# Time to speed up [closed]

Mr. X lives in A-City and wants to visit B-city. There are 1000 km between these 2 cities.

So, Mr. X takes his car and drives to go to B-City. At halfway of his ride (500 km), he realises that his average speed has been 100 km/h, from A-City to now.

What is the average speed that he has to drive from now to B-City, in order to have an average speed of 200 km/h on the full ride (from A-City to B-City) ?

• it is definitely a math question. – manshu Jun 9 '16 at 10:12
• This is totally a math question. -_- – Gaurav Agarwal Jun 9 '16 at 17:48

## 4 Answers

Needed speed for 2nd half:

infinity km/hour

Because:

200 km/hour for 1,000km trip = 5 hours which he has already spent driving!

5 hours driven at 100km per hour = 500km or halfway there. He has already driven 5 hours at the halfway point. To average 200km for the trip he needs to be at city B in 5 hours from the start which allows no more time for the 2nd half of the trip

• You should hide your answer, using >! at the beginning of the line. – user21939 Jun 9 '16 at 10:13
• Well, I knew it was easy, but only 2 minutes... :'( Well done anyway, give you the accepted mark when possible :) – Mayo Jun 9 '16 at 10:15

He should

teleport himself to B

Because

It took him 5 hours to drive the first half and if he wants an average speed of 200km/h he must drive the full ride in 5 hours

• Hey, I wanted to ask you a question since you have a lot of reputation and as it is my first puzzle : if someone posts the good answer but with no explanation, then someone posts the good answer with the explanation, and then the first one edits their answer with the explanation, which one of both am I supposed to give the "accepted" mark ? – Mayo Jun 9 '16 at 10:20
• @Menace In this case Greg was first and he explained his answer only a few seconds later so (I think) you should accept him. – Fabich Jun 9 '16 at 10:26
• @Menace great meta question! HERE is a meta.puzzling discussion on it. I'm sure there are more discussions about this topic there... – Ben Jun 9 '16 at 10:51
• @Menace This seems like a friendly community, so any decision you make is fine. I am brand new here and trying to figure out all the guidelines myself – Greg Hastings Jun 9 '16 at 10:55
• So I finally accepted Greg, thank you for all your answers, and thank you Lord of dark for your understanding, because you deserved the accepted mark too :) – Mayo Jun 9 '16 at 12:13

An alternative approach:

If Mr. X is allowed to turn around, and travel $k$ km back towards A-city, then turn around again and drive to B-city, it is possible, and there are an infinity of answers.

The distance to travel is $500+2k$ km which takes time $t$ hours. The total distance travelled will be $1000+2k$ km, taking $\dfrac{1000+2k}{200}=5+\dfrac{k}{100}$ hours at $200$km/h. Therefore Mr.X must travel the $500+2k$ km in $\dfrac{k}{100}$ hours to average a speed of $200$km/h for the whole journey.

For example, let $k=100$. If Mr.X covers the $700$km he sets himself in $1$ hour, he will have covered $1200$km in $6$ hours, averaging $200$km/h. His speed in this case is $700$km/h.

The general formula is $\dfrac{100(500+2k)}{k}$ km/h.

• Nice idea, +1 ! – Mayo Jun 9 '16 at 12:15

What is average? The average of a function $f:X \rightarrow Y$ over the interval $I$ is the integral of $f\mathbb{d}\mu$ over $I$, where $\mu$ is a measure on $Y$ and $X$ is a vector space over $\mathbb{R}$ or $\mathbb{C}$.

The naive solution to your problem is to describe the drivers speed as a function of time. So we have $f:time \rightarrow speed$ and the interval $I$ is the time from leaving the first city A to reaching city B.

This requires either infinite velocity (teleportation), or wasting a bunch of time and not going directly to city B. This relies on the fact that we are mapping to $speed$ and not $velocity$, for which going backwards or sideways doesn't help.

The traveller takes a right, and travels at 300 km an hour for 5 hours, then 300 km per hour back. Then head towards the city at 100 km/h.

The traveler goes 1000 km + 1500 km * 2 = 4000 km, and takes 5+5+5+5=20 hours,

thus going at an average of 200 km/h.

However, there is another way to approach it.

Instead of using the function $f:time \rightarrow speed$, use $f:Distance To City B -> speed$ and find its average. The result is still average speed, so it answers the question. So long as you approach monotonically, this is well defined.

In this case, simply go at 300 km/h.