SOLUTION: hi there, I asked the following question before and I had a solution given to me, I do not understand some of it though, some guidance would be great. s=u.(v-u)/a + 1/2a((v-u)/a

Algebra ->  Expressions -> SOLUTION: hi there, I asked the following question before and I had a solution given to me, I do not understand some of it though, some guidance would be great. s=u.(v-u)/a + 1/2a((v-u)/a      Log On


   

Question 93260: hi there,
I asked the following question before and I had a solution given to me, I do not understand some of it though, some guidance would be great.
s=u.(v-u)/a + 1/2a((v-u)/a)^2
it should end up as v^2=u^2+2as
first step was to multiply by 2a giving....
2as=2u(v-u)+(v-u)^2
expanding
2as= 2uv-2u^2+v^2-2vu+u^2
apparently by collecting terms this is equal to 2as=v^2-u^2, could someone please explain this last step for me
Found 2 solutions by stanbon, bucky:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2as= 2uv-2u^2+v^2-2vu+u^2
apparantly by collecting terms this is equal to 2as=v^2-u^2, could someone please explain this last step for me
---------------
2uv-2u^2+v^2-2vu+u^2
=2u(v-u) + [v^2-2uv+u^2]
= 2u(v-u) + [v-u]^2
=(v-u)[2u + v-u]
= (v-u)(v+u)
= v^2-u^2
=================
Cheers,
Stan H.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
You were given:
.
s=u%28v-u%29%2Fa+%2B+%281%2F2%29%2Aa%2A%28%28v-u%29%2Fa%29%5E2%29
.
You need to check the above and make sure this is the problem that you actually meant to
have explained to you because that is the one that was worked out in the explanation you
got.
.
First you were told to multiply both sides (all terms) by 2a. The purpose of this multiplication
was to eventually get rid of the denominators on the right side. It was performed as follows:
.
2a%2As+=+2a%2Au%28v-u%29%2Fa+%2B+%28%282a%2A1%29%2F2%29%2Aa%2A%28%28v-u%29%2Fa%29%5E2%29
.
On the right side in the second term multiply the "a" that is in the middle times the
numerator of the first factor 2a%2A1 which results in 2a%5E2 and makes
the problem
become:
.
2a%2As+=+2a%2Au%28v-u%29%2Fa+%2B+%28%282a%5E2%2A1%29%2F2%29%2A%28%28v-u%29%2Fa%29%5E2%29
.
Now in the second term on the right side square out the last factor of %28%28v-u%29%2Fa%29%5E2%29
by squaring both its numerator and denominator as follows:
.

.
You can now substitute the right side of this result in place of %28%28v-u%29%2Fa%29%5E2%29
in the problem, and the problem then becomes:
.

.
At this point you can cancel the factors of 2a in the numerators with corresponding
factors of 2a in the denominator as follows:
.

.
and you are left with:
.
2a%2As+=+2u%28v-u%29%2Bv%5E2+-+2uv%2Bu%5E2
.
Doing the distributed multiplication of the first term on the right side results in:
.
2a%2As+=+2uv+-+2u%5E2+%2Bv%5E2+-+2uv%2Bu%5E2
.
Notice that on the right side you have terms that can be combined. The 2uv and the -2uv
cancel each other and therefore, they disappear to leave you with:
.
2a%2As+=+-2u%5E2+%2B+v%5E2+%2B+u%5E2
.
Then the -2u^2 and the +u^2 combine to result in -u^2 and as a result the equation
becomes:
.
2a%2As+=+v%5E2+-+u%5E2+
.
and that's the result you are looking for. Hope this helps.