From Mechanics

Revision as of 12:54, 10 April 2010 (edit) Ian (Talk | contribs) (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>) ← Previous diff		Current revision (17:27, 11 May 2010) (edit) (undo) Ian (Talk | contribs)
(One intermediate revision not shown.)
Line 1:		Line 1:
-	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}}=3\textrm{.}75\ \text{m}</math>

---

Current revision

We investigate the distance after \displaystyle 10\ \text{s}.

Using \displaystyle s=ut+\frac{1}{2}a{{t}^{\ 2}} gives

\displaystyle s=0+\frac{1}{2}\times \left( 0 \textrm{.}075 \right)\times {{10}^{2}}=3\textrm{.}75\ \text{m}