There was a race between ship and a raft.
Ship's own speed =
v. Raft's own speed = 0, but when going down stream it moves at speed =p. (flow of the river)Rules of the race:
first go downstream 20km, then turn around and come back to the starting point. (total 40km)Ship completed full route (forth and back) in 7h.
Raft met ship, at 12km mark from the start. (ship already had turned around)What is the speed of the ship and the river flow? (v=?;p=?)
I got this from school book.
So my assumption was that solution should be quite straightforward.
But in my approach I get some cubic equation at one point, which seems too complicated compared to all other exercises in same book.
Is there some trivial solution that I don't see? What's the simplest way to approach it?
My naive approach.
Here is the equation system that I start with.
$ \begin{cases} 20/(v+p) + 20/(v-p) = 7 \\ 20/(v+p) + 8/(v-p) = 12/p \end{cases} $
Express v through p (or vice versa) and then use it in other equation.
1st equation 'simplifies' to $40v - 7v^2 +7p^2=0$.
Solve it for v, and then put that into the 2nd equation.
But expression of v it self is non-trivial and putting that into 2nd equation creates some even more complicated thing.
BTW. the answer is:
v=7, p=3
EDIT:
ok seems that simple path is just starting with 2nd equation..
(thanks @james15c; now I feel silly :) )
2nd equation simplifies to $28vp - 12v^2 = 0$,
which can be rewritten to $v (28p - 12v) = 0$
We can ignore solution $v=0$, so it leaves $28p - 12v = 0$,
in other words $v = 7p/3$
Make substitution into 1st equation $40(7p/3) - 7(7p/3)^2 +7p^2=0$
which simplifies to $p^2 - 3p=0$
So far so good. I would recommend that next you solve the second equation to simplify it. Not just put terms in. The terms should be a lot simpler and a much nicer substitution. I would start off by making the two fractions one, then multiply over the denominator... you do the rest.