The answer is 10 yards

[G-man]

Note that 30 mph is equivalent to 44/3 yds per second. Note also
that, since C went from 50 mph to 30 mph in the same time as B went
from 40 mph to 30 mph, C is deceleration at twice the rate of B.

One approach to the solution is to determine

(1) How far B and C traveled from the time they nearly crashed. The
key is that the distance traveled by each is their average speed times
the time they traveled. This is a function of the acceleration of
each. If B's velocity after the near collision (t = 0) is 44/3 - gt,
then C's velocity is 44/3 - 2gt. B's velocity will be 0 when t =
44/3g and C's will be 0 when t = 22/3g. Their average velocities are
simply half of their velocities at t = 0. The difference between the
distances they traveled is 45 yards. From this, you can determine g.

(2) Similarly, you can start over with a new t = 0 when they start
breaking and determine how far they travel in the time it takes them
to reduce their speeds to 30 mph = 44/3 yards per second. These
distances are also functions of g, which you can detemine from (1).
The difference in these distances turns out to be 10 yards, the answer
to the teaser. Details available upon request: dgitt...@alamo.edu