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Solution 19.7d

From Mechanics

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(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 (16:34, 11 October 2010) (edit) (undo)
(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 r from part c),

r(20)=2203−820+90i+43202−245203+20j=3930i−1347jDistance = 39302+13472=4154 m 

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