Kinematics with graphs (Position-time graphs)

General overview: When studying Physics (especially Kinematics), we may often encounter situations when the interpretation of a graph is needed. Many useful information regarding a motion can be obtained by correctly interpreting a motion graph. Thus, graphs represent another "language" of Physics which provides the necessary information besides the normal language of words and numbers.

There are some generally accepted rules while plotting a Kinematics graph. They are:

1- The horizontal axis always represents the time. Thus, it is called the time (t) axis. It can extend only to the right of the origin because the time is always positive.

2- In the vertical axis we can put one of the 3 other quantities studied in Kinematics (position, velocity or acceleration). They can be positive and negative, therefore the vertical axis can extend both above and below the origin.

3- Unlike in Maths, it is not important to use the same criteria of units segmentation, because in Kinematics graphs, the quantities represented in the horizontal and vertical axis are totally of different nature. For example, in the horizontal axis, 1 unit may represent 2 s and in the vertical axis 1 unit may represent 10 m.

4- Within the same axis, we must use the same criteria of units' segmentation. for example, It is not allowed that at the beginning 1 cm of the graph to represent 3m distance and then, within the same axis, 1 cm of the graph to represent 10 m of distance.

There are 3 types of graphs in Kinematics: a) position-time graphs,  b) velocity-time graphs  and c) acceleration-time graphs. Let's explain them more in detail.

Position-time graph

In this graph, the horizontal axis represents the time and the vertical axis the position. If the graph is linear, the motion is uniform, because for the same intervals of time we have the same changes in position. As you may know, the slope of a linear graph is constant. But as this motion represents a uniform motion (with constant velocity), we obtain a very important rule regarding the graphs in kinematics:

"The slope of a position-time graph gives the velocity"

In Maths, the value of the slope (k) is obtained by using the equation  \(k=\frac{\Delta{y}}{\Delta{x}}\) 

Here, \(k\) corresponds to velocity, \(\Delta{y}\) to the change in position and \(\Delta{x}\) to the time elapsed. Therefore, the above equation becomes


where \(\Delta{\vec{x}}\) is the displacement and \(\Delta{t}\) or simply t (as the initial time is almost always zero) is the time elapsed during the motion.

Example 1: Find the velocity for the motion described in the graph below.

Solution: We already know that the slope of the position-time graph gives the velocity. It is obvious that there would be enough to take the coordinates of only 2 random points of the graph in order to calculate the slope (as the graph is linear). Thus, for the 2 points taken in the graph above, we have the values: 

\(x_1=4m; x_2=8m; t_1=2s; t_2=4s\)

Therefore, \(\vec{v}=\Delta{\vec{x}}/\Delta{t} = (\vec{x_2}-\vec{x_1})/(t_2-t_1)=(8m-4m)/(4s-2s)=4m/2s=2m/s\)

The same result would be obtained for any couple of points studied. For example if you get the extremities of the graph \(x_1=0m; x_2=12m; t_1=0s; t_2=6s\) you will still obtain a value of 2m/s for the velocity because\((12m-0m)/(6s-0s)=12m/6s=2m/s\).


1- When the graph is horizontal, there is no motion as the position doesn't change. Thus, the object is at rest and its velocity is zero.

2- For any decreasing linear position-time graph, the slope (velocity) is negative. It means that the object is moving at constant velocity towards the negative direction of motion.                                                              

When the motion is not uniform, the position-time graph will not be anymore a straight line but a curved one (more exactly a parabola). For the uniformly accelerated motion, this parabola will have a positive coefficient (it will wave a fixed minimum but not a fixed maximum). For the same intervals of time, the object will have more and more displacement than before. On contrary, for the uniformly deccelerated motion, the parabola will have a negative coefficient. It will have a fixed maximum and for equal intervals of time the object will have less and less displacement than before. 


Example 2: For the motion given in the graph below determine: 

a) Acceleration \((\vec a)\)

b) Initial velocity \((\vec {v_0})\)

Solution: As you can easily see, the motion is uniformly accelerated one because the graph is a parabola. Thus, by knowing the equation of parabola \(y=mx+nx^2\) and bringing at mind the second equation of uniformly accelerated motion \(\Delta{x}=v_0t+{at^2 \over 2}\) we see a similarity between them, where \((y)\) of parabola represents the displacement \((\Delta{x})\)\((m)\) of parabola represents the initial velocity \((v_0)\) and \((n)\) of parabola represents half of acceleration \(({a \over 2})\).

Let's take 2 points of the graph and substitute to the equation of motion, for example [t1=2s; x1=1m] and [t2=4s; x2=4m].

Therefore we have 2 equations of motion that create a system of equations:

\(\begin{cases} 1=m \cdot 2 + n \cdot 2^2\\ 4=m \cdot 4 + n \cdot 4 \end{cases}\)

\(\begin{cases} 2m +4n=1\\ 4m+16n=4 \end{cases}\)

The second equation is simplified by 4. Thus, 

\(\begin{cases} 2m+4n=1\\ m+4n=1 \end{cases}\)

The second equation is multiplied by -1 and then both equations are added

\(\begin{cases} 2m+4n=1\\ -m-4n=-1 \end{cases}\)

Hence we obtain \(m=0\). Therefore, \(v_0=0 \) since m represents the initial velocity v0. 

Also we have for n:

\(4n=1\)   hence \(n=\frac{1}{4}\)

Since \(n=a/2\) we have for the acceleration:



The same procedure would be followed even if we had a uniformly deccelerated motion.