Solution 6.4a

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(New page: Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> gives)
Current revision (10:29, 7 April 2010) (edit) (undo)
 
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Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> gives
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Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as \ \</math>
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gives
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<math>\begin{align}
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& {{15}^{2}}={{5}^{2}}+2\times a\times 500 \\
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& 200=1000a \\
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& a=0\textrm{.}2 \text{ m}{{\text{s}}^{\text{-2}}}\\
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& \\
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\end{align}</math>

Current revision

Using \displaystyle {{v}^{\ 2}}={{u}^{\ 2}}+2as \ \ gives

\displaystyle \begin{align} & {{15}^{2}}={{5}^{2}}+2\times a\times 500 \\ & 200=1000a \\ & a=0\textrm{.}2 \text{ m}{{\text{s}}^{\text{-2}}}\\ & \\ \end{align}