Solution 19.5b
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(New page: Using the expression for <math>v</math> obtained in part a), <math>\begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...)
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Revision as of 15:50, 11 October 2010
Using the expression for \displaystyle v obtained in part a),
\displaystyle \begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{128} \right]_{0}^{40} \\ & \\ & =\left( 20\times 40-\frac{{{40}^{3}}}{128} \right)-0 \\ & \\ & =300\text{ m} \end{align}
