldn't get it from the equation if you didn't already know.

Anonymous

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Anonymous Thu 01 Aug 2019 18:28:19 No.10855222 Report

Quoted By:

>>10855211
I see you're point.
twf when brainlet

Anonymous

Anonymous Thu 01 Aug 2019 18:29:23 No.10855224 Report

Quoted By:

>>10855217
>>10855180
It's the same only expanded by eighty and described in words.

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