Re puzzle #2: An practical solution with minimum math is to accept that extreme accuracy to 3 places (see my post #3 above) is probably not overly important for the dipstick markings so most of the more complex circular geometric calculations can be bypassed. The tractor-trailer driver should know the tank capacity from the owner's manual. If not, the example tank volume is pi x 0.25m^2 x 1m = 0.19635 cubic meters = 196.35 liters. Round this to 196 liters to simplify, if desired convert to U.S. or Imperial gallons.
The driver places the half-full dipstick mark at 25 cm and drives the truck until the fuel level reads half or less. He drives to the truck stop and adds just enough fuel to measure 1/2 tank on his dipstick. Then he notes the pump reading and adds exactly 49 litres or 1/4 tank more fuel according to the pump reading. The fuel level is now 3/4 tank and can be marked on the dipstick, and the 1/4 tank mark will be the same distance from center towards the bottom end of his dipstick.
Sometimes the practical approach gives results good enough for the task at hand. In this case the fuel pump itself is used as the measuring instrument to calibrate the dipstick.

MY DISCUSSION: The book's editor and some high school math teachers might believe that answer, but it is a naive interpretation of the problem, and real engineers might beg to differ. Contractors, and in particular wallpaper and carpet installers, are a lot more clever than that. The contractor would most likely place a large number of minute undulations in the ribbon, such that considerable length could be taken up without the undulations having enough amplitude to be noticed by the job inspectors. I'll bet he would get the bonus. The real engineering puzzler is then to design a plan of such tiny undulations, such as a sinusoidal ripple. Try it, and post your answer!
When I saw the puzzler in the book as a kid, it angered me, because the simplistic answer given seriously discouraged deep thinking and creativity, which after all is what we are hopefully learning in school. Instead, it promoted overly simplified smartass answers to complex problems (maybe it was intended to groom kids for political careers?) That's probably the reason I still remember it after all these years. The puzzle as presented here, by stipulating that the string be levitated over the surface, removes ambiguity from the answer, and still teaches the geometry lession, but also doesn't encourage creative thinking.

Problem #1 is a rephrasing of a slightly different statement of this puzzler that appeared in a book of science experiments and teasers that was popular decades ago when I was in high school. But the way they posed the problem then leads to a more interesting discussion:
BOOK PROBLEM: A contractor is hired to fabricate a very thin ribbon to fit around the equator of a perfectly spherical planet exactly 20 km in diameter. He is to get a bonus if he fabricates the ribbon to exactly the correct length. When the contractor installs the ribbon, he finds it is a meter too long. He remains confident that he will receive the bonus, because he believes the extra 1 meter will be easily taken up in slack along the over 60 km circumference of the planet, without anyone noticing. Does the contractor get the bonus?
BOOK ANSWER: No! To take up the 1 m slack, the ribbon would have to be held over 6 inches (15.915... cm) above the surface of the planet, all around, regardless of the size of the planet. (Curiously, the book mixed imperial and metric units, just this way.)
But is this right???

It can be done with or without calculus, but I'm pretty sure it can't be done without a numerical solver. This is because you need to solve for an unknown that is both inside and outside an arcsin function in the final equation.

Ooooh, that's clever -- I thought you were wrong but I just tried working it out for 100 hours (99 for the second car) and 1,000 hours (999 for the second car) and I see what you mean -- Max

If both cars left at the same time, the second will always have traveled twice as far as the first since it is moving twice as fast. Since in this problem it leaves an hour later, it will never have travled twice as far as the first car, but the ratio of distances will approach two asymptotically.

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