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I saw an interesting question during the review, so I wrote down the thinking process.
There is a point-to-point link with a length of $20000\ km$, the rate of data transmission is $1\ kbit/s$, and the data to be sent is $100\ bit$. Data travels at $2\times10^8\ m/s$ on this link. Assuming we can see the bits transmitted on the wire, try to draw the bits on the wire as we see them (draw two graphs, one when $100\ bit$ has just been sent, and one after $0.05\ s$ has passed).
First we consider the time to send $100\bit$. It can be calculated according to the formula,
$$
t = \frac{100\ bit}{1\ kbit/s} = 0.1\ s
$$
If the bit is imagined as a “particle” moving in the link, then after $0.1\ s$, the last particle of this bunch of bits has been sent out of the device and is located at the beginning of the link. It is pointed out in the title that the propagation speed of the data on this link is $2×10^8\ m/s$ , then the distance that the first particle sent at the beginning has moved is
$$
s_1 = 0.1\ s \times 2\times10^8\ m/s = 2\times10^7\ m=2000\ km
$$
It just so happens that the length of the link is also $20000\ km$, that is, the bit particles sent out continuously, the first one is at the end of the link, and the last one is at the beginning of the link. Then the link is full of bit particles at this time.
In the link, the speed at which bit particles move is determined by the propagation speed. Then after $0.05\ s$, the distance traveled by the last particle is
$$
s_2 = 0.05\ s \times 2\times10^8\ m/s = 2\times10^7\ m=1000\ km
$$
It is half of the link length $20000\ km$. At this time, the last particle is located in the center of the link, and half of the previous particles have left the end of the link one after another. Then we can draw a rough picture, which means when $100\ bit$ has just been sent and after $0.05\ s$.
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