The displacement-time curve of two objects A and bare shown here then

In this explainer, we will learn how to use displacement–time graphs and interpret the slope of the curve as the velocity of the body.

We start by recalling that a scalar is a quantity that has magnitude (size) but no direction, whereas a vector is a quantity that has both magnitude and direction. This distinction helps us to understand the difference between distance and displacement.

Distance is a scalar quantity. The distance that an object travels is the length of the path that the object takes from the starting point of its motion to the endpoint of its motion.

Displacement is a vector quantity. The displacement of an object has a magnitude equal to the shortest distance between the starting point and the endpoint of the motion of the object, and the direction of the motion must be specified.

Therefore, although the distance traveled by an object and its displacement can be measured using the same units, such as kilometres or metres, the displacement must always have its direction specified too. Furthermore, distances are always positive, while displacements involving motion in a straight line (one-dimensional motion) can be positive or negative.

In parallel to the definitions for distance and displacement, we have the following definitions for speed and velocity.

Speed is a measure of how fast an object moves. It is a scalar quantity. Speed can be measured as a distance per unit of time, such as kilometres per hour (km/h) or metres per second (m/s). The speed of an object does not tell us the direction in which it travels.

Velocity is the rate of change of displacement. It is a vector quantity. Velocity can be measured as a displacement per unit of time. The units of measurement are the same as those for speed but with a direction specified, such as kilometres per hour in an easterly direction or metres per second in the positive direction of travel.

Having built up these concepts, we are now in a position to explore displacement–time graphs, our main focus in this explainer. We can represent the motion of an object by using a displacement–time graph. These graphs always have the displacement from a given point represented on the vertical axis and time represented on the horizontal axis.

It is easiest to learn about these graphs by studying some specific examples. In the next three graphs, the displacement of an object from a given point in metres is plotted against the time in seconds.

In the above displacement–time graph, the horizontal line starting at 4 on the vertical axis shows that the displacement of the object remains constant over time, at 4 m from the given point.

On a displacement–time graph, the gradient represents velocity. The fact that the gradient of a horizontal line is zero shows that the velocity is zero, and hence the object is stationary.

In general, any horizontal sections of a displacement–time graph represent the time intervals for which the given object is stationary.

Our second displacement–time graph shows the motion of an object in a straight line, away from its starting point (positive direction of travel) and then back toward its starting point (negative direction of travel).

Here, the upward-sloping straight line from the origin to the point (20,100) tells us that this object starts at a given point and then travels 100 metres directly away from this point in the first 20 seconds, as shown below.

Recall that on a displacement–time graph, the gradient represents the velocity. The positive gradient of the line shows that the displacement from the starting point increases at a constant rate over time, so during this period, the object is moving with a constant velocity in the positive direction. The velocity is given by the gradient, which is changeindisplacementtimetakenms=100−020−0=10020=5/.

Next, the downward-sloping straight line from (20,100) to (40,50) tells us that in the next 40−20=20seconds, the object reverses its direction of motion and travels 100−50=50metres directly back toward its starting point, as shown below.

The negative gradient of the line shows that the displacement from the starting point decreases at a constant rate over time, so during this period, the object is moving with a constant velocity in the negative direction. The velocity is given by the gradient, which is changeindisplacementtimetakenms=50−10040−20=−5020=−2.5/.

In general, any upward- or downward-sloping straight-line sections of a displacement–time graph represent the given object’s motion in the time intervals for which it is traveling at a constant velocity. An upward slope (and thus a positive gradient) shows a positive velocity; a downward slope (and thus a negative gradient) shows a negative velocity.

Our third displacement–time graph shows the motion of an object away from its starting point in a straight line in the positive direction.

This time, the curve from the origin to (8,25) tells us that this object starts at the given point and then travels 25 metres directly away from this point in 8 seconds. The constantly increasing gradient of the curve shows that the displacement increases at an increasing rate over time. Therefore, the velocity is constantly increasing, so the object is accelerating.

In general, any curved sections of a displacement–time graph represent the given object’s motion in the time intervals for which it is accelerating or decelerating. A common misconception is that any curves on a displacement–time graph imply that the given object is following a curved trajectory, but in fact, they tell us nothing about the shape of the trajectory.

Finally, we need two important formulas that will enable us to calculate the average speed and average velocity of an object after reading off some relevant information from displacement–time graphs.

The average speed of an object is equal to the total distance traveled divided by the time taken: averagespeedtotaldistancetraveledtimetaken=.

