Question

Hey There,

I've got quite a simple algebra question, but I just can't quite sort it. I need to solve 1/2at^2 -vt + x = 0 for t, but I'm not sure what to do with that x value.

Any help greatly appreciated.

Cheers

Answer 1

Have you considered the quadratic formula? a, v, and x are numbers. t = ?? Typically, only t >= 0 matters.

Answer 2

you know that a quadratic is in the form of

ax^2 +bx +c = 0

so in your exercise case the c signed by this c

this is all

Answer 3

that is not algebra...

Answer 4

it is physics

Answer 5

merely a translation of the equation:
x = x) + v0t + 1/2 at^2

Answer 6

This is a legitimate algebra problem. Many application problems are quite acceptable in this forum.

No need to translate the equation. Just solve it.

Answer 7

@tkhunny not really, if he is trying to solve for t, he is trying to find the time of something, yes it is algebra, but it is sourced from physics

Answer 8

@siberman if i am correct that it is physics, tell me what you know and dont know
there are many equations you can use to find time, this is just one of them

Answer 9

@jackthegreatest If you are not going to help the student, please go elsewhere where you WILL help. Solving a quadratic equation is well within the mandate for the Mathematics forum.

@siberman Please listen to jhonyy9. Change your equation to that standard form and apply the Quadratic Formula.

Answer 10

at^2 -2vt +2x = 0 solve it for t

D = 4v^2 -8x = 4(v^2 -2x)

t_1,2 = (8 +/- sqrt D )/2 = ?

Answer 11

@tkhunny i am helping, you never know the possibilities
i am sure that you, as a mod, have been tracking all the other people i have helped
so no i am not trying to not help
instead, i am trying to help at a higher level

Answer 12

XD wow

Answer 13

sorry but i not going just to help other askers

and than you check i ve wrote there above solved this quadratic

Answer 14

@OtherWorldly sorry but i not can sending to you messages just in this way - so this was one happy way when you wrote there above this xD - so for this i give you medal - ok.

Answer 15

XD oh yeah i forgot sorry and lol ok

Answer 16

@jackthegreatest I have no concern for your other behavior, only that your classification on this thread was inappropriate. I am glad that you help elsewhere.

Answer 17

Hi, thanks everyone for your help. The quadratic formula was exactly what I needed, completely blanked on that. In looking at the solution they take another step which I can't quite reconcile algebraically. I've attached an image, it's the v^2 inside the sqrt that's throwing me. Thanks Again!

Answer 18

physics is applied math

Answer 19

Further to my previous post, this is where I've managed to get to, I just need to get that v under the 2ax !

Answer 20

1 thing i noticed is you have an extra negative sign inside sqrt
From quadratic formula ----> b^2 - 4ac = (-v)^2 - 2ax = v^2 - 2ax
squaring the "-v" makes it positive v^2

Other than that it looks good :)
For the final step, use the fact that
\[v = \sqrt{v^2}\]
\[\rightarrow \frac{v}{a}(1 \pm \frac{\sqrt{v^2 -2ax}}{v}) = \frac{v}{a}(1 \pm \frac{\sqrt{v^2 -2ax}}{\sqrt{v^2}}) = \frac{v}{a}(1 \pm \sqrt{1 -\frac{2ax}{v^2}})\]

Answer 21

@dumbcow

(\dfrac{Big}{Fraction}) Produces \((\dfrac{Big}{Fraction})\)

\left(\dfrac{Big}{Fraction}\right) Produces \(\left(\dfrac{Big}{Fraction}\right)\)

Jus tone step on the road to making things pretier.