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Solution 8.6b

From Mechanics

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Line 1: Line 1:
Here we use
Here we use
-
<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>
+
<math>\mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}}</math>
 +
 
 +
in the first part.
According to the text
According to the text
Line 7: Line 9:
<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
-
and part a) gave
+
<math>\mathbf{v}=6\mathbf{i}-8\mathbf{j}</math>
-
 
+
-
<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}</math>
+
We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
-
At <math>t=10+40=50</math> we obtain
+
At <math>t=10</math>
-
<math>\mathbf{r}=(\mathbf{i}+2\mathbf{j}) \times 50+\frac{1}{2}(\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j} )\times{{50}^{\ 2}}</math>
+
<math> \mathbf{r}=\frac{1}{2}(\mathbf{i}+2\mathbf{j}+6\mathbf{i}-8\mathbf{j})t=(3\textrm{.}5\mathbf{i}-3\mathbf{j}) \times 10=35\mathbf{i}-30\mathbf{j}=(7\mathbf{i}-6\mathbf{j}) \times 5</math>
 +
The distance is the magnitude of this vector
 +
<math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math>
-
The distance is the magnitude of this vector
+
Thus during the first part the boat has travelled a distance 3.6 m.

Revision as of 19:54, 14 April 2010

Here we use

r=21(u+v)t+r0

in the first part.

According to the text

u=i+2j

v=6i8j

We assume the starting point is the origin so that r0=0.

At t=10

r=21(i+2j+6i8j)t=(3.5i3j)10=35i30j=(7i6j)5

The distance is the magnitude of this vector

72+62=13=3.6 m 

Thus during the first part the boat has travelled a distance 3.6 m.