This is a scalar quantity measured as a distance per unit of time, for example, kilometres per hour (km/h) or metres per second (m/s).

The average velocity of an object is equal to the displacement from its starting point divided by the time taken: averagevelocitydisplacementfromstartingpointtimetaken=.

Since displacement is a vector quantity, average velocity is also a vector quantity and so may be positive or negative. Its magnitude, which is a scalar, is measured as a distance per unit of time and is therefore always positive.

To illustrate the difference between these two quantities, we can refer back to our second displacement–time graph above.

To calculate the average speed, we need to divide the total distance traveled by the time taken. The object traveled 100 m in the first 20 seconds and 50 m in the next 20 seconds. Hence, its average speed is given by the following: averagespeedtotaldistancetraveledtimetakenms==100+5020+20=15040=3.75/.

To calculate the average velocity, we need to divide the total displacement from the starting point by the time taken. The object traveled 100 m in the positive direction in the first 20 seconds and 50 m in the negative direction in the next 20 seconds. Hence, its average velocity is given by averagevelocitydisplacementfromstartingpointtimetakenms==100−5020+20=5040=1.25/.

Note that the first and third displacement–time graphs above would have their average velocity equal in magnitude to their average speed.

In the first graph, the given object is stationary, so the distance traveled is 0, as is the magnitude of the object’s displacement. Therefore, when calculating the average speed or the average distance, we would be taking a value of 0 and dividing it by the time taken, which clearly gives an answer of 0 in both cases.

In the third graph, the given object simply moves directly away from its starting point in the positive direction, so its motion has only one stage. It travels 25 m in 8 seconds, so its average speed is given by averagespeedtotaldistancetraveledtimetakenms==258=3.125/.

Again, the magnitude of its displacement is equal to its distance traveled, so we would get exactly the same result if we substituted the values for the displacement from the starting point and the time taken into the formula for average velocity.

Let us now look at an example where we can practice calculating the average velocity from information given in a displacement–time graph.

The graph represents the relation between the displacement of a particle and the time for which it moves in a straight line.

Fill in the blank: The average velocity of the whole trip =/ms.

Answer

Recall that the average velocity of an object is equal to the displacement from its starting point divided by the time taken: averagevelocitydisplacementfromstartingpointtimetaken=.

Recall also that on a displacement–time graph, the gradient represents the velocity. By analyzing the gradients shown on the above graph, we can spot that the motion of the particle has three stages.

First, in the time interval from 𝑡=0 to 𝑡=3, the positive gradient of the straight line from (0,0) to (3,6) tells us that the particle has a constant velocity, moving directly away from its starting point by 6 m in the positive direction. Therefore, its change in displacement in the first 3 seconds is 6 m.

Second, in the time interval from 𝑡=3 to 𝑡=5, the horizontal straight line (with gradient zero) from (3,6) to (5,6) tells us that the particle has a velocity of zero and thus is stationary. Therefore, its change in displacement over these 2 seconds is 0.

Third, in the time interval from 𝑡=5 to 𝑡=10, the positive gradient of the straight line from (5,6) to (10,16) tells us that the particle again has a constant velocity, moving directly away from its starting point by a further 16−6=10m in the positive direction. Therefore, its change in displacement in the final 5 seconds is 10 m.

To calculate the average velocity for the whole trip, we take the results for the three separate stages and apply the formula as follows: averagevelocityofthewholetripdisplacementfromstartingpointtimetakenms==6+0+103+2+5=1610=1.6/.

We conclude that the final sentence of the question should read: “The average velocity of the whole trip = 1.6 m/s.”

In the above example, we used a displacement–time graph to help us calculate the average velocity over an entire journey. In the next example, we will need to calculate the average velocity at different stages of a journey.

This displacement–time graph shows the journey of a cyclist along a straight road. She cycles in a straight line for 15 minutes and then stops for 10 minutes. Finally, she cycles back to her starting point in 25 minutes, traveling in a straight line.

Calculate the average velocity for each stage of the journey in kilometres per hour.

Answer

Recall that the average velocity of an object is equal to the displacement from its starting point divided by the time taken: averagevelocitydisplacementfromstartingpointtimetaken=.

In this question, we are asked to calculate the average velocity, in kilometres per hour, for each stage of the cyclist’s journey shown in the above displacement–time graph. Recall that on a displacement–time graph, the gradient represents the velocity, so we can see that this journey splits into three separate stages: from 𝑂 to 𝐴, then from 𝐴 to 𝐵, and finally from 𝐵 to 𝐶.

