Solution 18.1a
From Mechanics
(Difference between revisions)
(New page: <math>v=\frac{ds}{dt}={{t}^{2}}-\frac{{{t}^{3}}}{20}</math>) |
|||
| Line 1: | Line 1: | ||
| - | <math>v=\frac{ds}{dt}={{t}^{2}}-\frac{{{t}^{3}}}{ | + | <math>v=\frac{ds}{dt}={{t}^{2}}-\frac{{{t}^{3}}}{15}</math> |
Current revision
\displaystyle v=\frac{ds}{dt}={{t}^{2}}-\frac{{{t}^{3}}}{15}
