Solution 8.9b

From Mechanics

(Difference between revisions)
Jump to: navigation, search
(New page: We use <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> with <math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> <math>\mathbf{a}=-10\mathbf{j}</math> <math...)
Current revision (15:40, 19 April 2010) (edit) (undo)
 
Line 1: Line 1:
-
We use <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> with
+
We use <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ </math> with
<math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
<math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
Line 10: Line 10:
Simplifying <math>\mathbf{v}=15\mathbf{i}+18\mathbf{j}-30\mathbf{j}=15\mathbf{i}-12\mathbf{j} </math>
Simplifying <math>\mathbf{v}=15\mathbf{i}+18\mathbf{j}-30\mathbf{j}=15\mathbf{i}-12\mathbf{j} </math>
 +
 +
The speed <math>S</math> is the magnitude of this velocity.
 +
 +
<math>S=\sqrt{{{15}^{2}}+{{\left( -12 \right)}^{2}}}=\sqrt{225+144}=19\textrm{.}2\ \text{m}{{\text{s}}^{-1}}</math>

Current revision

We use \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t \ with

\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}

giving \displaystyle \mathbf{v}=15\mathbf{i}+18\mathbf{j}+(-10\mathbf{j} )\times 3

Simplifying \displaystyle \mathbf{v}=15\mathbf{i}+18\mathbf{j}-30\mathbf{j}=15\mathbf{i}-12\mathbf{j}

The speed \displaystyle S is the magnitude of this velocity.

\displaystyle S=\sqrt{{{15}^{2}}+{{\left( -12 \right)}^{2}}}=\sqrt{225+144}=19\textrm{.}2\ \text{m}{{\text{s}}^{-1}}

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