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...)
Current revision (17:40, 27 March 2011) (edit) (undo)
 
Line 2: Line 2:
<math>\begin{align}
<math>\begin{align}
-
& s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\
+
& s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{320} \right)}dt \\
& \\
& \\
-
& =\left[ 20t-\frac{{{t}^{3}}}{128} \right]_{0}^{40} \\
+
& =\left[ 20t-\frac{{{t}^{3}}}{960} \right]_{0}^{40} \\
& \\
& \\
-
& =\left( 20\times 40-\frac{{{40}^{3}}}{128} \right)-0 \\
+
& =\left( 20\times 40-\frac{{{40}^{3}}}{960} \right)-0 \\
& \\
& \\
-
& =300\text{ m}
+
& =733\text{ m}
\end{align}</math>
\end{align}</math>

Current revision

Using the expression for \displaystyle v obtained in part a),

\displaystyle \begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{320} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{960} \right]_{0}^{40} \\ & \\ & =\left( 20\times 40-\frac{{{40}^{3}}}{960} \right)-0 \\ & \\ & =733\text{ m} \end{align}