A particle moves such that its position vector $\vec{r}(t)=\cos \omega t \hat{i}+\sin \omega t \hat{j}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\vec{v}(t)$ and acceleration $\vec{a}(t)$ of the particle:

(A) $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed towards the origin.

(B) $\vec{v}$ and $\vec{a}$ both are parallel to $\vec{r}$

(C) $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed away from the origin.

(D) $\vec{v}$ and $\vec{a}$ both are perpendicular to $\vec{r}$

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Kinematics

A particle moves such that its position vector

r (t) = cos ω t \hat{i} + sin ω t \hat{j}

where

ω

is a constant and

t

is time. Then which of the following statements is true for the velocity

v (t)

and acceleration

a (t)

of the particle:

(A) $v$ is perpendicular to $r$ and $a$ is directed towards the origin.✓ Correct
(B) $v$ and $a$ both are parallel to $r$
(C) $v$ is perpendicular to $r$ and $a$ is directed away from the origin.
(D) $v$ and $a$ both are perpendicular to $r$

Answer: (A)

Differentiating

r = cos ω t \hat{i} + sin ω t \hat{j}

gives

v = ω (- sin ω t \hat{i} + cos ω t \hat{j})

and

a = - ω^{2} (cos ω t \hat{i} + sin ω t \hat{j}) = - ω^{2} r

. Since

a = - ω^{2} r

, the acceleration points back along

r

toward the origin, and

v \cdot r = ω (- sin ω t cos ω t + cos ω t sin ω t) = 0

, so the velocity is perpendicular to

r

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option