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Can some one please prove that Vav=(Vi+Vf)/2 is valid only for constant acceleration?
No, because it isn't valid only for constant acceleration. There are other velocity functions that also yield this average velocity.Can some one please prove that Vav=(Vi+Vf)/2 is valid only for constant acceleration?
It's posible for Vav=(Vi+Vf)/2 to be true for variable acceleration as well. Start off with a graph of velocity (y axis) versus time (x axis). Then the area below the horizontal line that goes from {t0, Vav} to {t1, Vav} = Vav x (t1 - t0) = the distance traveled. Then note that any line of any shape with the same amount of area under the line from {t0, Vi} to {t1, Vf} would also have the same average velocity as constant acceleration.Can some one please prove that Vav=(Vi+Vf)/2 is valid only for constant acceleration?