Ian: New page: Using the result for the position vector $\mathbf{r}$ from part c), $\begin{align} & \mathbf{r}(20)=\left( \frac{{{20}^{3}}}{2}-8\times 20+90 \right)\mathbf{i}+\left( \fra...$

Ian — Mon, 11 Oct 2010 16:34:58 GMT

New page: Using the result for the position vector <math>\mathbf{r}</math> from part c), <math>\begin{align} & \mathbf{r}(20)=\left( \frac{{{20}^{3}}}{2}-8\times 20+90 \right)\mathbf{i}+\left( \fra...

New page

Using the result for the position vector <math>\mathbf{r}</math> from part c),

<math>\begin{align}
& \mathbf{r}(20)=\left( \frac{{{20}^{3}}}{2}-8\times 20+90 \right)\mathbf{i}+\left( \frac{3\times {{20}^{2}}}{4}-\frac{5\times {{20}^{3}}}{24}+20 \right)\mathbf{j}=3930\mathbf{i}-1347\mathbf{j} \\
& \\
& \text{Distance }=\text{ }\sqrt{{{3930}^{2}}+{{1347}^{2}}}=4154\text{ m} \\
\end{align}</math>

Solution 19.7d - Revision history

Ian: New page: Using the result for the position vector \mathbf{r} from part c), \begin{align} & \mathbf{r}(20)=\left( \frac{{{20}^{3}}}{2}-8\times 20+90 \right)\mathbf{i}+\left( \fra...

Ian: New page: Using the result for the position vector $\mathbf{r}$ from part c), $\begin{align} & \mathbf{r}(20)=\left( \frac{{{20}^{3}}}{2}-8\times 20+90 \right)\mathbf{i}+\left( \fra...$