Answer 18.1a

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(New page: <math>v={{t}^{2}}-\frac{{{t}^{3}}}{20}</math>)
Current revision (16:51, 27 March 2011) (edit) (undo)
 
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<math>v={{t}^{2}}-\frac{{{t}^{3}}}{20}</math>
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<math>v={{t}^{2}}-\frac{{{t}^{3}}}{15}</math>

Current revision

\displaystyle v={{t}^{2}}-\frac{{{t}^{3}}}{15}