However, notice that, on this graph, the displacement from the starting point in kilometres is plotted against the time taken in minutes. This means we have a choice of either calculating the three velocities in kilometres per minute and then converting them to kilometres per hour or converting the times in minutes into hours before calculating the velocities in kilometres per hour. We will work through the first method in detail.

We know that the stage of the journey from 𝑂 to 𝐴 takes 15 minutes. Moreover, the positive gradient of the straight line from (0,0) to (15,4) tells us that over this time interval, the cyclist moves directly away from her starting point, traveling 4 km in the positive direction. This stage of the journey is shown below.

Therefore, the cyclist’s change in displacement over these 15 minutes is 4 km. Applying the formula, we get the following: averagevelocityfromtodisplacementfromtotimetakenkmmin𝑂𝐴=𝑂𝐴=415=0.2̇6/.

To convert a value in kilometres per minute to kilometres per hour, we multiply it by 60 (the number of minutes in an hour). In this case, we get the following: averagevelocityfromtokmh𝑂𝐴=0.2̇6×60=16/.

The stage of the journey from 𝐴 to 𝐵 takes 10 minutes, and we know that the cyclist is not in motion throughout this stage. This fact is confirmed by the horizontal straight line (with zero gradient) from (15,4) to (25,4), telling us that during this stage the cyclist has zero velocity and hence is stationary. This stage of the journey is shown below.

Therefore, the cyclist’s change in displacement over these 25−15=10minutes is 0. Since we calculate the average velocity by dividing the change in displacement by the time taken, this immediately implies that the cyclist’s average velocity from 𝐴 to 𝐵 is 0.

Lastly, the stage of the journey from 𝐵 to 𝐶 takes 25 minutes. Moreover, the negative gradient of the straight line from (25,4) to (50,0) tells us that over this time interval, the cyclist moves directly toward her original starting point, traveling 4 km in the negative direction. This stage of the journey is shown below.

Therefore, the cyclist’s change in displacement over these 50−25=25minutes is −4 km. Applying the formula, we get the following: averagevelocityfromtodisplacementfromtotimetakenkmmin𝐵𝐶=𝐵𝐶=−425=−0.16/.

To convert this value to kilometres per hour, we multiply it by 60, which gives us the following: averagevelocityfromtokmh𝐵𝐶=−0.16×60=−9.6/.

Notice that if we had first converted the times in minutes for the different stages of the journey into hours and then carried out the velocity calculations, we would have obtained the following results.

The stage of the journey from 𝑂 to 𝐴 takes 15 minutes, which is 1560=0.25hours, so averagevelocityfromtodisplacementfromtotimetakenkmh𝑂𝐴=𝑂𝐴=40.25=16/.

The stage of the journey from 𝐴 to 𝐵 takes 10 minutes, which is 1060=0.1̇6hours. However, as the change in displacement over this time interval is 0, and we calculate the average velocity by dividing the change in displacement by the time taken, then converting the time from minutes to hours still gives the average velocity from 𝐴 to 𝐵 as 0.

Finally, the stage of the journey from 𝐵 to 𝐶 takes 25 minutes, which is 2560=0.41̇6hours, so averagevelocityfromtodisplacementfromtotimetakenkmh𝐵𝐶=𝐵𝐶=−40.41̇6=−9.6/.

As expected, all three average velocities agree with our original results.

Now, we compare the average velocity and the average speed for a given journey.

This displacement—time graph shows the motion of a particle in a straight line. It travels away from its starting point for 15 minutes and then returns in 30 minutes.

1. Calculate the average speed for the whole journey in kilometres per hour.
2. Calculate the average velocity for the whole journey in kilometres per hour.

Answer

In this question, the given journey has two stages, which we will refer to as the outward and return stages.

We are asked to calculate the average speed and average velocity, in kilometres per hour, for the whole journey. However, notice that, on this graph, the displacement from the starting point in kilometres is plotted against the time taken in minutes. Therefore, we will convert the relevant times into hours before doing any calculations to ensure that our answers have the correct units.

The outward stage takes 15 minutes, which is 1560=0.25hours.

Recall that on a displacement–time graph, the gradient represents the velocity. The upward-sloping straight line from (0,0) to (15,12) tells us that the particle has a constant velocity, moving directly away from its starting point by 12 km in the positive direction. Therefore, its change in displacement in the first 0.25 hours is 12 km.

