Solution 8.10c
From Mechanics
m (New page: We use the result from part a). <math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{...) |
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<math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j}) \times 40+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}) \times{{40}^{\ 2}}+10\mathbf{k}=328\mathbf{i}+336\mathbf{j}+10\mathbf{k} \ </math>, for any time <math>t</math>. | <math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j}) \times 40+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}) \times{{40}^{\ 2}}+10\mathbf{k}=328\mathbf{i}+336\mathbf{j}+10\mathbf{k} \ </math>, for any time <math>t</math>. | ||
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| + | The car starts at the origin and finishes at <math>328\mathbf{i}+336\mathbf{j}</math>. | ||
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| + | The magnitude of this position vector is the distance <math>D\ </math> travelled. | ||
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| + | <math>D=\sqrt{{{328}^{2}}+{{336}^{2}}}=\sqrt{107584+112896}=\sqrt{220480}=470\ \text{m}</math> | ||
Current revision
We use the result from part a).
\displaystyle \mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{k}\ , for any time \displaystyle t.
At the end when \displaystyle t=40 also from part a), we get
\displaystyle \mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j}) \times 40+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}) \times{{40}^{\ 2}}+10\mathbf{k}=328\mathbf{i}+336\mathbf{j}+10\mathbf{k} \ , for any time \displaystyle t.
The car starts at the origin and finishes at \displaystyle 328\mathbf{i}+336\mathbf{j}.
The magnitude of this position vector is the distance \displaystyle D\ travelled.
\displaystyle D=\sqrt{{{328}^{2}}+{{336}^{2}}}=\sqrt{107584+112896}=\sqrt{220480}=470\ \text{m}
