A contradictory thought process of a simple problem

Question

Lets say there are 2 point particles A and B separated by a distance $x_{0}$ whose velocities are $v_{1}$ and $v_{2}$ along the positive X-axis respectively such that $v_{1}>v_{2}$

We can clearly see that at some time, A will over take B.

Now we consider the exact scenario in a different way.

A covers $x_{0}$ in $t_{0}s$. In the $t_{0}s$, B covers a distance $x_{1}$. A covers $x_{1}$ in $t_{1}s$ and so on. Now every time A reaches B's old position, B has already moved forward from there.

After a while, $x_{n}$ is very small but still not 0. Then how exactly does A overtake B?

Answer 1

The time taken to overtake by the first method is obviously

$$\frac{x_o}{v_1-v_2}$$

By the second method, the time taken is given by the infinite series: $$\frac{x_o}{v_1} \left(1 + \frac{v_2}{v_1}+ \frac{v_2^2}{v_1^2} ... \right ) = \frac{x_o}{v_1 (1-v_2/v_1)}= \frac{x_o}{v_1-v_2} $$

Where is the paradox? An infinite series can give a finite answer.

Answer 2

If A is faster than B, we know that eventually, A will overtake B. x_n will indeed never be 0, but it is because we are only considering things before A overtakes B. Suppose A's speed is twice that of B and it took a time t₀ for A to come to initial position of B. For next x₂ distance, it will take t₀/2, for x₃ time interval will be t₀/4 and so on. After a while, x_n will be low, but the time interval taken by A to cover x_n will also be very low. The sum of time interval will be: t₀(1 + 1/2 + 1/4 + 1/8 ....) and so on till infinite terms. This sum is actually finite(Its sum of infinite GP, you can look it up) and total time taken will be 2t₀. The problem in your approach was that you were considering only distance and not time. Therefore you never reached the instant 2t₀, where A overtakes B.