A particle moves such that its position vector r(t) = cos ω t i + sin ω t 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 and a both are perpendicular to r (b) v and a both are parallel to r (c) v is perpendicular to r and a is directed towards the origin (d) v is perpendicular to r and a is directed away from the origin

Answer: (c). v = ω(-sin ω t i + cos ω t j), and v· r = 0, so v ⊥ r a = -ω² r, directed towards the origin So the answer is (c).

A particle moves such that its position vector r(t) = cos ω t i + sin… | JEE Main 2020 (8 Jan, Shift 2)

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Motion in a Plane (2-D Kinematics)

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:

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Answer: (c)

$\vec v = \omega(-\sin\omega t\,\hat i + \cos\omega t\,\hat j)$ , and $\vec v\cdot\vec r = 0$ , so $\vec v \perp \vec r$
$\vec a = -\omega^2\vec r$ , directed towards the origin
So the answer is (c).

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Motion in a Plane (2-D Kinematics) question from JEE Main 2020 (8 Jan, Shift 2)

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