Solution 6.8b
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(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>)
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Revision as of 12:54, 10 April 2010
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( 0.075 \right)\times {{10}^{2}}
