It is true that there are infinitely many solutions (or zero
solutions) (x, y, z) in any case.
But what we are interested here is S=x+y+z.
Apply y = S-x-z, you get:
S/Vp+x(1/Vd-1/Vp)+z/(1/Vu-1/Vp) = T1
S/Vp+x(1/Vu-1/Vp)+z/(1/Vd-1/Vp) = T2
Adding the two, you get
2S/Vp +
You could also get a unique solution if the car has speed of 72 63 56
in downhill, plain and uphill respectively.
I think the speed Vd, Vp, Vu was chosen so that 2Vp = Vd + Vu.
But for unique solution, it ought to be 2/Vp = 1/Vd + 1/Vu.
Under this condition, we can get the unique S=x+y+z:
the solution seems to be simple.
Just try to imagine what is happening
You have a road with downhill and uphill.
So if u travel 5 km uphill and then 5 km on plain and then 5 km on downhill
then time taken
by you will be equal to 15 km on the plain road(that is solely due avg of
speed of downhill
So if u travel 5 km uphill and then 5 km on plain and then 5 km on
downhill then time taken
by you will be equal to 15 km on the plain road
This is not the truth.
5/72 + 5/64 + 5/56 - 15/64 = 5/72+5/56-10/64 = 10/63-10/64 0
(that is solely due avg of speed of downhill and uphill is = speed
No. Two linear equations in three unknowns will always yield many
solutions (or zero solutions). These are essentially plane
equations. Two planes intersect in a line (unless they are
parallel). You might get a de facto unique solution for some values
of Vu, Vd, Vu, T1, T2 from the constraints
Actually the solution is unique. The middle part with the Ys is the
same and therefore can be omitted out. Now you are left with
2 equations and 2 unknowns.
I used time in minutes and I have x = 1.28, z = 0.30476 units (y can
be found out).
I guess the trick was 1. to write the equations that
This isn't right. Dropping both y terms is the same as setting y to
zero. The answer you get is correct, but there are many others as has
been said.
You could get a unique solution if the route were constrained to be
monotonic (level and up or else level and down).
On Sep 14, 4:28 pm,
