24. I FIGURE EX4.24 shows the angular-position-versus-time graph for a particle moving in a circle. What is the particle's angular velocity at (a) $t=1\text{s}$, (b) $t=4\text{s}$, and (c) $t=7\text{s}$?

The angular velocity of the particle is,

$\omega =\frac{{\theta }_1-{\theta }_0}{t_1-t_0}$

At $t_{{}^1}=1\text{s}$the angular position of the particle is${\theta }_1=2\pi$

At $t_0=0\text{s}$the initial angular position of the particle is${\theta }_0=0$

The angular velocity of the particle is,

$\omega =\frac{\left(2\pi \text{rad}\right)-0\text{rad}}{1\text{s}-0\text{s}}=6.28\text{rad}/\text{s}$

The angular velocity of the particle at$t=1\text{s}$ is$6.28\text{rad}/\text{s}$

The angular velocity of the particle is,

$\omega =\frac{{\theta }_4-{\theta }_2}{t_4-t_2}$

The particle at $t_4=4\text{s}$, the angular position is ${\theta }_4=4\pi$

The particle at $t_2=2\text{s}$, the initial angular position is${\theta }_2=4\pi$

The angular velocity of the particle is,

$\omega =\frac{\left(4\pi \text{rad}\right)-\left(4\pi \text{rad}\right)}{4\text{s}-2\text{s}}=0\text{rad}/\text{s}$

Thus, the angular velocity of the particle at$t=4\text{s}$ is$0\text{rad}/\text{s}$ .

The angular velocity of the particle is,

$\omega =\frac{{\theta }_7-{\theta }_6}{t_7-t_6}$

At $t_7=7\text{s}$time the angular position of the particle is${\theta }_7$

At $t_6=6\text{s}.$time the initial angular position of the particle is${\theta }_6$

The angular velocity of the particle is,

$\omega =\frac{\left(0\text{rad}\right)-\left(2\pi \text{rad}\right)}{7\text{s}-6\text{s}}=-2\pi \text{rad}/\text{s}=-6.28\text{rad}/\text{s}$

Thus, the angular velocity of the particle at $t=7\text{s}$is$-6.28\text{rad}/\text{s}$