Work, Energy and Power NEET PYQ | Previous Year Questions PDF with Solutions

Velocity is perpendicular to $\vec{r}$ and acceleration is directed towards the origin.
Velocity is perpendicular to $\vec{r}$ and acceleration is directed away from the origin.
Velocity and acceleration both are perpendicular to $\vec{r}$ .
Velocity and acceleration both are parallel to $\vec{r}$ .

(1)

Given, $\vec{r} = \cos ω t \hat{x} + \sin ω t \hat{y}$

$∴ \vec{v} = \frac{d \vec{r}}{d t} = - ω \sin ω t \hat{x} + ω \cos ω t \hat{y}$

$\vec{a} = \frac{d \vec{v}}{d t} = - ω^{2} \cos ω t \hat{x} - ω^{2} \sin ω t \hat{y} = - ω^{2} \vec{r}$

Since position vector $(\vec{r})$ is directed away from the origin, so, acceleration $(- ω^{2} \vec{r})$ is directed towards the origin.

Also,

$\vec{r} \cdot \vec{v} = (\cos ω t \hat{x} + \sin ω t \hat{y}) \cdot (- ω \sin ω t \hat{x} + ω \cos ω t \hat{y})$

$= - ω \sin ω t \cos ω t + ω \sin ω t \cos ω t = 0$

$\Rightarrow \vec{r} ⊥ \vec{v}$

(1)

Two vectors $\vec{A}$ and $\vec{B}$ are orthogonal to each other, if their scalar product is zero i.e. $\vec{A} \cdot \vec{B} = 0$

Here, $\vec{A} = \cos ω t \hat{i} + \sin ω t \hat{j}$

and $\vec{B} = \cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j}$

$∴$ $\vec{A} \cdot \vec{B} = (\cos ω t \hat{i} + \sin ω t \hat{j}) \cdot (\cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j})$

$= \cos ω t \cos \frac{ω t}{2} + \sin ω t \sin \frac{ω t}{2} = \cos (ω t - \frac{ω t}{2})$

But $\vec{A} \cdot \vec{B} = 0$ (as $\vec{A}$ and $\vec{B}$ are orthogonal to each other)

$∴$ $\cos (ω t - \frac{ω t}{2}) = 0$

$\cos (ω t - \frac{ω t}{2}) = \cos \frac{π}{2} or ω t - \frac{ω t}{2} = \frac{π}{2}$

$\frac{ω t}{2} = \frac{π}{2} or t = \frac{π}{ω}$

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