Solution 8.8b
From Mechanics
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<math>\mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}</math> | <math>\mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}</math> | ||
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| + | <math>\mathbf{r}=120\mathbf{i}+180\mathbf{j} +450\mathbf{i}+900\mathbf{j} = 570\mathbf{i}+1080\mathbf{j}</math> | ||
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| + | As the starting point is at the origin, the distance is the magnitude of this position vector. | ||
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| + | <math>\sqrt{{{570}^{2}}+{{108}^{2}}}=\sqrt{324900+1166400}=\sqrt{1491300}=1221\ \text{m}</math> | ||
Current revision
We must find the position of the particle after 30 s.
Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with \displaystyle \ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}} which was calculated in part a), \displaystyle \ \mathbf{u}=4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{\ \mathbf{r}}_{0}}=0 gives,
\displaystyle \mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}
\displaystyle \mathbf{r}=120\mathbf{i}+180\mathbf{j} +450\mathbf{i}+900\mathbf{j} = 570\mathbf{i}+1080\mathbf{j}
As the starting point is at the origin, the distance is the magnitude of this position vector.
\displaystyle \sqrt{{{570}^{2}}+{{108}^{2}}}=\sqrt{324900+1166400}=\sqrt{1491300}=1221\ \text{m}
