Ian at 14:45, 21 May 2010

2010-05-21T14:45:30Z

←Older revision		Revision as of 14:45, 21 May 2010
Line 4:		Line 4:

	<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>		<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>
		+
		+	gives

	<math>\begin{align}		<math>\begin{align}
	& {{0}^{2}}={{50}^{2}}+2\times a\times 300 \\		& {{0}^{2}}={{50}^{2}}+2\times a\times 300 \\
-	~~& \\~~	+	& a=-\frac{2500}{600}=-4\textrm{.}167\ \text{m}{{\text{s}}^{-2}} \\
-	& a=-\frac{2500}{600}=-4.167\ \text{m}{{\text{s}}^{-2}} \\	+
	\end{align}</math>		\end{align}</math>
		+
		+	Then <math>F=ma\ </math> gives
		+
		+	<math>F=1000\times 4\textrm{.}167=4167\ \text{N}</math>

Ian: New page: We first obtain the acceleration $a$ . Using the kinematic equation (see section 6) ${{v}^{\ 2}}={{u}^{\ 2}}+2as$ $\begin{align} & {{0}^{2}}={{50}^{2}}+2\tim...$

2010-05-21T14:39:15Z

New page: We first obtain the acceleration <math>a</math>. Using the kinematic equation (see section 6) <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> <math>\begin{align} & {{0}^{2}}={{50}^{2}}+2\tim...

New page

We first obtain the acceleration <math>a</math>.

Using the kinematic equation (see section 6)

<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>

<math>\begin{align}
& {{0}^{2}}={{50}^{2}}+2\times a\times 300 \\
& \\
& a=-\frac{2500}{600}=-4.167\ \text{m}{{\text{s}}^{-2}} \\
\end{align}</math>

Solution 10.2 - Revision history

Ian at 14:45, 21 May 2010

Ian: New page: We first obtain the acceleration a. Using the kinematic equation (see section 6) {{v}^{\ 2}}={{u}^{\ 2}}+2as \begin{align} & {{0}^{2}}={{50}^{2}}+2\tim...

Ian: New page: We first obtain the acceleration $a$ . Using the kinematic equation (see section 6) ${{v}^{\ 2}}={{u}^{\ 2}}+2as$ $\begin{align} & {{0}^{2}}={{50}^{2}}+2\tim...$