Q: A particle moves such that its position vector $\displaystyle \vec{r(t)} = cosωt \hat{i} + sinωt \hat{j} $ ; where ω is constant and t is time . Then which of the following statement is true for the velocity $\vec{v(t)$ and acceleration $\vec{a(t)}$ of the particle ?
(a) $\displaystyle \vec{v}$ and $\vec{a}$ both are parallel to $\vec{r}$
(b) $\displaystyle \vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed away from origin .
(c) $\displaystyle \vec{v}$ and $\vec{a}$ both are perpendicular to $\vec{r}$
(d) $\displaystyle \vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed towards the origin .
Ans: (d)
Sol: $\displaystyle \vec{r(t)} = cosωt \hat{i} + sinωt \hat{j} $
$\displaystyle \vec{v(t)} = \frac{d \vec{r}}{dt} = – ω sin ωt \hat{i} + ω cos ωt \hat{j} $
$\displaystyle \vec{a(t)} = \frac{d \vec{v}}{dt} = – ω^2 cos ωt \hat{i} – ω^2 sin ωt \hat{j} $
$\displaystyle \vec{a(t)} = – ω^2 ( cos ωt \hat{i} + sin ωt \hat{j}) $
$\displaystyle \vec{a(t)} = – ω^2 \vec{r(t)} $
Negative sign indicates that acceleration is directed towards origin .
$\displaystyle \vec{v(t)}.\vec{r(t)} = (- ω sin ωt \hat{i} + ω cos ωt \hat{j}).(cosωt \hat{i} + sinωt \hat{j}) $
$\displaystyle \vec{v(t)}.\vec{r(t)} = – ω sin ωt \; cosωt + ω sin ωt \; cosωt $
$\displaystyle \vec{v(t)}.\vec{r(t)} = 0 $
$\vec{v(t)}$ is perpendicular to $\vec{r(t)}$