Solution 8.6b

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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...)
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<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
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and part a) gave
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<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>.
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At <math>t=10+40=50</math> we obtain
At <math>t=10+40=50</math> we obtain
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<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>
The distance is the magnitude of this vector
The distance is the magnitude of this vector

Revision as of 19:35, 14 April 2010

Here we use

\displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}

According to the text

\displaystyle \mathbf{u}=\mathbf{i}+2\mathbf{j}

and part a) gave

\displaystyle \mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}

We assume the starting point is the origin so that \displaystyle {{\mathbf{r}}_{0}}=0.

At \displaystyle t=10+40=50 we obtain

\displaystyle \mathbf{r}=(\mathbf{i}+2\mathbf{j}) \times 50+\frac{1}{2}(\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j} )\times{{50}^{\ 2}}


The distance is the magnitude of this vector

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