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

Revision as of 15:09, 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 \textrm{.}075 \right)\times {{10}^{2}}=37 \textrm{.}5\ \text{m}

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