You're in a car going 100KPH and your destination is 100KM

>>9137911
about 5

>>9137912
Close enough to answer the question, not close enough for copying homework. This is a good answer.

>>9137911
That is [math]\sum_{i=1}^{100} 1/i[/math] hours. According to wolfram, that's [math]\frac{14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}[/math] hours, or about 5 hours, 11 minutes, and 14 seconds.

>>9137918
it's not homework, i thought of it when I was driving today, and I thought it was an interesting question but I suck at math.

explain your answer so I can learn something

>>9137924
And this is the bad answer.

>>9137925
>explain your answer so I can learn something
Okay!

During the first kilometer, you are driving 100 km/h. That means it takes you 1/100 hours to travel that kilometer.

During the second kilometer, you are driving 99 km/h, meaning it takes you 1/99 hours for the second kilometer.

The last kilometer at 1 km/h takes you one hour, of course.

Completing the pattern gives you 1/100 + 1/99 + 1/98 + ... + 1/3 + 1/2 + 1/1 hours, or [math]\sum_{i=1}^{100} 1/i[/math] in mathematical notation. Which you can compute to a specific number.

>>9137911
100 kilomeaters

>>9137911
Sorry bro I'm not a communist I only read imperial units

What happenes if the decrease in speed is a continous function?

Id est for example, if the decrease from 100kmh to 99kmh is parametrized to a first order polynomial?

>>9138496
Then the problem involved an integral sign instead of a capital sigma.