Thus the for the given velocity versus time graph the average acceleration of the object during segment B is 0 m/s/s.Hence, option A is the correct option.
Average acceleration- Acceleration is defined as the rate of change of acceleration. Average acceleration for an interval can be defined as the rate of change of velocity for that interval per time.
Let [tex]\Delta v[/tex] be the change in velocity and [tex]\Delta t[/tex] be the change in time in a given time interval. Then average acceleration is,
[tex]a= \dfrac{\Delta v}{\Delta t}[/tex]
At segment B, It starts at 4.5 seconds and ends at 10 seconds. The total time interval for this segment is 5.5 seconds.Therefore,
[tex]\Delta t=5.5[/tex]
Now the change in velocity for segment B. In segment B the velocity does not change at any point and keeps the value of 20 m/sec at the time interval.
According to the definition of the average acceleration when an object is changing its velocity (whether by a constant amount or a varying amount) then it is an accelerating object. And an object with a constant velocity is not accelerating.
Thus the for the given velocity versus time graph the average acceleration of the object during segment B is 0 m/s/s.Hence, option A is the correct option.
For more about the acceleration, follow the link below,
https://brainly.com/question/1479537
