High-schooler here trying to work through an interesting problem:
A boy races an ant, giving the latter a head start. The motion of the two bodies can be decomposed according to the following steps:
Step #1: The boy runs to the ant's starting point while the ant crawls forward.
Step #2: The boy advances to where the ant was at the end of Step #1 while the ant goes yet further.
Step #3: The boy advances to where the ant was at the end of Step #2 while the ant goes yet further.
Step #4: The boy advances to where the ant was at the end of Step #3 while the ant goes yet further.
And so on and so forth.
Given that the boy is running at 8 m/s, the ant is crawling at 0.08 m/s and the latter has an 80-metre head start, calculate the time taken for the boy to overtake the ant.
One may deduce that this lends itself to a geometric series. Since the boy does overtake the ant, the series must be convergent. How do we go about calculating the common ratio r and solving the rest of the problem?
The relative velocity (in m/s) between the boy and the ant is ... $$v_r = 8 - 0.08 = 8\times \frac{99}{100} $$
So the time for the boy to catch up to the ant is ... $$T = \frac{80}{v_r} =\frac{1000}{99}=10.\dot 1\dot0s$$
We can use geometric series to show the the sequence of times converges to the same value. $$t_1 = \frac{80}8 = 10s$$ at which time the ant has now moved on by $0.8$m
so if $t_2$ represents the time between steps #1 and #2... $$t_2 = \frac{0.8}8 = 0.1s$$
the common ratio for this sequence is $$r = \frac{t_2}{t_1}= 0.01$$ The total time can now be calculated using the formula for geometric series...
$$T = \sum_{n=1}^{\infty}t_n= t_1\times \frac 1{1-r}= \frac{10s}{0.99}= 10.\dot 1\dot0s$$