By going through this article you will get a deep knowledge about the position-time graphs, how to draw the p-t graph, what information do they provide regarding the motion of the object, understand about the different shapes of p-t graph, finding the slope on a p-t graph, drawing the velocity-time graph from a position-time graph etc.

## Understanding Position-Time graph

First, I am going to discuss the different ways to represent the change in position over time. One of the way to represent it is by using a table.

For example, consider the table 1 in which the time is mentioned in the left column, let’s say it’s in seconds, and the position in the right column, let’s say it’s in meters. From the table it is understood that for the time zero, one and two seconds, the position is at three metres, for the third and fourth seconds the position is  at negative two metres, for the fifth and sixth seconds the position is at zero,  and then the position is at  two, one, negative three and two metres for the time seven, eight, nine and ten seconds respectively.

Now this way of representation is somewhat useful, but it’s a little bit difficult to visualize. Also, this way of representation does not tell us what is happening in between these time intervals. That is, this table does not tell us the position of the object at time quarter of a second or half of a second.  So for that, we can construct a graph, representing the time on the horizontal axis and position over the vertical axis, known as the position-time graph.  The corresponding position-time graph for the details provided in the above table can be drawn as shown below.

Usually, the X-axis is taken as the time axis and some instant of time is chosen as the origin and is assigned the value zero of time. A convenient unit of time like a second, minute, hour, day or year is chosen. The events that occurred before the origin of time are assigned negative values, while events after this are assigned positive values of the time.

Similarly, the position of a moving body can be represented graphically say, along with the Y–axis. A convenient unit of length like a centimetre, metre or kilometre etc. is chosen to represent the position. The position of the body above origin is taken as positive and that below origin is taken as negative. The position of the moving body is represented by a real number xt or x(t) at any instant t. Then, xt or x(t) is called as the position coordinate. It can be a positive, a negative or a zero number. As said earlier, the graph between the position of an object at different instants of time and the corresponding time is called its position-time graph (p-t graph).

Again look at figure 1. For the time up to two seconds, there is no change in the position and the position of the object is constant at three metres and can be represented as a horizontal straight line. So a horizontal straight line in a position-time graph represents no motion or the object is stationary. As the object moves from the second to the third second, the position went from three to negative two at a constant rate and is represented again by a straight line, and similarly, the position for remaining time intervals can be drawn as shown in the graph. So the graph constructed here is known as a position-time graph, and from this graph, you can immediately get an understanding of how the object’s position has changed over time.

Similarly, we can represent the movement of a loaded spring, the movement of the needle of a sewing machine,  a body thrown vertically upwards, etc. graphically by plotting the position-time graph.

### Position time graph for a stationary object

As said earlier, a graph is drawn by taking time along x-axis and position along the y-axis. For a stationary object, its position will not change with time. That is, its position co-ordinate xt is a constant and has the same value x0 as at t = 0. So the graph of a stationary object is a straight line (AB) parallel to the x-axis as shown in the above figure 2.

## What does a position-time graph tell you about the motion of the object?

There is more than just a position from a p-t graph, instead, it provides a lot of information.

Let us discuss in detail about the things that can be obtained from the position-time graph with the help of the example shown in figure 3.

### Displacement from p-t graph:

So one of the information you can get from the p-t graph is the displacement ( Δx ) of the object.
Displacement = final position – initial position.

The displacement between any two time instants can be found out from the p-t graph.
For the example given in figure 3, the displacement between 2 seconds and 5 seconds can be calculated as follows. Here, the final position would be the position of the object at 5 second which is -3 metres and initial position would be the position of the object at 2 seconds which is 4 metres.
Therefore displacement between 2 to 5 seconds  = –3 – 4  = –7.

Similarly, the displacement for the total time shown on the graph = final position (which is the position of the object at 11 seconds) – initial position ( which is the position of the object at 0 seconds) = 4 – 4 = 0. So here the object started and ended at the same point and hence the total displacement is zero.

### Distance from p-t graph:

Next, we can find out the total distance or distance between any two points from the p-t graph shown in example figure 3. Remember, the total distance travelled will be the sum of all the path links travelled ( remember, all the path links should be taken as positive as the distance is always positive and no need to consider its direction).

So, here the total distance travelled = 0 m (there was no distance travelled between 0 to 3 seconds) + 7 m (between 3 seconds and 5 seconds, the object moves from four to negative three metres, i.e., a distance of 7 metres(remember distance is always positive and no need to consider direction)) +  7 m (between 5 seconds and 11 seconds, the object moves back from negative three to four metres, i.e., a distance of 7 metres again) = 14 metres.
That means a total distance of 14 metres was travelled by the object for the whole trip.  Similarly, the distance between any two points can be found out from the p-t graph.

### Average velocity from p-t graph:

Next, we can figure out the average velocity. We know that average velocity is the displacement per time.

