Ytdyklly

A car is moving at a constant speed 40 km/h along a straight road which heads towards a wall.A fly flying at a constant speed of 100 km/h, starts from wall the towards the car at a instant when the car is 20 km away, flies until it reaches the car and comes back to the wall at the same speed.It continues to fly between the car and the wall till the car reach the wall. How many trips has it made between the car and the wall?

I don't expect a brute force because I already did that. Some thing like arithmetic/geometric/harmonic progression will satisfy my curiosity.

Suppose the fly is at the wall. And suppose this will be the last trip or partial trip of the fly. Suppose the car is $h$ miles away.

The fly and and car have a combined speed of $140 frac {km}{hr}$ so the fly reaches the car in $frac h{140}$ hours. In that time the car has traveled $40frac h{140} = frac 27h$ and is now $h-frac 27h = frac 57h$ from the wall. So the fly heads back to the wall.

As the trip back is just as far this takes $frac h{140}$ hours and the car has traveled another $frac 27h$ and is now $frac 37h$ from the wall. [1]

So the fly starts another trip, contradicting that this was his last. So the fly never makes a last trip an instead there are an infinite number of trips.

Figuring out how far the fly flies is a matter of noting the car is on a straight path and travels $20km$ at $40 kmh$ so this takes $30$ minutes. The fly no matter how many times (infinitely many) it zigs will travel at $100kmh$. So in $30$ minutes it flies $50 km$.

If one wishes to set this up as an infinite sum.....

Each trip the fly flies $frac {10}7$ of the distance the car was away. And each trip the car is $frac {3}{7}$ of the distance it was before. So the distance the fly travels is $sum_{k=0}^{infty} frac {10}7*(frac {3}{7})^k*20$ which if I did this correctly is

$frac {10}7*20(sum_{k=0}^{infty} (frac {3}{7})^k)= frac {200}{7}frac 1{1-frac {3}{7}} = frac {200}{7}frac {7}{4}= 50$km.

[1](for the record, in this time, $frac 1{70}h$ hours, the car has traveled $frac 47h$ and the fly has travelled $frac {10}7h$).

It has made infinitely many trips. Every trip will be shorter than the last, but the fly will always reach the wall before the car, so it will always have room for one more trip. And one more. And one more.

The total length the fly flies is 50km, as the car crashes into the wall exactly 30 minutes after the whole experiment started.

Simple way to calculate it: It takes the car 30 minutes to travel the 20km to the wall at 40km/hr. The fly, traveling at 100km/hr will travel 50km in those same 30 minutes.

Edit: There will be an infinite number of trips. It's similar to how Zeno's paradox works where the trips get shorter and shorter and eventually take an infinitely small amount of time. But all those infinitely small trips end up as a finite amount of distance.

Arthur's answer is very good, but here is another (equally valid) way of visualizing the problem:

Let us graph the positions of the wall, fly, and car over time. The wall doesn't move, so it is represented by a horizontal line. The car starts at some distance away from the wall but moves towards it at a constant speed until it hits the wall -- so we have a line that intersects the wall's line. Now for the interesting part:

The fly's path starts off as a line with greater slope than the car's line, until it hits the wall's line. The fly's speed remains the same, but it is going in the opposite direction. So now the path continues as if the wall's line was a mirror and it was reflected. When the fly hits the car, the same thing happens -- the fly's path is reflected and it continues towards the wall again (one caveat: when it bounces off the car's line, the angles of incidence and reflection are not equal so it isn't behaving exactly as light would).

So we can see that the fly's path continues bouncing up and down, always at the same or opposite slope.

The final thing to notice to grasp the intuition is that this diagram we have constructed is self-similar. If we zoom in so that the second bounce with the wall is where the first bounce used to be, we have the same exact diagram as before. I won't prove this, but if you draw it out, you can see it intuitively. Essentially, this means that no matter how close the car gets to the wall, we can zoom in and see more bounces for the fly.