The following question and answer is taken from careerbless
A naughty bird is sitting on top of the car. It sees another car approaching it at a distance of 12 km. The speed of the two cars is 60 kmph each. The bird starts flying from the first car and moves towards the second car,reaches the second car and come back to the first car and so on. If the speed at which bird flies is 120kmph then
1)the total distance travelled by the bird before the crash is?
2)the total distance travelled by the bird before it reaches the second car for the second time is?
3)the total number of times that the bird reaches the bonnet of the second car is(theoretically)?
I am clear with the explanation available in the mentioned site for 1 and 2.
For the third part, i.e., the total number of times that the bird reaches the bonnet of the second car is(theoretically), answer given is infinite times with the explanation proving the infinite sequence.
But we know it cannot go on infinite times. It has to be a finite number if we look at it practically. How to explain this?
I am puzzled because the explanation looks convincing (infinite times) whereas we know it cannot be infinite times. Please help in understanding this contradiction.
EDIT: Adding the explanation(below) from careerbless as advised by @DylanSp for clarify and my aim is to understand why it looks as infinite times(theoretically) whereas we know it is finite practically or where my understanding is wrong. (it was a detailed explanation and that is why I have not added initially)
(3) infinite times
As explained for the previous case,
The bird reaches the second car in $\dfrac{12}{180}=\dfrac{1}{15}$ hour for the first time.
In this time, the cars together covers a distance of $\dfrac{1}{15}×120=8$ km and therefore the distance between the cars becomes 12-8=4 km.
The bird reaches back the first car in $\dfrac{4}{180}=\dfrac{1}{45}$ hour.
In this time, the cars together covers a distance of $\dfrac{1}{45}×120=\dfrac{8}{3}$ km and therefore the distance between the cars becomes $4-\dfrac{8}{3}=\dfrac{4}{3}$ km.
Now the bird flies to the second car for the second time. It takes $\dfrac{\left(\dfrac{4}{3}\right)}{180}=\dfrac{1}{135}$ hour for this.
In this time, the cars together covers a distance of $\dfrac{1}{135}×120=\dfrac{8}{ 9}$ km and therefore the distance between the cars becomes $\dfrac{4}{3}-\dfrac{8}{9}=\dfrac{4}{9}$ km.
The bird reaches back the first car in $\dfrac{\left(\dfrac{4}{9}\right)}{180}=\dfrac{1}{405}$ hour.
In this time, the cars together covers a distance of $\dfrac{1}{405}×120=\dfrac{8}{ 27}$ km and therefore the distance between the cars becomes $\dfrac{4}{9}-\dfrac{8}{27}=\dfrac{4}{27}$ km.
Now the bird flies to the second car for the third time. It takes $\dfrac{\left(\dfrac{4}{27}\right)}{180}=\dfrac{1}{1215}$ hour for this.
so on.
Sine this goes on repeatedly, the bird reaches the bonnet of the second car infinite times(theoretically)