That is, Average velocity, Vavg =  Δx / Δt = |x2 – x1| / |t2 – t1| , where x1 and x2 are the initial and final positions respectively for their respective time t1 and t2.

For example, let’s find the average velocity between 3 second and 5 seconds in figure 3. Here the final position is -3 metres at time 5 seconds and the initial position is 4 metres at time 3 seconds.
Vavg =  Δx / Δt = |x2 – x1| / |t2 – t1| = |-3 – 4| / |5 – 3| = |-3.5| = 3.5 m/s, and the direction is same as that of displacement.
Similarly, the total average velocity for this graph = total displacement/time = 0/11 = 0 m/s, means the total average velocity for the entire motion shown in this graph is zero because the object has no displacement here.

### Average Speed from the p-t graph:

Similarly, we can find out the average speed also from the position versus time graph. We know that the average speed is defined as the distance per time.

Consider figure 3, the total average speed for the whole 11 seconds = Total distance covered / Total time.
We already found that the total distance covered by the object was 14 meters.
So total average speed = 14 m /11 s = 1.272 m/s.
That is, the object was moving at 1.272 meters per second, on average. That was its average speed. Similarly, we can find out the average speed between any two points.

### Instantaneous velocity from a p-t graph:

Now the most important part is we can find out the instantaneous velocity from the p-t graph.  To find the instantaneous velocity from the p-t graph, you need to look at the slope. The slope of the position-time graph will give us the velocity in that direction.  Here, we had a horizontal position-time graph and so the slope will give us the velocity in the x-direction. That is, the average slope along a line that doesn’t change its slope will give us the average velocity in that line and the instantaneous slope of the graph will give us the instantaneous velocity.

How can we find the instantaneous slope from a graph?  The lines in the graph are all straight lines. Therefore, the average slope between any two points on these lines is equal to the instantaneous slope at any point on that line.

The instantaneous slope at any point of a straight line can find out using the equation, slope = (x2 – x1)/(t2 – t1), where x1 and x2 are the displacements at the first and second point, t1 and t2 are the respective time at the first and second point.

For the example shown in figure 4, let’s find the instantaneous velocity of the object at 4 seconds. Take any two points in that line and name it as point 1 and point 2. Any points can be chosen in this straight line, I choose these two points just because of convenience as I know exactly where they are at. So reading from the graph, at point 1 the horizontal position is 4 metres at time 3 seconds and at point 2 the position is  –3 metres at time 5 seconds.

The same will get if you choose the other two points in this line PQ. So we get the slope of the line PQ as -3.5 m/s. So that was the instantaneous velocity at four seconds. i.e., Vinst @ 4sec = -3.5 m/s. Negative because the object was moving backwards. i.e., the object was moving 3.5 metres backwards in every second between 3 to 5 seconds, that means the object was going 3.5 m/s on average between 3 to 5 seconds. And since PQ is straight, this was the rate the object was moving at any moment between 3 and 5 seconds.  That is, the instantaneous velocity at 3.5 second or 4.5 seconds etc. is also –3.5 m/s.  i.e., -3.5 m/s for the whole line QR.

Similarly, the instantaneous velocity at 2 seconds or anywhere between 1 second and 3 seconds can be found out by taking the slope of the line PQ. i.e.,  Vinst @ 4 sec = Slope of the line PQ = (4 – 4)/(3 – 0) = 0 m/s.

Similarly, the instantaneous velocity at 9 seconds or anywhere between 5 seconds and 11 seconds can be found out by taking the slope of the line RS. i.e.,  Vinst @ 9 sec = Slope of the line RS = (4 – -3)/(11 – 5) = 7/6 = 1.167 m/s. Positive because the object was moving forward.

### Instantaneous Speed from p-t graph:

Lastly, we can figure out the instantaneous speed at a point from the p-t graph. The magnitude of instantaneous velocity at a point gives the instantaneous speed at that point, assuming the body is having motion in only one direction.

i.e.,  Instantaneous speed, Sinst = |Vinst|.

For example, the instantaneous speed at 4 seconds is positive 3.5 metres/second. i.e., Sinst @ 4sec = |Vinst @ 4sec |= |-3.5| = 3.5 m/s.
This would be the instantaneous speed at 4 seconds or any time between three to five seconds.
Similarly, the instantaneous speed at 7 seconds, Sinst @ 7 sec = |Vinst @ 7 sec |= |1.167| = 1.167 m/s. This would be the instantaneous speed at 4 seconds or any time between five to eleven seconds.

So, that’s it. Let’s recap quickly. The value of the horizontal p-t graph gives the horizontal position of the object and the slope of the horizontal p-t graph gives the velocity in the x-direction. The average slope gives the average velocity in that line and the instantaneous slope gives the instantaneous velocity. Also, the magnitude of the instantaneous velocity gives the instantaneous speed.

