Solution 18.5a
From Mechanics
(Difference between revisions)
(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>) |
(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>) |
Current revision
\displaystyle \begin{align} & v=\frac{ds}{dt}=2kt-5{{t}^{2}} \\ & v(0)=2k\times 0-5\times {{0}^{2}}=0 \\ \end{align}
