Answer to Question #90496 in Mechanics | Relativity for Mirabella

Question

A rabbit run across a parking lot. The coordinate of the rabbits as a function of time t are x=0.3t square +7.2t square+28 Y=0.22t square-9.1t+80 with the in seconds and x and y in meters. At the =15s, what is the rabbits position vector r in unit vector notation and as magnitude and an angle

Answer 1

$$x= 0.3t^2+7.2t^2+28 = 7.5t^2+28\\ y=0.22t^2-9.1t+80$$

$$\vec{r}(t)=x(t) \cdot \vec{i} +y(t)\cdot \vec{j}\\ \vec{r}(15) = (7,5\cdot 15^2+28)\cdot\vec{i}+(0,22\cdot 15^2-9.1 \cdot 15 +80)\vec{j}=\\ = 1715.5 \cdot\vec{i}-7\cdot\vec{j}$$

$$\lvert \vec{r}(15) \rvert = \sqrt{1715.5^2+7^2} = 1715.51 m \\ \phi(15) = tan^{-1}(\frac{y(15)}{x(15)}) = tan^{-1}(\frac{-7}{1715.5}) = -0.23 deg$$