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HomePuzzleIs The Common Velocity Actually Appropriate? – Thoughts Your Choices

Is The Common Velocity Actually Appropriate? – Thoughts Your Choices


This downside was posted to Reddit They Did The Math, and surprisingly many individuals had been confused by its wording.

Richard drives from his house to work at a median velocity of fifty miles per hour. Later he drives from his work to his house at a velocity of 60 miles per hour. What’s his common velocity over the 2 journeys?

The poster put 55 mph as the reply, however the appropriate reply was given as 54.5 mph. Is that reply actually appropriate?

As traditional, watch the video for an answer.

Is The Common Velocity Actually Appropriate?

Or maintain studying.
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“All shall be effectively should you use your thoughts to your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport concept and arithmetic. MindYourDecisions now has over 1,000 free articles with no advertisements due to group help! Assist out and get early entry to posts with a pledge on Patreon.

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Reply To Is The Common Velocity Actually Appropriate?

(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll appropriate them, thanks).

Briefly, sure the reply is basically appropriate.

Many individuals took the easy common of speeds as (50 + 60)/2 = 55 mph. However this isn’t correct, because the time spent on the slower velocity of fifty mph is longer than the time spent at 60 mph. The right common velocity accounts for this time distinction, and you want to use the harmonic imply of the speeds. This works out to:

2/(1/50 + 1/60)
= 54.5454…

The reply of 54.5 mph is definitely appropriate (rounded to 1 decimal place).

Here’s a extra in-depth derivation. Let D be the space betweeen work and residential.

(picture)

The common velocity for the journey is given by:

common velocity = (complete distance)/(complete time)

The entire distance is D + D = 2D. The time to go to work is D/50, and the time to return is D/60. The entire time is the sum of the 2 occasions. So the common velocity is:

common velocity
= (complete distance)/(complete time)
= 2D/(D/50 + D/60)
= 2/(1/50 + 1/60)

A pleasant trick is to multiply the numerator and denominator by 50(60), as a result of this eliminates the fractions within the denominator as 50(60)/50 = 60 and 50(60)/60 = 50. The expression then simplifies to:

2(50)(60)/(60 + 50)
= 6000/110
= 60/11
= 54.5454…

So certainly the common velocity is about 54.5 mph.

Discover for speeds of fifty and 60 we discovered the common velocity was:

2/(1/50 + 1/60)

If the speeds are generalized as a and b, the common velocity when touring two equal distances is then:

2/(1/a + 1/b)

Einstein riddle

The shocking nature of common speeds intrigued even the good physicist Albert Einstein.

The psychologist Max Wertheimer was a buddy of Albert Einstein. In 1934, he wrote a letter to Einstein with the next mind teaser.

Is The Common Velocity Actually Appropriate? – Thoughts Your ChoicesIs The Common Velocity Actually Appropriate? – Thoughts Your Choices

An outdated automotive must go up and down a hill. Within the first mile–the ascent–the automotive can solely common 15 miles per hour (mph). The automotive then goes 1 mile down the hill. How briskly should the automotive go down the hill with the intention to common 30 mph for your complete 2 mile journey?

Let’s use the formulation for common velocity we simply derived.

common velocity = 2/(1/a + 1/b)

We’d like the common to be 30, and let the ascent velocity be a = 15 and the unknown descent velocity be b.

30 = 2/(1/15 + 1/b)
30 = 2/(1/15 + 1/b) (15b/15b)
30 = 30b/(b + 15)
30(b + 15) = 30b
30b + 450 = 30b
450 = 0

This can be a contradiction, so this equation has no resolution! What’s happening? Begin from the start equation:

2/(1/15 + 1/b)
= 30b/(b + 15)

The restrict as b goes to infinity is 30, so we will see we want an impossibly infinitely quick velocity to common 30 mph for your complete journey.

One other method to see why that is inconceivable is to backwards. In an effort to common 30 mph, how lengthy should the automotive take for your complete journey? That is:

time journey = (2 miles)/(30 mph) = 1/15 hour = 4 minutes

How lengthy does the automotive take to go up the hill?

time going up = (1 mile)/(15 mph) = 1/15 hour = 4 minutes

So we attain an issue. The automotive wants to finish your complete journey in 4 minutes, however it has already used that point going up the hill. In an effort to common 30 mph for your complete journey, it has to take 0 minutes going downhill. That may imply the automotive travels 1 mile instantaneously–which isn’t attainable! The automotive must be infinitely quick, however we all know that no object can go sooner than the velocity of sunshine.

Thus this can be a trick query–it isn’t attainable, irrespective of how briskly the automotive goes, for it to common 30 mph for your complete journey.

Einstein himself was tricked at first, writing, “not till calculating did I discover there isn’t a time left for the way in which down!”

I actually like this downside as a result of it’s straightforward to grasp, and but it’s difficult sufficient that even Einstein wanted to work it out fastidiously. Averaging speeds is a tough idea!

Reddit They Did The Math
https://www.reddit.com/r/theydidthemath/feedback/1kdh0gi/request_is_this_really_the_correct_answer/

Einstein
Gerd Gigerenzer e-book “Danger Savvy: The right way to Make Good Choices” tells the story of the mind teaser.
H/T Farnham Road weblog
https://fs.weblog/2014/06/einstein-wertheimer-car-problem/

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