Question

EAPERT
EXPERT CET MOCK TEST-02
1
A particle moves so that its position vector varies with time as
P= A coswt i + A sinwtj. Then the angle between the position ve
particle and the velocity at any time is...
1) zero
2) 45°
3) 180°
4) 90°​

Answer 1

$\Large\boxed{\sf{\quad(4)\quad\!90^o\quad}}$

The position vector is given as,

$\longrightarrow\sf{\overrightarrow{\sf{P}}=A\cos(\omega t)\ \hat i+A\sin(\omega t)\ \hat j}$

We know velocity vector is the first derivative of the position vector with respect to time.

$\longrightarrow\sf{\overrightarrow{\sf{v}}=\dfrac{d\overrightarrow{\sf{P}}}{dt}}$

$\longrightarrow\sf{\overrightarrow{\sf{v}}=\dfrac{d}{dt}\left[A\cos(\omega t)\ \hat i+A\sin(\omega t)\ \hat j\right]}$

$\longrightarrow\sf{\overrightarrow{\sf{v}}=-A\omega\sin(\omega t)\ \hat i+A\omega\cos(\omega t)\ \hat j}$

The magnitude of the position vector is,

$\longrightarrow\sf{\left|\overrightarrow{\sf{P}}\right|=\sqrt{(A\cos(\omega t))^2+(A\sin(\omega t))^2}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{P}}\right|=\sqrt{A^2\cos^2(\omega t)+A^2\sin^2(\omega t)}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{P}}\right|=\sqrt{A^2\left(\cos^2(\omega t)+\sin^2(\omega t)\right)}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{P}}\right|=A}$

And the magnitude of the velocity vector is,

$\longrightarrow\sf{\left|\overrightarrow{\sf{v}}\right|=\sqrt{(-A\omega\sin(\omega t))^2+(A\omega\cos(\omega t))^2}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{v}}\right|=\sqrt{A^2\omega^2\sin^2(\omega t)+A^2\omega^2\cos^2(\omega t)}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{v}}\right|=\sqrt{A^2\omega^2\left(\sin^2(\omega t)+\cos^2(\omega t)\right)}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{v}}\right|=A\omega}$

Let $\theta$ be the angle between $\overrightarrow{\sf{P}}$ and $\overrightarrow{\sf{v}}.$ Then,

$\longrightarrow\sf{\theta=\cos^{-1}\left(\dfrac{\overrightarrow{\sf{P}}\cdot\overrightarrow{\sf{v}}}{\left|\overrightarrow{\sf{P}}\right|\cdot\left|\overrightarrow{\sf{v}}\right|}\right)}$

$\longrightarrow\sf{\theta=\cos^{-1}\left(\dfrac{\left(A\cos(\omega t)\ \hat i+A\sin(\omega t)\ \hat j\right)\cdot\left(-A\sin(\omega t)\ \hat i+A\cos(\omega t)\ \hat j\right)}{A\cdot A\omega}\right)}$

$\longrightarrow\sf{\theta=\cos^{-1}\left(\dfrac{A\cos(\omega t)\cdot -A\sin(\omega t)+A\sin(\omega t)\cdot A\cos(\omega t)}{A^2\omega}\right)}$

$\longrightarrow\sf{\theta=\cos^{-1}\left(\dfrac{-A^2\sin(\omega t)\cos(\omega t)+A^2\sin(\omega t)\cos(\omega t)}{A^2\omega}\right)}$

$\longrightarrow\sf{\theta=\cos^{-1}\left(\dfrac{0}{A^2\omega}\right)}$

$\longrightarrow\sf{\theta=\cos^{-1}(0)}$

$\longrightarrow\sf{\underline{\underline{\theta=90^o}}}$