Solution 18.5a
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(New page: <math>\begin{align} & v=\frac{ds}{dt}=2kt-5{{t}^{2}} \\ & v(0)=2k\times 0-5\times {{0}^{2}}=0 \\ \end{align}</math>)
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\displaystyle \begin{align} & v=\frac{ds}{dt}=2kt-5{{t}^{2}} \\ & v(0)=2k\times 0-5\times {{0}^{2}}=0 \\ \end{align}