## Position-time graph for uniform motion

A particle is said to move with uniform velocity if it makes equal displacements in equal intervals of time. i.e., the velocity remains constant throughout the motion. The position-time graph in uniform motion will be a straight line without any curve or bend as shown in the figure and the slope of the p-t graph gives the velocity of an object moving with uniform velocity.

• A positive slope on a position-time graph means that the object’s position from the origin is increasing with respect to time.
• A negative slope on a position-time graph means that the object’s position moves towards the origin and in some cases, it pasts the origin.
• A zero slope on a position-time graph means that the object is at rest or stationary.

## Position-time graph for non-uniform motion

For non-uniform motion, the object is changing its velocity throughout the motion. i.e., the object covers an unequal distance in equal intervals of time. The position-time graph of a body moving with non-uniform velocity will be curved and may have many changes in its direction.

We know, velocity is the gradient (slope) of the position-time graph. For a body moving with non-uniform velocity, the velocity of the body is always changing, which means the gradient of the curve is always changing. The gradient of the curve is changing means the velocity is changing. The gradient or slope of a curve at any point is given by the gradient(slope) of the tangent at that point. A tangent is a straight line which touches the curve at one point only. The slope or gradient of a curve is different at each point on the curve.

• The position-time graph of a slowing down object is a curve as shown below in figure 8. In this case, the gradient is getting flattered and flatter so the velocity is getting slower and slower. By taking the tangent of the curve at t= 1,2.3 and 4 seconds, we can observe that the gradient value becomes smaller as time increases.
• Now consider an object which is moving faster and faster. If you apply a similar line of thought, since the velocity will be higher and higher, the position-time graph will also be a curve and the gradient of the curve will be steeper and steeper.

## Position Time graph to Velocity Time graph

By looking at the position-time graph, not only you can figure out exactly what the object is doing but you should be able to draw a velocity versus time graph from a position versus time graph.

Consider the position-time graph given in the below figure 10.  By looking at the motion of the body from 0 to 3 second in the p-t graph given, it is clear that the object is at position 4 and it is not moving. i.e., the object is stationary for 0 to 3 seconds. Therefore, its slope is 0 and hence the velocity is 0 m/s. So mark 0 m/s in the velocity-time graph (fig. 10 bottom portion) for time 0 to 3 second.

Similarly, it is already found in the previous section (figure 4, section 2.5) that the object is moving at a constant velocity of –3.5 m/s for time 3 sec to 5 seconds(Portion QR in the pt graph). So mark it also on the velocity-time graph.

Similarly, the object is moving at a constant velocity of 1.5 m/s (found in section 2.5, figure 4) for the time 5 seconds to 11 seconds(Portion RS in the pt graph) and mark that too in the velocity-time graph (as shown in figure 10) and you will get the corresponding velocity-time graph for the position-time graph.

You can see that the object waited for a period of time, then it moves at a constant velocity of -3.5 m/s in the negative direction, and then it went in the opposite direction not as fast, as you can see that the slope of the line is not as steep as it was over the previous portion. So that’s how you would go from a position versus time graph to a velocity versus time graph and the easy way to do that is by just figuring out the slope. Because if you figure out the slope of the position versus time graph that will always tell you the velocity versus time graph.

Similarly, if the body is moving at non-uniform velocity, the position in the p-t graph will be a curved line and then the slope can be found out by drawing the tangents on the curve and this will give the instantaneous velocity and by marking that velocities you can draw the velocity-time graph.

## P-t graph Solved Examples

1. Describe the motion of the object from t = 0 to 7 seconds.

The direction of motion of the object can find out by finding the starting and ending the displacement of the object. Here the initial position is 7.5 metres and the final position is 0 metres. So the object is moving in the negative direction

By comparing the gradient of the graph at t = 0 s and t = 7 s, we can find out whether the object is speeding up or slowing down. If the gradient is getting flatter, the object is slowing down and if the gradient is getting steeper, the object is speeding up.

The tangent of the curve can be drawn as shown in the figure. From the figure, it is clear that the gradient is getting flattered and flatter, which means the object is slowing down.

So, here the object is moving in the negative direction and is getting slower as time increases from 0 to 7 seconds

2. Describe the motion of the object from t = 0 to 7 seconds.

Here also, we can describe the motion of the object in the same way as said above in the previous problem. Here the initial position is 6.5 metres and the final position is 0 metres. So here also, the object is moving in the negative direction

Similarly, by comparing the gradient of the graph at t = 0 s and t = 7 s, we can find out whether the object is speeding up or slowing down.

The tangent of the curve can be drawn as shown in the figure. From the figure, it is clear that the gradient is getting steeper and steeper as time increases, which means the object is moving faster and faster.

