SOLUTION: I'm stuck. I am supposed to solve for t and the original equation is: x=vt-(1/2)at^2 Here is what I have done so far. 2x=2vt-at^2 2x=t(2v-at) t=2x/(2v-at) But since I am sup

Algebra ->  Inequalities -> SOLUTION: I'm stuck. I am supposed to solve for t and the original equation is: x=vt-(1/2)at^2 Here is what I have done so far. 2x=2vt-at^2 2x=t(2v-at) t=2x/(2v-at) But since I am sup      Log On


   

Question 49749: I'm stuck. I am supposed to solve for t and the original equation is:
x=vt-(1/2)at^2
Here is what I have done so far.
2x=2vt-at^2
2x=t(2v-at)
t=2x/(2v-at)
But since I am supposed to solve for t how do I get the other t out or am I done?
Thanks!!
Andrea
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, your first few steps are fine. Remember, when you solve an equation for a specified variable, in this case it's t, you have to write the equation so that the variable t in on one side of the equals sign and everything else is on the other side. If you have a t on both sides of the equation, the job won't be finished. Let's pick it up where you got it right so far:
2x+=+2vt-at%5E2 Now this begins to look much like a quadratic equation in t, so let's put it into the standard form for quadratic equations: at%5E2%2Bbt%2Bc+=+0
at%5E2-2vt%2B2x+=+0 This is the same equation you had only it's now in the standard form. Now you can solve this quadratic equation using the quadratic formula: t+=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a In your equation: a = a, b = -2v, and c = 2x
t+=+%28-%28-2v%29%2B-sqrt%28%28-2v%29%5E2+-+4%28a%29%282x%29%29%29%2F2%28a%29 Simplify this.
t+=+%282v%2B-sqrt%284v%5E2-8ax%29%29%2F2a
t+=+%282v%2B-sqrt%284%28v%5E2-2ax%29%29%29%2F2a
t+=+%282v%2B-2sqrt%28v%5E2-2ax%29%29%2F2a
As with any quadratic equation, there are two solutions:
t+=+%28v%2Fa%29%2B%28sqrt%28v%5E2-2ax%29%29%2Fa and
t+=+%28v%2Fa%29-%28sqrt%28v%5E2-2ax%29%29%2Fa