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Solution 8.6b

From Mechanics

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Current revision (12:31, 27 March 2011) (edit) (undo)
 
Line 8: Line 8:
<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>
<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>
-
<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}-50\mathbf{j}</math>
+
<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>
The next stage has
The next stage has
-
<math>t=40</math>, <math>\mathbf{a}=0</math>, <math>{{\mathbf{r}}_{0}}=35\mathbf{i}+70\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.
+
<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.
Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get
Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get
-
<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-50\mathbf{j})=275\mathbf{i}-355\mathbf{j}</math>
+
<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>
This is the boat´s final position.
This is the boat´s final position.
Line 23: Line 23:
-
<math>\sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=450\ \text{m}</math>
+
<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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