Question

A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn.
The coordinates(meters) of the rabbit's position as function of time t(seconds) are given by

$x\left(t\right)=-\frac{t^2}{2}+5t+20Andy\left(t\right)=t^2-10t+30$

(a)At t=10s, what is the rabbot's position vector $\overrightarrow{r}$ in unit-vector notation and in magnitude-angle notation?

• A. $20\hat{i}-30\hat{j};20\sqrt{13}\&ta{n}^{-1}\left(\frac{3}{2}\right)$
• B. $30\hat{i}-20\hat{j};10\sqrt{13}\&ta{n}^{-1}\left(\frac{-2}{3}\right)$
• C. $20\hat{i}+30\hat{j};10\sqrt{13}\&ta{n}^{-1}\left(\frac{3}{2}\right)$
• D. $30\hat{i}+20\hat{j};10\sqrt{3}\&ta{n}^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is C

$20\hat{i}+30\hat{j};10\sqrt{13}\&ta{n}^{-1}\left(\frac{3}{2}\right)$

After 10 seconds, to find the rabbit's position, put t=10 for x(t) and y(t)
$x=\frac{-100}{2}+5\times 10+20$
=20 m
On $y=t^2-10t+30$
$=10\times 10-10\times 10+30$
=30 m
So position in vector notation = $20\hat{i}+30\hat{j}$
Magnitude = $\sqrt{(20{)}^2+(30{)}^2}=\sqrt{400+900}=10\sqrt{13}$
Angle =$ta{n}^{-1}\left(\frac{y}{x}\right)=ta{n}^{-1}\left(\frac{30}{20}\right)=ta{n}^{-1}\left(\frac{3}{2}\right)$