Answer:
b) 68,9 km/h a) picture
Explanation:
In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.
Using this formula we pass time to hours:

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.
Using the values of the speed we calculate the distance in each interval

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).
Finally, we compute the average speed with the distance over time:

Soft target by impact and its contribution to indirect bone fractures.
Answer:
Starts on Saturday, June 1
and ends on
Saturday, November 30
Explanation:
Answer:
I would love to help but I don't know I'm so sorry
Answer:
A) OA, AB, BC
B) 25m/s^2
C) see explanation
D) 25
E) Rest
Explanation:
From the Velocity time graph shown:
The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.
Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.
-ve slope = BC
B) Acceleration of body in path OA.
Acceleration = change in Velocity / time
Acceleration = (150 - 0) / 6
Acceleration = 150/6 = 25m/s^2
C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).
D) Length of BC
BC corresponds to the distance moved, that velocity / time
Velocity = 150 ; time = 6
Therefore Distance (BC) = 150/6 = 25
E.) Velocity =0 ; Hence body is at rest