The displacement is a vector that is calculated by subtracting the final and initial positions of an object that has made a movement.
We know that initially the object was at the origin and moved to the -8m position.
Suppose that the movement was made on the x-axis. So:
d = [-8 - (0)]x
d = -8x
Regarding the second question.
When the acceleration is constant then the speed changes at a constant rate. Therefore the graph of velocity vs. time will look like a line with a negative slope (if the acceleration is negative) or a line with a positive slope (if the acceleration is positive)
For example. Suppose that the acceleration of gravity is 10 ![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
Now suppose you drop an object from a building.
At the initial moment (second 0) the speed is 0 m/s.
After 1 second the speed of the object will be 10 m/s.
After 2 seconds the speed of the object will be 20 m/s.
After 4 seconds the speed will be 40 m/s.
The graph of this example is shown in the attached figure. Note that it is a straight line.
We're given the velocity at two points, and the time between the mentioned two points.
What we'll be calculating is the average acceleration (not to be confused with instantaneous acceleration).
Given two points, lets call them initial and final, the average acceleration is calculated as:
![a= \frac{v_f-v_0}{t}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7Bv_f-v_0%7D%7Bt%7D)
In which
![v_f](https://tex.z-dn.net/?f=v_f)
is the final velocity (in your problem equal to 1m/s),
![v_0](https://tex.z-dn.net/?f=v_0)
is the initial velocity (in your problem equal to 16 m/s), and
![t](https://tex.z-dn.net/?f=t)
the time between two points (in your problem equal to 2s).
So, we plug-in our known values:
The answer is -7.5m/s^2.