Answer:[tex]\hat{v}=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}[/tex]
Explanation:
Given
Time Period [tex]T=6.5 s[/tex]
Position Vector [tex]\vec{r}=6\hat{i}-6\hat{j}[/tex]
Since it is moving in a circular path therefore magnitude of position vector will give the radius of circular path
[tex]|r|=\sqrt{6^2+6^2}[/tex]
[tex]|r|=6\sqrt{2}[/tex]
[tex]distance=velocity\times time[/tex]
[tex]6\sqrt{2}=velocity\times 6.5[/tex]
[tex]velocity=\frac{6\sqrt{2}}{6.5}[/tex]
[tex]v=1.305[/tex]
this velocity is at angle of [tex]\tan \theta =\frac{-6}{6}[/tex]
[tex]\theta =-45^{\circ}[/tex]
velocity [tex]\vec{v}=v\cos \theta +v\sin \theta [/tex]
[tex]\vec{v}=1.305\cos (-45)+1.305\sin (-45)[/tex]
Therefore unit velocity vector
[tex]\hat{v}=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}[/tex]
