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Average Velocity

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How do you find the average velocity of the position function $s (t) = 3 t^{2} - 6 t$ $s(t)=3t^2-6t$ on the interval from $t = 2$ $t=2$ to $t = 5$ $t=5$ ?

Average velocity is defined as total displacement/ total time taken for that

Given, $s = 3 t^{2} - 6 t$ $s=3t^2 -6t$

So,displacement in between $2 s$ $2s$ and $5 s$ $5s$ is $s = 3 {[t^{2}]}_{2}^{2} - 6 {[t]}_{2}^{5} = 3 (25 - 4) - 6 (5 - 2) = 45 m$ $s=3[t^2]_2^5 -6[t]_2^5=3(25-4)-6(5-2)=45m$

So,average velocity = $\frac{45}{5 - 2} = 15 m s^{- 1}$ $45/(5-2)=15 ms^-1$

Jane · 3 · Mar 17 2018

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