Not sure if I understand, please help?

answers:
Philomel say: 1. D
2. D

sojsail say: RealPro s definition of acceleration is incorrect. If V starts at 0, and after some unknown time is 10 m/s, that is a velocity change of 10 m/s. But that is not a measure of acceleration. To calculate acceleration, you need to know how quickly it made that change.
Velocity has units of m/s. 10 m/s means that each second you have gone another 10 meters. You need to understand why the units of acceleration have the seconds squared.
The acceleration due to gravity is 9.8 m/s^2.
That means that each second, the downward velocity increases from an initial velocity, Vi, to Vi+9.8 m/s.
So let us assume you drop a ball from the roof.
Vi = 0.
After 1 second, V = Vi + 9.8 m/s = 9.8 m/s
After another second, V = 9.8 m/s + 9.8 m/s = 19.6 m/s
After another second, V = 19.6 m/s + 9.8 m/s = 29.4 m/s
And so on.
The units of acceleration can be written (m/s)/s. (I am sure that you will agree that algebraically, it is equivalent to m/s^2.) Perhaps (m/s)/s would emphasize that the m/s are changing by the indicated amount each second.
The formula that does that for you is V = U + a*t

Andrew Smith say: I can see your problem. You don't understand about the different scanning for different materials.
Your question is in black and in white. Yet you have scanned it as a colour document.
This makes it impossible to read.
Now given that you have been studying your physics you would know that
a) velocity is the slope of a displacement time graph
and
b) an acceleration is a change in velocity.
So there can only be an acceleration if the slope is ALTERING.
Clearly the slope is altering in both region 2 and region 4.
In the subsequent question, the acceleration is positive if the velocity is increasing. This is true even if the velocity starts off as a negative value.
So you only need to look at C and D
In D the slope is negative ( meaning a negative velocity ) but the slope is becoming steeper so that the velocity is becoming more negative. ie a negative acceleration.
In C the slope is positive ( meaning a positive velocity) but the slope is becoming larger so the positive velocity is increasing. ie a positive acceleration.

RealPro say: Of course you don't, it's not even the right subject!
The definition of acceleration is so super complicated it's gonna blow your socks off.
Acceleration = change in velocity.
When the positiontime graph is a straight line, that means there's no acceleration (because velocity is the slope and a straight line has constant slope).
When the graph is a curvy line, that means there is acceleration
(Increasing slope means positive acceleration and decreasing means negative)

Dr W say: "D" is the correct answer for both questions
in the first question, the object is speeding up in interval II and slowing down in interval IV.
in the 2nd question, plots A and B are constant velocity (zero acceleration), plot "C" is + acceleration D is 
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velocity = Δd / Δt = slope of position vs time plot. If that slope is constant (and not zero.. meaning not a horizontal line) the object is moving with constant velocity. since acceleration = Δ v / Δt.. if the position vs time plot is linear, then velocity is constant and acceleration = 0

Chasel say: tmfxivlt

Byron say: lozvkajt
