Ian: New page: From part a) $v=2kt-5{{t}^{2}}$ Thus $\begin{align} & 0=2k\times 20-5\times {{20}^{2}} \\ & 0=40k-2000 \\ & k=\frac{2000}{40}=50 \\ \end{align}$

Ian — Wed, 06 Oct 2010 18:22:59 GMT

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>

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>

Solution 18.5b - Revision history

Ian: New page: From part a) v=2kt-5{{t}^{2}} Thus \begin{align} & 0=2k\times 20-5\times {{20}^{2}} \\ & 0=40k-2000 \\ & k=\frac{2000}{40}=50 \\ \end{align}

Ian: New page: From part a) $v=2kt-5{{t}^{2}}$ Thus $\begin{align} & 0=2k\times 20-5\times {{20}^{2}} \\ & 0=40k-2000 \\ & k=\frac{2000}{40}=50 \\ \end{align}$