|
Um.. im getting extra credit for the following question, if answered would be lovely 
So what my teacher is asking, is the process between one equation to another
(Final Velocity ^ 2) = (Initial Velocity ^ 2) + 2a x Delta X
a = Acceleration
Thats the equation that I found, and these are the possible equations that could be the answer
DeltaX = (Initial Velocity x Delta Time) + (1/2a Delta Time squared)
Hard to write it on the post forum like this..
But if answered, i will give you a fat e-hug
post if i didnt explain anything throughly
:edit: thanks for the problem
I need help finding the PROCESS
|
Calgary25989 Posts
Your original equation makes no sense unless 0 = 2a dX
|
your 1st equation is wrong
it's vf^2 = vi^2 + 2a(x)
vf = final velocity
vi = initial velocity
x= displacement
|
Big Five equations of motion, given constant a, to=0, xo=0
assumptions:
(vf+vo)/2=vavg=(xf-xo)/t
(vf-vo)/t=aavg
1. vavgt=x
2. vf=vo+at
3. vf^2=vo^2+2ax
4. x=vot+.5at^2
5. x=vft-.5at^2
|
let's see here, you can do it w/ calculus or just algebra. i'll show you the algebra way... if i can remember it correctly. this might be wrong, lol
Vf^2 = Vi^2 * 2a(x)
x = 1/2(Vi + V)([V-Vi]/a)
if t = (V-Vi)\a then x = 1/2(Vi + V)([V-Vi]/a) = 1/2(Vi + Vi + V)t
if V = at then x= 1/2(Vi + Vi + at)t
x = Vi*t + 1/2a * t^2
ok i messed up somewhere, probably between 3-5... not sure exactly
|
Could you guys write the entire procedure in the way that I did it, sort of
Doghunter, i dont really understand that one 
|
uhh for mine
Vf = final velocity (i usually use just V as final velocity, so i think i put that in there a few times)
Vi = Initial velocity
a = acceleration
x = displacement (delta x)
t = time (delta t)
i'm not gonna go retype that the way you did it though t.t
edit: not sure if this is what you meant, but... yeah
|
United States24745 Posts
I honestly can't tell what you are asking. I'm assuming you want to know how to derive the three key kinematics equations using algebra.
Acceleration is defined as the change of velocity with respect to time, so a= delta(v) / t
1) Vf = Vi + at
But (Vf+Vi)/2 = Vavg and d=Vavg*t so
(Vf+Vi)/2 = d/t
Combing with 1) gives
(Vf+Vi)/2 = d/[ (Vf-Vi)/a ]
d = .5 * (Vf+Vi) * (Vf-Vi) *1/a
d = .5/a * (Vf^2 + Vi^2)
2) Vf^2 = Vi^2 + 2ad
Combining 1 and to by setting Vf=Vf:
Vi+at = sqrt(Vi^2 + 2ad)
Vi^2 + a^2*t^2 + 2Vi*at = Vi^2 + 2ad
a^2*t^2 + 2Vi*at = 2ad
at^2 + 2Vi*t = 2d
3) d = vit+.5at^2
edit: wooh just did that from memory without looking at anything and was nervous I wouldn't remember how to do it on the spot
|
he needs to get from 2a(x) + Vi^2 = Vf^2 to x = Vit + 1/2 at^2
nvm, you edited in the last step
|
United States24745 Posts
On October 16 2008 09:58 Nitrogen23 wrote:
he needs to get from 2a(x) + Vi^2 = Vf^2 to x = Vit + 1/2 at^2
nvm, you edited in the last step
I just showed that. I assume you posted this before my first edit.
edit: ah ok
|
I honestally dont know which one to write, so i wont do it at all
Thanks for your efforts though,
/hug
|
right down micronesia's, i'd trust him more.
oh wow, i spelled write wrong... god damn
|
Hi guys, maybe i can help:
we know that
a=(Vf-Vi)/t
then
Vf=Vi+at
Vf^2=(Vi+at)^2=Vi^2 + (at)^2 +2atVi
then: Vf^2 - Vi^2 = (at)^2 +2atVi........ (1)
now we have:
Vf^2=Vi^2 + 2ax
then
x=(Vf^2 - Vi^2)/2a.............(2)
now we replace (1) in (2)
x=[(at)^2 +2atVi]/2a
finally:
x = Vit + a(t^2)/2
Is this what you need???
|
I see someone didn't read the rest of the thread
|
On October 16 2008 10:16 renezerg wrote:
Hi guys, maybe i can help:
we know that
a=(Vf-Vi)/t
then
Vf=Vi+at
Vf^2=(Vi+at)^2=Vi^2 + (at)^2 +2atVi
then: Vf^2 - Vi^2 = (at)^2 +2atVi........ (1)
now we have:
Vf^2=Vi^2 + 2ax
then
x=(Vf^2 - Vi^2)/2a.............(2)
now we replace (1) in (2)
x=[(at)^2 +2atVi]/2a
finally:
x = Vit + a(t^2)/2
Is this what you need???
a bit late, but good work ^.^
|
On October 16 2008 10:52 alphafuzard wrote:Show nested quote +On October 16 2008 10:16 renezerg wrote:
Hi guys, maybe i can help:
we know that
a=(Vf-Vi)/t
then
Vf=Vi+at
Vf^2=(Vi+at)^2=Vi^2 + (at)^2 +2atVi
then: Vf^2 - Vi^2 = (at)^2 +2atVi........ (1)
now we have:
Vf^2=Vi^2 + 2ax
then
x=(Vf^2 - Vi^2)/2a.............(2)
now we replace (1) in (2)
x=[(at)^2 +2atVi]/2a
finally:
x = Vit + a(t^2)/2
Is this what you need???
a bit late, but good work ^.^
i see ,, is just that im quite slow at typing ^^
|
|
|
|
|
|
EPT
