Solution 8.10b

From Mechanics

(Difference between revisions)
Jump to: navigation, search
Line 11: Line 11:
giving,
giving,
-
<math>\mathbf{v}=8\ \mathbf{i}+8 \ \mathbf{ j}+((0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}))\times 40 =8\textrm{.}4\mathbf{i}+8\textrm{.}8\mathbf{j}+12\mathbf{k}</math>
+
<math>\mathbf{v}=8\ \mathbf{i}+8 \ \mathbf{ j}+((0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}))\times 40 =8\textrm{.}4\mathbf{i}+8\textrm{.}8\mathbf{j}+4\mathbf{k}</math>
-
This is the ''velocity'' of the aeroplane. The ''speed'' is the magnitude of the velocity.
+
This is the ''velocity'' of the aeroplane. The ''speed'' is the magnitude of the velocity,
 +
 
 +
<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>

Revision as of 10:28, 20 April 2010

We use the equation \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t \ where,

\displaystyle \mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}^{\text{-2}}}

and from part a)

\displaystyle \mathbf{u}=8\ \mathbf{i}+8 \ \mathbf{ j}\text{ m}{{\text{s}}^{\text{-1}}} \ and

\displaystyle t=40 \ \text{s} .

giving,

\displaystyle \mathbf{v}=8\ \mathbf{i}+8 \ \mathbf{ j}+((0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}))\times 40 =8\textrm{.}4\mathbf{i}+8\textrm{.}8\mathbf{j}+4\mathbf{k}

This is the velocity of the aeroplane. The speed is the magnitude of the velocity,

\displaystyle \sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Solution_8.10b"