Suppose $$ s(t) $$ is the position of an object moving along a line at time $$ t \geq 0 $$. What is the average velocity between the times $$ t=a $$ and $$ t=b $$ ? A. $$ v_{\mathrm{av}}=\frac{s(a)+s(b)}{a-b} $$ B $$ v_{\mathrm{av}}=\frac{s(a)-s(b)}{b-a} $$ C $$ v_{a v}=\frac{s(b)-s(a)}{b-a} $$ OD. $$ v_{a v}=\frac{s(b)+s(a)}{b-a} $$

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The correct answer is C. $$ v_{\mathrm{av}} = \frac{s(b) - s(a)}{b - a} $$ The average velocity $$ v_{\mathrm{av}} $$ of an object over the interval from time $$ t = a $$ to time $$ t = b $$ is defined as the change in position divided by the change in time. Mathematically, this is expressed as: $$ v_{\mathrm{av}} = \frac{s(b) - s(a)}{b - a} $$ Thus, the correct answer is: C. $$ v_{\mathrm{av}} = \frac{s(b) - s(a)}{b - a} $$

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