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Constant Acceleration -- Equation for Vavge
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[QUOTE="RPinPA, post: 6269245, member: 651116"] They're just giving it to you as an unproven theorem. Even though average velocity is emphatically NOT defined as the average of the start and end velocities in general, in the case of uniform acceleration they are the same. You can prove it graphically as another answer suggests. You can also derive it algebraically from the equations of uniform acceleration. ##v = v_0 + at \Rightarrow at = v - v_0## ##d = v_0t + (1/2)at^2## So ##v_{avg} = d/t = v_0 + (1/2)at = v_0 + (1/2)(v - v_0) = (1/2)(v + v_0)## [/QUOTE]
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Constant Acceleration -- Equation for Vavge
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