Question

A particle is in uniform circular motion about the origin of an xy coordinate system, moving counter-clockwise with a period of 6.50 s. At one instant, its position vector (from the origin) is r with arrow = (6.00 m)i hat − (6.00 m)j. At that instant, what is its velocity in unit-vector notation?

Answer 1

Answer:$\hat{v}=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}$

Explanation:

Given

Time Period $T=6.5 s$

Position Vector $\vec{r}=6\hat{i}-6\hat{j}$

Since it is moving in a circular path therefore magnitude of position vector will give the radius of circular path

$|r|=\sqrt{6^2+6^2}$

$|r|=6\sqrt{2}$

$distance=velocity\times time$

$6\sqrt{2}=velocity\times 6.5$

$velocity=\frac{6\sqrt{2}}{6.5}$

$v=1.305$

this velocity is at angle of $\tan \theta =\frac{-6}{6}$

$\theta =-45^{\circ}$

velocity $\vec{v}=v\cos \theta +v\sin \theta$

$\vec{v}=1.305\cos (-45)+1.305\sin (-45)$

Therefore unit velocity vector

$\hat{v}=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}$