Solution 8.9a
From Mechanics
(Difference between revisions)
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- | Here we use | + | Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with |
- | <math> | + | <math>{{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{i} \text{ m}</math> |
- | + | <math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> | |
- | <math> | + | <math>\mathbf{a}=-10\mathbf{j}</math> |
- | + | <math>t=3\text{ s}</math> | |
- | + | ||
- | <math> | + | |
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Revision as of 15:06, 19 April 2010
Here we use \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with
\displaystyle {{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{i} \text{ m}
\displaystyle \mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}
\displaystyle \mathbf{a}=-10\mathbf{j}
\displaystyle t=3\text{ s}