Solution 19.3c

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(New page: <math>\begin{align} & s(10)=\int_{0}^{10}{v}dt \\ & \\ & =\int_{0}^{10}{\left( 8t-\frac{2{{t}^{2}}}{5} \right)dt} \\ & \\ & =\left[ 4{{t}^{2}}-\frac{2{{t}^{3}}}{15} \right]_{0}^{10} ...)
Current revision (09:36, 11 October 2010) (edit) (undo)
 
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Using the result for the velocity obtained in part a)
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<math>\begin{align}
<math>\begin{align}
& s(10)=\int_{0}^{10}{v}dt \\
& s(10)=\int_{0}^{10}{v}dt \\

Current revision

Using the result for the velocity obtained in part a)

\displaystyle \begin{align} & s(10)=\int_{0}^{10}{v}dt \\ & \\ & =\int_{0}^{10}{\left( 8t-\frac{2{{t}^{2}}}{5} \right)dt} \\ & \\ & =\left[ 4{{t}^{2}}-\frac{2{{t}^{3}}}{15} \right]_{0}^{10} \\ & \\ & =\left( 4\times {{10}^{2}}-\frac{2\times {{10}^{3}}}{15} \right)-0 \\ & \\ & =267\text{ m} \end{align}