Solution 8.4b

From Mechanics

(Difference between revisions)
Jump to: navigation, search
(New page: Here we have <math>\mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}...)
Current revision (14:18, 14 April 2010) (edit) (undo)
 
Line 5: Line 5:
Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> at <math>\ \ t=10</math> gives
Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> at <math>\ \ t=10</math> gives
-
Using <math>\mathbf{r}=(4\mathbf{i}– 8\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j}) \times {{10}^{\ 2}}+6\mathbf{i}+2\mathbf{j}</math>
+
<math>\mathbf{r}=(4\mathbf{i}– 8\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j}) \times {{10}^{\ 2}}+6\mathbf{i}+2\mathbf{j}</math>
<math>\mathbf{r}=56\mathbf{i}– 63\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}</math>
<math>\mathbf{r}=56\mathbf{i}– 63\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}</math>

Current revision

Here we have \displaystyle \mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \ \displaystyle \mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \ and \displaystyle \ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}.

Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} at \displaystyle \ \ t=10 gives

\displaystyle \mathbf{r}=(4\mathbf{i}– 8\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j}) \times {{10}^{\ 2}}+6\mathbf{i}+2\mathbf{j}

\displaystyle \mathbf{r}=56\mathbf{i}– 63\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Solution_8.4b"