Velocity shrinks. What's the longest distance ?

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Homework Statement


A man walks with velocity 12 km/h during the first hour. At the second hour, the velocity decreases to 1/3 of it, and it goes on for next hours. What's the longest distance the man travelled?

2. The attempt at a solution

I think that the function of velocity with respect to time is y = ##12\cdot \left(\frac{1}{3}\right)^x##
Then, I integrate it, so it becomes ##Y = -\frac{4\cdot 3^{\left(1-x\right)}}{\ln \left(3\right)}##
The distance is maximum when the velocity is zero.
But, the velocity never gets zero.

The answer in the book just treat the problem as an infinite sequence problem.
It just use S=a/(1-r)=12/[1-(2/3)]=12/(2/3)=12(3/2)=18 km

But, it makes me confused.
Why does it treat it as infinite sequence? The distance unit is km, and velocity unit is km/h..
How come the distance become the sum of velocities?
Is the answer in my book wrong?
I think it's impossible since there must be a change in a graph, and we need to integrate it.
I think the book is just treating the exact hour, so it's not a graph, but not-connected dots.
So, what's the answer?
 
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For the question as posed, the velocity versus time graph is a step function. It is level for the first hour at 12 km/h and level for the second hour at 4 km/h and so on. Yes, this involves infinite acceleration once every hour. But you are expected to ignore that and concentrate on the distance travelled. Can you see that (reading the problem this way) that the distance traveled in the first hour is 12 km?

Your solution of fitting a smooth exponential to the velocity at the beginning of each hour is nicer because it eliminates the need for the infinite accelerations. Good job! Unfortunately, it does not fit the intent of the problem.

One clue that I see to the intended interpretation is "the man walks with velocity 12 km/h during the first hour". The word "during" suggests that the velocity applies throughout the hour. An alternate reading could be that the instantaneous velocity was 12 km/h at some time during the first hour. But that interpretation eliminates the possibility of a definite answer. Another clue is "at the second hour the velocity decreases to 1/3 of it". That suggests an instantaneous change in velocity.

A [pedantic] mathematician might point out that the distance traveled never attains its upper bound and that, accordingly, there is no "longest distance travelled". But in a physics class, that fine point would not be probed, so that cannot be the intended response.
 
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The answer is, it is a geometric series. Nowhere does it say his speed slowly decreases. I always used to get caught out with stuff like this. I remember one: "six poles are standing in a circle". I assumed there were somewhere inside the circle but they were on the boundary.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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