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Re: Solution to a problem!!



Note that the water velocity 4km/hr is "close" to the boat's water speed
(6km/hr). Because of this:
He heads UPSTREAM at an angle T (straight across would be T=0)
in t1 hrs he lands a distance x DOWNSTREAM. He then sprints the distance
x in t2 hrs. This gives:

1) 1 = 6*t1*cos(T)
2) x = 4*t1 - 6*t1*sin(T)
3) x = 10*t2

Note these equations are valid only for positive x , ie: for sin(T) </=
2/3. The equations for larger T (or for a downstream angle) do not
generate a minimum time.

Equate (2) and (3). From the result and (1) construct:

t1+t2 = 7/(30*cos(T)) - tan(T)/10 ; from d(t1+t2)/dT = 0 => sin(T)
= 3/7 => T=.443rad=25.38 deg, giving t1 = .184hr ; t2 = .026hr ;
t1 +t2 = .21 hr = 12.6 minutes.

This was done in the wee hours; there may be errors! A worthy problem,
complete with pitfalls!

-Bob

Bob Sciamanda
Physics, Edinboro Univ of PA (ret)
trebor@velocity.net
http://www.velocity.net/~trebor

-----Original Message-----
From: James Harris <jharris@top.monad.net>
To: phys-l@atlantis.uwf.edu <phys-l@atlantis.uwf.edu>
Date: Tuesday, October 27, 1998 8:09 PM
Subject: Solution to a problem!!



Dear Physics People,

I gave the following problem as an extra credit exercise on a test for
my
Honors physics.
What would you say the solution was? I believe it is from Tipler
Physics. I
don't have an answer key for the book so I am not sure. A couple of kids
got
into a pretty good debate (until the soccer coach arrived and chewed
them
out) over what the solution is or was.
Jim
jharris@monad.net or jharris@newpisgah.keene.edu
Teacher: Monadnock Regional High School
Adjunct Faculty: Keene State College, Chemistry Department

EXTRA CREDIT Jack/Jill is strolling along the bank of a river 1 km
wide
when the most beautiful/handsome girl/boy he/she has ever seen
materializes
on the shore directly opposite him/her (perhaps she/he was beamed
down?).
Fearing that she/he will disappear before he/she has a chance to
establish
face-to-face communication, he/she quickly devises a plan to reach the
opposite shore in the shortest possible time. In the wildest of
coincidences, there is a rowboat beached on the shore right in front of
him/her. Jack/Jill is ecstatic, because it so happens that he/she's an
expert oarsperson. (He/she rows on the crew for an ivy league school.)
He/she knows that he/she can row at a speed of 6 km/h in still water,
and
he/she estimates--as he/she sprints for the boat at 10 km/h--that the
river
current has a speed of 4 km/h. Now, besides being an athlete and an
excellent judge of river velocities, Jack/Jill is also an accomplished
physics student. During his/her sprint to the boat, he/she computes the
path
he/she must take from his/her side of the river to reach the girl/boy on
the
opposite side in the shortest possible time. In general, his/her path
includes a diagonal trip across the river followed by a sprint along the
opposite shore to reach his/her goal. (Note that Jack/Jill has a
standard
sprinting speed of 10 km/h.) Assuming that Jack/Jill did the physics
correctly, in what direction did he/she head the boat and how long did
it
take him/her to establish first contact?