Solution 8.9a

From Mechanics

(Difference between revisions)
Jump to: navigation, search
Line 1: Line 1:
-
Here we use the equation <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> with
+
Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with
-
<math>\mathbf{u}=<math>4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and <math>\mathbf{v}=24\mathbf{i}+46\mathbf{j\text{ m}{{\text{s}}^{\text{-1}}}}</math>.
+
<math>{{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{i} \text{ m}</math>
-
Also <math>t=20\text{ s}</math> whichh gives,
+
<math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
-
<math>24\mathbf{i}+46\mathbf{j}=4\mathbf{i}+6\mathbf{j}+\mathbf{a} \times 20</math>
+
<math>\mathbf{a}=-10\mathbf{j}</math>
-
So,
+
<math>t=3\text{ s}</math>
-
 
+
-
<math>20\mathbf{a}=20\mathbf{i}+40\mathbf{j}</math>
+
-
giving,
+
-
 
+
-
<math>\mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}}</math>
+

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}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Solution_8.9a"