Solution 18.5b
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(New page: From part a) <math>v=2kt-5{{t}^{2}}</math> Thus <math>\begin{align} & 0=2k\times 20-5\times {{20}^{2}} \\ & 0=40k-2000 \\ & k=\frac{2000}{40}=50 \\ \end{align}</math>)
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From part a)
\displaystyle v=2kt-5{{t}^{2}}
Thus
\displaystyle \begin{align} & 0=2k\times 20-5\times {{20}^{2}} \\ & 0=40k-2000 \\ & k=\frac{2000}{40}=50 \\ \end{align}
