Here is a problem I ran into in "Martin Gardner's New Mathematical Diversions from Scientific American". It is simple enough that you can describe it while talking with friends at a break point or during a drive...
A group of cadets is marching in a 50m x 50m square formation. Their pet dog starts from the middle of the back (A in figure below), runs to B at the front, turns back instantaneously, and runs to A in the rear again. In the meantime, the formation has moved forward by 50m. Assuming the dog runs at constant speed, how far did it travel?
My first guess was that the dog runs 50m + 50m forward, and then 50m back, so 150m. But of course this is an overstimate, since actually, it's turning back from some point well before 100m ahead.
(Read on only if you have tried it now...)
If the dog's speed is v m/s and the formations is u m/s, then while running forward, the dog's speed is v-u with respect to the formation, so it runs forward for 50/v-u seconds. Then it runs back for 50/v+u seconds. In the meanwhile, the formation has taken 50/u seconds to march 50m.
SO - the sum of these two times is 50/u. You can now eliminate 50 and and u and solve for v/u, which will give you the answer since 50/u . v is the distance he runs.
So - what do you think - will it come down to a quadratic equation?
Let me know what you get!