Exhibit 10-2 The figure below shows a graph of angular velocity versus time for a man bicycling around a circular track. Use this exhibit to answer the following question(s). -Refer to Exhibit 10-2.What is his average angular acceleration,in rad/s²,in the period from t = 6 min to t = 8 min? A)0 B) $- \frac { \pi } { 90 }$ C) $- \frac { \pi } { 30 }$ D) $+ \frac { \pi } { 30 }$ E) $+ \frac { \pi } { 90 }$

Question

Quizplus · Accepted Answer

The angular acceleration is the slope of the angular velocity-time graph. From t = 6 min to t = 8 min, the angular velocity decreases from 2π rad/s to π rad/s. The change in angular velocity is -π rad/s. The change in time is 2 min, or 120 s. Therefore, the average angular acceleration is:$$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{-π rad/s}{120 s} = - \frac { \pi } { 30 } rad/s^2$$

Exhibit 10-2 the Figure Below Shows a Graph of Angular −π90- \frac { \pi } { 90 }−90π​

Exhibit 10-2 the Figure Below Shows a Graph of Angular $- \frac { \pi } { 90 }$