Solution 19.4c

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(New page: Using the result from part b) for the velocity, <math>\begin{align} & s(20)=\int_{0}^{20}{v}dt \\ & \\ & =\int_{0}^{20}{\left( \frac{{{t}^{2}}}{5}+8 \right)dt} \\ & \\ & =\left[ \fr...)
Current revision (13:52, 11 October 2010) (edit) (undo)
(New page: Using the result from part b) for the velocity, <math>\begin{align} & s(20)=\int_{0}^{20}{v}dt \\ & \\ & =\int_{0}^{20}{\left( \frac{{{t}^{2}}}{5}+8 \right)dt} \\ & \\ & =\left[ \fr...)
 

Current revision

Using the result from part b) for the velocity,

\displaystyle \begin{align} & s(20)=\int_{0}^{20}{v}dt \\ & \\ & =\int_{0}^{20}{\left( \frac{{{t}^{2}}}{5}+8 \right)dt} \\ & \\ & =\left[ \frac{{{t}^{3}}}{15}+8t \right]_{0}^{20} \\ & \\ & =\left( \frac{{{20}^{3}}}{15}+8\times 20 \right)-0 \\ & \\ & =693\text{ m} \end{align}