Solution 19.5a - Revision history

Ian at 17:29, 27 March 2011

Ian — Sun, 27 Mar 2011 17:29:08 GMT

←Older revision		Revision as of 17:29, 27 March 2011
Line 14:		Line 14:
	& 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>

Ian at 15:43, 11 October 2010

Ian — Mon, 11 Oct 2010 15:43:29 GMT

←Older revision		Revision as of 15:43, 11 October 2010
Line 18:		Line 18:
	& k=\frac{15}{320}=\frac{3}{64}		& k=\frac{15}{320}=\frac{3}{64}
	\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
		+
		+	<math>v=20-\frac{3{{t}^{2}}}{128}</math>

Ian: New page: $\begin{align} & v=\int{adt} \\ & \\ & =\int{\left( -kt \right)dt} \\ & \\ & =-\frac{k{{t}^{2}}}{2}+c \\ & \\ & t=0,\ v=20\ \Rightarrow \ c=20 \\ & \\ & v=20-\frac{k{{t}^{...$

Ian — Mon, 11 Oct 2010 15:37:35 GMT

New page: <math>\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}^{...

New page

<math>\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} \\
& \\
& 320k=15 \\
& \\
& k=\frac{15}{320}=\frac{3}{64}
\end{align}</math>

Solution 19.5a - Revision history

Ian at 17:29, 27 March 2011

Ian at 15:43, 11 October 2010

Ian: New page: \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}^{...

Ian: New page: $\begin{align} & v=\int{adt} \\ & \\ & =\int{\left( -kt \right)dt} \\ & \\ & =-\frac{k{{t}^{2}}}{2}+c \\ & \\ & t=0,\ v=20\ \Rightarrow \ c=20 \\ & \\ & v=20-\frac{k{{t}^{...$