Here's an analogy from Computer Networking: A Top-Down Approach (6th ed.), written by James F. Kurose and Keith W. Ross, found on page 38:
Consider a highway that has a tollbooth every 100 kilometers.
You can think of the highway segments between
tollbooths as links and the toll- booths as routers. Suppose that cars
travel (that is, propagate) on the highway at a rate of 100 km/hour
(that is, when a car leaves a tollbooth, it instantaneously acceleratesrates
to 100 km/hour and maintains that speed between tollbooths).
Suppose next that 10 cars, traveling together as a caravan, follow
each other in a fixed order. You can think of each car as a bit and
the caravan as a packet. Also suppose that each tollbooth services
(that is, transmits) a car at a rate of one car per 12 seconds, and
that it is late at night so that the caravan’s cars are the only cars
on the highway. Finally, suppose that whenever the first car of the
caravan arrives at a tollbooth, it waits at the entrance until the
other nine cars have arrived and lined up behind it. (Thus the entire
caravan must be stored at the tollbooth before it can begin to be
forwarded.) The time required for the tollbooth to push the entire
caravan onto the highway is (10 cars)/(5 cars/minute) = 2 minutes.
This time is analogous to the transmission delay in a router. The time
required for a car to travel from the exit of one tollbooth to the
next tollbooth is 100 km/(100 km/hour) = 1 hour. This time is
analogous to propagation delay. Therefore, the time from when the
caravan is stored in front of a tollbooth until the caravan is stored
in front of the next tollbooth is the sum of transmission delay and
propagation delay—in this example, 62 minutes.