To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.
PART A)
The time taken to travel a distance of 250km with a speed of 95km/h is
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{250km}{95km/h}[/tex]
[tex]t = 2.63h[/tex]
Time taken for the lunch is
[tex]t = 1h[/tex]
The time taken travel a distance of 250km with a speed of 55km/h
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{250}{55}[/tex]
[tex]t = 4.54h[/tex]
The total time taken is
[tex]t = t_{outgoing}+t_{lunch}}t_{return}[/tex]
[tex]t = 2.63+1+4.54[/tex]
[tex]t = 8.17h[/tex]
The average speed is the ratio of total distance and total time
[tex]v = \frac{250+250}{8.17}[/tex]
[tex]v = 61.15km/h[/tex]
PART B)
As the displacement is zero the average velocity is zero.
The average speed is 61.12 km/hr and the average velocity is zero and this can be determined by using the kinematics equation.
Given :
A complete round-trip in which the outgoing 250 km is covered at 95 km/h followed by a 1.0-hour lunch break, and then return 250 km is covered at 55 km/h.
The time is taken for outgoing 250 Km with the speed of 95 km/hr is given by:
[tex]\rm t = \dfrac{d}{t}[/tex]
[tex]\rm t = \dfrac{250}{95}[/tex]
t = 2.63 hrs
To total time taken including lunch for outgoing is:
t = 2.63 + 1
t = 3.63 hrs
Now, the time taken for returning 250 Km with the speed of 55 km/hr is given by:
[tex]\rm t = \dfrac{250}{55}[/tex]
t = 4.55 hrs
So, the total time taken to complete the round trip is:
t = 4.55 + 3.63
t = 8.18 hrs
Now, the average speed is given by:
[tex]\rm v = \dfrac{250+250}{8.18}[/tex]
v = 61.12 km/hr
B) When the displacement is zero the average velocity becomes zero.
For more information, refer to the link given below:
https://brainly.com/question/10581158
