The figure(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle, starting from θ0=0 rad at t=0s. Draw the angular-position-versus-time graph. Include an appropriate scale on both axes.

Question

contestada

Respuesta :

Аноним Аноним · Answer 1 · 2021-10-09T14:07:23+02:00

Answer:

As position is the product of velocity and time, the area under the velocity-time curve will be the position.

Between 2 and 4 s, the area increases linearly to 20(4) = 80

Between 4 and 6 s, the additional area is 0(2) = 0

Between 6 and 8 s the area increases linearly by -10(2) = -20

which is a decrease.

Answer Link

batolisis batolisis · Answer 2 · 2021-10-20T15:33:00+02:00

In the angular-position vs time graph; The angular position is derived from the area under the velocity vs time graph

Attached below is the angular position vs time graph

Given that : Angular position = Angular Velocity * time

As Time increases from 0 to 4 secs the area increases from 0 to 80 rad

As Time increases from 4 to 6 secs the area remains constant i.e. 0 rad

As Time increases from 6 to 8 secs the area under the graph decreases 80 to 60 rad

Hence we can conclude that the Angular position of the graph is derived from the area under the velocity vs time graph.

Learn more : https://brainly.com/question/23832769