Question

A particle is moving in $x-y$ plane. The positive vector of the particle at time $t$ is $r=b(1-\cos \omega t)\hat{i}+b\sin \omega t\hat{j}$, where $b$ and $\omega$ are positive constants. The angle between acceleration and velocity at instant $t$ is

• A. $0$
• B. $\pi /2$
• C. None of these
• D. $\pi /4$

Answer 1

The correct option is B $\pi /2$

Given, displacement
$\overrightarrow{r}=b(1-\cos \omega t)\hat{i}+b\sin \omega t\hat{j}$
Velocity,
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=b\omega \sin \omega t\hat{i}+b\omega \cos \omega t\hat{j}$
Acceleration,
$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=b{\omega }^2\cos \omega t\hat{i}-b{\omega }^2\sin \omega t\hat{j}$
$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{v}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{v}\right|}$
$\overrightarrow{a}.\overrightarrow{v}=(b{\omega }^2\cos \omega t\hat{i}-b{\omega }^2\sin \omega t\hat{j}).(b\omega \sin \omega t\hat{i}+b\omega \cos \omega t\hat{j})$
$\Rightarrow \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{v}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{v}\right|}=0$
$\theta =\frac{\pi }{2}$