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Revision as of 15:43, 11 October 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (17:29, 27 March 2011) (edit) (undo) Ian (Talk | contribs)
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	& 5=20-\frac{k\times {{40}^{2}}}{5} \\		& 5=20-\frac{k\times {{40}^{2}}}{5} \\
	& \\		& \\
-	& 320k=15 \\	+	& 800k=15 \\
	& \\		& \\
-	& k=\frac{15}{320}=\frac{3}{64}	+	& k=\frac{15}{800}=\frac{3}{160}
	\end{align}</math>		\end{align}</math>
	Note that substituting for <math>k</math> and <math>c</math> in the expression for <math>v</math> we get		Note that substituting for <math>k</math> and <math>c</math> in the expression for <math>v</math> we get
-	<math>v=20-\frac{3{{t}^{2}}}{128}</math>	+	<math>v=20-\frac{3{{t}^{2}}}{320}</math>

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Current revision

\displaystyle \begin{align} & v=\int{adt} \\ & \\ & =\int{\left( -kt \right)dt} \\ & \\ & =-\frac{k{{t}^{2}}}{2}+c \\ & \\ & t=0,\ v=20\ \Rightarrow \ c=20 \\ & \\ & v=20-\frac{k{{t}^{2}}}{2} \\ & \\ & \text{ Using}\quad t=40,\ v=5\quad \text{gives} \\ & \\ & 5=20-\frac{k\times {{40}^{2}}}{5} \\ & \\ & 800k=15 \\ & \\ & k=\frac{15}{800}=\frac{3}{160} \end{align}

Note that substituting for \displaystyle k and \displaystyle c in the expression for \displaystyle v we get

\displaystyle v=20-\frac{3{{t}^{2}}}{320}