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Solution 8.6b

From Mechanics

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(New page: Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math> According to the text <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> We assume th...)
Current revision (12:31, 27 March 2011) (edit) (undo)
 
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Here we use
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We assume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>
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<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>
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The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math>{{\mathbf{r}}_{A}}</math>.
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According to the text
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Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with,
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<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
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<math>t=10</math>, <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and from part a)
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<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>
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<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}-30\mathbf{j}</math>
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We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
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The next stage has
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At <math>t=10+40=50</math> we obtain
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<math>t=40</math>, <math>\mathbf{a}=0</math>, <math>{{\mathbf{r}}_{0}}=35\mathbf{i}-30\mathbf{j}</math>, and <math>\mathbf{u}=6\mathbf{i}-8\mathbf{j}</math> as the final position and velocity of the first stage is the initial position and velocity of the second stage.
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Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get
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<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-30\mathbf{j})=275\mathbf{i}-350\mathbf{j}</math>
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This is the boat´s final position.
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The distance from the starting point is the magnitude of this vector,
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The distance is the magnitude of this vector
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<math>\sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=445\ \text{m}</math>

Current revision

We assume the starting point is the origin. This means r0=0

The boat first accelerates to a point A say. We first must calculate the position of this point rA.

Using r=ut+21at 2+r0 with,

t=10, u=i+2j and from part a) a=0.5i−j ms−2

rA=(i+2j)10+21(0.5i−j)10 2+0=35i−30j

The next stage has

t=40, a=0, r0=35i−30j, and u=6i−8j as the final position and velocity of the first stage is the initial position and velocity of the second stage.

Using once again r=ut+21at 2+r0 we get

r=(6i−8j)40+0+(35i−30j)=275i−350j

This is the boat´s final position.

The distance from the starting point is the magnitude of this vector,


2752+−3552=445 m 

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