Solution 6.8b - Revision history

Ian at 17:27, 11 May 2010

Ian — Tue, 11 May 2010 17:27:34 GMT

←Older revision		Revision as of 17:27, 11 May 2010
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-	<math>s=0+\frac{1}{2}\times \left( 0 \textrm{.}075 \right)\times {{10}^{2}}=37 \textrm{.}5\ \text{m}</math>	+	<math>s=0+\frac{1}{2}\times \left( 0 \textrm{.}075 \right)\times {{10}^{2}}=3\textrm{.}75\ \text{m}</math>

Ian at 15:09, 10 April 2010

Ian — Sat, 10 Apr 2010 15:09:14 GMT

←Older revision		Revision as of 15:09, 10 April 2010
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-	We investigate the distance after <math>10\ \text{s}</math>.~~Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives~~	+	We investigate the distance after <math>10\ \text{s}</math>.

		+	Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives

-	<math>s=0+\frac{1}{2}\times \left( 0.075 \right)\times {{10}^{2}}</math>	+
		+	<math>s=0+\frac{1}{2}\times \left( 0 \textrm{.}075 \right)\times {{10}^{2}}=37 \textrm{.}5\ \text{m}</math>

Ian: New page: We investigate the distance after $10\ \text{s}$ .Using $s=ut+\frac{1}{2}a{{t}^{\ 2}}$ gives $s=0+\frac{1}{2}\times \left( 0.075 \right)\times {{10}^{2}}$

Ian — Sat, 10 Apr 2010 12:54:29 GMT

New page: We investigate the distance after <math>10\ \text{s}</math>.Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives <math>s=0+\frac{1}{2}\times \left( 0.075 \right)\times {{10}^{2}}</math>

New page

We investigate the distance after <math>10\ \text{s}</math>.Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives

<math>s=0+\frac{1}{2}\times \left( 0.075 \right)\times {{10}^{2}}</math>

Solution 6.8b - Revision history

Ian at 17:27, 11 May 2010

Ian at 15:09, 10 April 2010

Ian: New page: We investigate the distance after 10\ \text{s}.Using s=ut+\frac{1}{2}a{{t}^{\ 2}} gives s=0+\frac{1}{2}\times \left( 0.075 \right)\times {{10}^{2}}

Ian: New page: We investigate the distance after $10\ \text{s}$ .Using $s=ut+\frac{1}{2}a{{t}^{\ 2}}$ gives $s=0+\frac{1}{2}\times \left( 0.075 \right)\times {{10}^{2}}$