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It's the takeoff safety speed. The speed at which the aircraft may safely become airborne with one engine inoperative.

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Q: What does v2 mean during take off?
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If acceleration=0, than according to "d=v2*t-1/2a*t^2" changed to get v2 is "v2=d+1/2a*t^2/t", but with acceleration=0, would v2=d V2=end velocity, d=distance, a=acceleration, t=time?

The equations of motion that relate velocity, distance, time and acceleration for the specific case of "constant acceleration" can be written as follow, acceleration a = (v2 - v1)/t from which v2 = v1 + at The distance covered during t time d = vav x t, where vav refers to average velocity in the process from v1 to v2. For the case of constant acceleration vav = (v1 + v2)/2. Substituting in d we get d = (v1 + v2)/2 x t from which, v2 = 2d/t - v1 If we take the constant acceleration to be zero, a = 0, you can see that the second equation we wrote becomes, v2 = v1 (There is no acceleration), so our equation for the distance d becomes, d = v1 x t = v2 x t


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