From Mechanics

Revision as of 15:09, 10 April 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (17:27, 11 May 2010) (edit) (undo) Ian (Talk | contribs)
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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>

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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}