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Question-108727




Question Number 108727 by Don08q last updated on 18/Aug/20
Answered by Dwaipayan Shikari last updated on 18/Aug/20
x(t)=t^3 −15t^2 +72t+3  v_(avg) =((x(8)−x(0))/8)=((512−15.64+72.8+3−3)/8)=16(m/s)
$${x}\left({t}\right)={t}^{\mathrm{3}} −\mathrm{15}{t}^{\mathrm{2}} +\mathrm{72}{t}+\mathrm{3} \\ $$$${v}_{{avg}} =\frac{{x}\left(\mathrm{8}\right)−{x}\left(\mathrm{0}\right)}{\mathrm{8}}=\frac{\mathrm{512}−\mathrm{15}.\mathrm{64}+\mathrm{72}.\mathrm{8}+\mathrm{3}−\mathrm{3}}{\mathrm{8}}=\mathrm{16}\frac{{m}}{{s}} \\ $$$$ \\ $$

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