Processing Math: Done
Solution 19.5b
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Ian (Talk | contribs)
(New page: Using the expression for <math>v</math> obtained in part a), <math>\begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...)
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Revision as of 15:50, 11 October 2010
Using the expression for
040
20−3t2128
dt=
20t−t3128
040=20
40−403128−0=300 m
