Solution 18.5b

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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}