What equation of motion is this formula derived off of?

Which equation of motion ( first, second or third ) is this equation "t = 鈭?2d/a)" derived/made out of? I'm thinking it is derived off of the third equation of motion "2aS = Vf^2-Vi^2".

I would really like to know how to make this equation out of the main equation it is derived off of aswell. Step by step that is. With full detail.



Thanks.What equation of motion is this formula derived off of?one of your equations of motion (remember these are only valid under constant acceleration) is



x = 1/2 g t^2 + v0 t + x0



if v0 and x0 are 0 (at rest initially, and by choice of coordinate system, respectively)



x = 1/2 g t^2, solve for t



t=sqrt(2x/g)What equation of motion is this formula derived off of?I'll use the standard SUVAT variables.



S = vT; where S is distance (your d), v is average speed, and T is time traveled.



Average speed = (V + U)/2 and V = U + AT. Therefore, we have...



S = (V + U)/2 * T = (U + AT + U)/2 * T = UT + 1/2 AT^2 which is the complete equation.



In your case, if we assume the initial speed U = 0, we have S = 1/2 AT^2 and T = sqrt(2S/A) QED.



BTW starting with 2aS = Vf^2-Vi^2 is the wrong place to start. You start with good old distance = average speed X time traveled, which is the most basic of all the kinematic relationship.What equation of motion is this formula derived off of?Why don't U try squaring both sides of the:

t = 鈭?d/a

and get

t虏 2d/a

then multiply both sides by "a" to get

at虏 = 2d

now divide both sides by "2" and now...does THAT equation look familiar? :%26gt;)