Question

THe position vector of a particle is r = (acoswt)i^ + (asinwt)j^. Find the velocity of the particle and the angle between the position and velocity vector?

Answer 1

we know, velocity is the rate of change of position.

e.g., $v=\frac{dr}{dt}$

now, differentiating r with respect to time, t .

$v=\frac{dr}{dt}=\frac{d}{dt}(acos\omega t)\hat{i}+\frac{d}{dt}(asin\omega t)\hat{j}$

$v=-a\omega(sin\omega t)\hat{i}+a\omega(cos\omega t)\hat{j}$

hence, velocity, $\vec{v}=a\omega(-sin\omega t\hat{i}+cos\omega t\hat{j})$

now, angle between velocity and position of particle.

$cos\theta=\frac{\vec{v}.\vec{r}}{|\vec{v}|.|\vec{r}|}$

so, $\vec{v}.\vec{r}=(-a\omega sin\omega t\hat{i}+a\omega cos\omega t\hat{j}).(acos\omega t\hat{i}+asin\omega t\hat{j})$

= $-a^2\omega sin\omega t. cos\omega t+ a^2\omega cos\omega t. sin\omega t$

= 0

so, cos$\theta$ = 0 = cos90°

$\theta=90^{\circ}$

hence, angle between velocity and position is 90°

Answer 2

Answer:

perpendicular to distance of positon vector