So, here the object is moving in the negative direction and its velocity is increasing as time increases from 0 to 7 seconds.

3. The position-time graph representing the motion of an object is given below.

a)Find the velocity of the object in the first 3 seconds.
b)Find the velocity in between 3 second and 6 seconds.
c)Find the velocity in between 6 second and 7 seconds.
d)Find the velocity in between 7 second and 9 seconds.
e)Find the velocity in between 9 second and 11 seconds.

a)
” alt=”” aria-hidden=”true” />” alt=”” aria-hidden=”true” /> \Large \begin{aligned}Velocity\;of\;the\;object\;in\;first\;3\;seconds,\;V_{1} & = \frac{x_{2}-x_{1}}{t_{2}-t_{1}} \\&= \frac{30-0}{3-0} \\&= 10\;m/s \end{aligned}

b)
” alt=”” aria-hidden=”true” />” alt=”” aria-hidden=”true” /> \Large \begin{aligned}Velocity\;in\;between\;3\;second\;and\;6\;seconds,\;V_{2} & = \frac{x_{2}-x_{1}}{t_{2}-t_{1}} \\&= \frac{30-30}{6-3} \\&= 0\;m/s \end{aligned}

c)
” alt=”” aria-hidden=”true” />” alt=”” aria-hidden=”true” /> \Large \begin{aligned}Velocity\;in\;between\;6\;second\;and\;7\;seconds,\;V_{3} & = \frac{x_{2}-x_{1}}{t_{2}-t_{1}} \\&= \frac{10-30}{7-6} \\&= -20\;m/s \end{aligned}

d)
” alt=”” aria-hidden=”true” />” alt=”” aria-hidden=”true” /> \Large \begin{aligned}Velocity\;in\;between\;7\;second\;and\;9\;seconds,\;V_{4} & = \frac{x_{2}-x_{1}}{t_{2}-t_{1}} \\&= \frac{10-10}{9-7} \\&= 0\;m/s \end{aligned}

e)
” alt=”” aria-hidden=”true” />” alt=”” aria-hidden=”true” /> \Large \begin{aligned}Velocity\;in\;between\;9\;second\;and\;11\;seconds,\;V_{5} & = \frac{x_{2}-x_{1}}{t_{2}-t_{1}} \\&= \frac{0-10}{11-9} \\&= -5\;m/s \end{aligned}

4. Consider the position-time graph given below and answer the following questions where A, B, C, D, E, F and G are the different objects in motion.

a) Among the objects travelled, which objects are accelerating?
b)Which objects are maintaining a constant velocity?
c) Which of the objects are in stationary
d)Which among the objects travelled are changing its direction.
e)Which object is travelling faster
f) Which object is moving slower
g) Which among the objects travelled has the greatest acceleration

a) Here object D is accelerating.

Changing velocity means an accelerating object. We know, the slope of the position-time graph will give us the velocity, and an accelerating object will have a changing slope.
Here, the p-t graph of object D is a curve which means it is having a changing slope and hence object D is accelerating. The p-t graph of all the other objects is a straight line and hence they are not accelerating.

b) Objects A, C, E and F are maintaining a constant velocity.

We know that, if the slope of the p-t graph for an object is constant, then its velocity is also constant.
Here objects A, C E and F are having a constant slope and hence they are maintaining a constant velocity.

c)Here, objects A and F are stationary.

We know, an object is stationary means the slope of its position-time graph is zero or its pt graph is a straight line parallel to the x-axis.
Here the graph of objects A and F are parallel to the x-axis and hence they are stationary.

d) Here, object B and G changes its direction.

An object changes its direction if it changes from a positive velocity to a negative velocity or vice versa. In other words, it changes from a positive slope to a negative slope or vice versa in a position-time graph.
Here, the slope of the object B changes from positive to negative and slope of object G changes from negative to a positive value. Hence both B and G are changing its direction.

e)Here, object C is travelling faster.

We know, object with the highest speed or greatest magnitude of velocity is travelling faster. That means, the object with the highest slope in a position-time graph is travelling faster.
Here, object C is having the greatest slope among the objects travelled and hence object C is travelling faster.

f) Here, object E is travelling slower.

Similarly, an object with the lowest speed or smallest magnitude of velocity is travelling slower. That means, the object with the smallest slope in a position-time graph is travelling faster.
Here, object E is having the lowest slope among the objects travelled and hence object E is travelling slower.

g) Object D has the greatest acceleration.

We know, accelerated motion in a position-time graph is represented by a curved line and here object D is the only object with a curved line. That is, object D is the only object with acceleration.

I hope the information in this article helps you to get a brief idea regarding the concept and different shapes of position-time graphs. Also, this article explored the different quantities that can be found out from the p-t graph and gives a clear idea of drawing a velocity-time graph from the p-t graph.