Solution 19.7d
From Mechanics
(Difference between revisions)
(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( \fra...) |
Current revision
Using the result for the position vector \displaystyle \mathbf{r} from part c),
\displaystyle \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}
