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Constant Acceleration -- Equation for Vavge
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[QUOTE="mfig, post: 6269281, member: 116528"] These expressions are equivalent. Perhaps it will help you to see how. Recall that: ##a=constant \implies v(t)=at+v_0 \implies x(t) = \frac{at^2}{2} + v_0t + x_0 \tag{1}## Now the first expression for the average velocity is: ##\frac{x_2-x_1}{t_2-t_1}\tag{2}## If we use the third expression in (1) for the x values in (2), then this gives: ##\frac{\frac{a t_2^2}{2} + v_0t_2 + x_0 -\left(\frac{a t_1^2}{2} + v_0t_1 + x_0\right)}{t_2-t_1}\tag{3}## This simplifies to: ##\frac{\frac{a }{2}(t_2^2 - t_1^2) + v_0(t_2-t_1)} {t_2-t_1}\tag{4}## Now consider the second expression you give for the average velocity: ##\frac{v_1 + v_2}{2}\tag{5}## If we use the second expression in (1) for the velocity values in (5), then we have: ##\frac{at_1 +v_0+ at_2 + v_0}{2} = \frac{a(t_1 + t_2)+ 2v_0}{2} = \frac{a(t_1 + t_2)+ 2v_0}{2} \cdot \frac{t_2-t_1}{t_2-t_1} = \frac{\frac{a }{2}(t_2^2 - t_1^2) + v_0(t_2-t_1)} {t_2-t_1}\tag{6} ## So the expressions (4) and (6) are equivalent, and therefore they both represent the average velocity over some time interval given constant acceleration. HTH [/QUOTE]
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Constant Acceleration -- Equation for Vavge
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