91Ó°ÊÓ

In the given time from 0 s to 4s, the angular velocity of the particle is$20\text{rad}/\text{s}$

The angular position of the particle is

$\begin{array}{l}\mathrm{Δ}{\mathrm{θ}}_1={\mathrm{Ó\neg }}_1\text{ }\mathrm{Δ}{t}_1\\ \mathrm{Δ}{\mathrm{θ}}_1=(20\text{rad}/\text{s})(4\text{s}−0\text{s})\\ \mathrm{Δ}{\mathrm{θ}}_1=80\text{rad}\end{array}$

The angular velocity of the particle is constant but time is changing so the angular position of the particle will increase from$0\text{  â¶ÄŠr²¹»å}$ to$80\text{  â¶ÄŠr²¹»å}$ linearly.

For 4s to 8s, the angular velocity of the particle is$0\text{  â¶ÄŠr²¹»å}/\text{s}$

The angular position of the particle is

$\begin{array}{l}\mathrm{Δ}{\mathrm{θ}}_2={\mathrm{Ó\neg }}_2\mathrm{Δ}{t}_2\\ \mathrm{Δ}{\mathrm{θ}}_2=(0\text{rad}/\text{s})(6\text{s}−4\text{s})=0\text{rad}\end{array}$

The angular velocity of the particle is zero, so the angular position of the particle will remain same at the initial position of $80\text{  â¶ÄŠr²¹»å }$. So the angular position is linear flat.

From 6s to 8s , the angular velocity of the particle is $−10\text{rad}/\text{s}$

$\begin{array}{l}\mathrm{Δ}{\mathrm{θ}}_3={\mathrm{Ó\neg }}_3\mathrm{Δ}{t}_3\\ \mathrm{Δ}{\mathrm{θ}}_3=(−10\text{rad}/\text{s})(8\text{s}−6\text{s})\\ \mathrm{Δ}{\mathrm{θ}}_3=−20\text{rad}\end{array}$

The angular velocity of the particle is constant but time is changing so the angular position of the particle will decrease from 80 rad to 60 rad linearly in time form 8 s to 6 s.

The following graph shows the graph of angular position versus time for a particle moving in circular path.