The return stage takes 30 minutes, which is 3060=0.5hours.

The downward-sloping straight line from (15,12) to (45,0) tells us that the particle has a constant velocity, moving directly back to its starting point 12 km in the negative direction. Therefore, its change in displacement in the remaining 0.5 hours is −12 km.

Part 1

For part 1, recall that the average speed of an object is equal to the total distance traveled divided by the time taken: averagespeedtotaldistancetraveledtimetaken=.

The particle traveled 12 km in the first 0.25 hours and 12 km in the remaining 0.5 hours. Hence, its average speed is given by the following: averagespeedtotaldistancetraveledtimetakenkmh==12+120.25+0.5=240.75=32/.

Part 2

For part 2, recall that the average velocity of an object is equal to the displacement from its starting point divided by the time taken: averagevelocitydisplacementfromstartingpointtimetaken=.

The particle traveled 12 km in the positive direction in the first 0.25 hours and 12 km in the negative direction in the next 0.5 hours. Hence, its average velocity is given by the following: averagevelocitydisplacementfromstartingpointtimetakenkmh==12−120.25+0.5=00.75=0/.

If we reflect upon part 2, it should be obvious why we get an answer of zero. After all, the particle’s journey started and ended in the same place, so its overall displacement is zero. Dividing this value by the time taken to get the average velocity necessarily gives a zero answer.

In our next example, we show how to read off gradient information from a displacement–time graph to work out the times when an object is at rest.

The displacement–time graph given below describes the motion of an object.

At what time is the object at rest (velocity = zero)?

Answer

Recall that on a displacement–time graph, the gradient represents the velocity. If the gradient is zero, then the velocity is zero. This means that the object is stationary, which is the same as saying it is at rest.

Only the horizontal sections of a displacement–time graph have a gradient of zero, so these represent the time intervals for which the given object is at rest.

By examining the displacement–time graph above, we can see that the only horizontal section occurs between the times 𝑡=2 and 𝑡=3. We conclude that the object is at rest for the time interval 2<𝑡<3.

We can also use displacement–time graphs to find the maximum height reached by a projectile. Here is an example of this type.

This is the displacement–time graph for a ball that is thrown vertically up in the air and falls back down to its starting point.

1. Find the maximum height reached by the ball.
2. Find the time at which the ball reaches its maximum height.

Answer

Recall that displacement–time graphs always have the displacement from a given point represented on the vertical axis and time represented on the horizontal axis.

Part 1

In this case, we know that a ball is thrown upward, with its height above the ground plotted on the vertical axis. Therefore, we can work out the maximum height reached by the ball by finding the highest point on the curve and reading across to the vertical axis. As shown in the diagram below, this value is 5 m.

Part 2

To find the time at which the ball reaches its maximum height, we read down from the highest point to the horizontal axis. As shown above, this value is 1 second.

Another way to tackle this problem is by realizing that when the ball reaches its maximum height, it will instantaneously be at rest (because it has finished rising and is about to start falling).

Recall that on a displacement–time graph, the gradient represents the velocity. If the gradient is zero, then the velocity is zero, which means that the given object is at rest. Inspecting the graph, we see that there is only a single point on the curve where the gradient is zero, namely, (1,5). Again, this implies that the ball reaches a maximum height of 5 m after 1 second.

In our final example, we learn how to spot when the velocity of an object is positive by using a displacement–time graph.

This displacement–time graph describes an object’s motion.

Which of the following statements is true?

1. The velocity is positive at the time interval 0<𝑡<2.
2. The velocity is positive at the time interval 3<𝑡<4 only.
3. The velocity is positive at the time interval 1<𝑡<3.
4. The velocity is positive at the time interval 0<𝑡<1 only.
5. The velocity is positive at the time intervals 0<𝑡<1 and 3<𝑡<4.

Answer

Recall that on a displacement–time graph, the gradient represents the velocity. Therefore, to work out when the velocity is positive, we need to inspect the given displacement–time graph to find the sections of the curve that have a positive gradient.

In the time interval 0<𝑡<1, the curve slopes upward, so the gradient is positive.

At 𝑡=1, the gradient is 0.

In the time interval 1<𝑡<3, the curve slopes downward, so the gradient is negative.

At 𝑡=3, the gradient is 0.

In the time interval 3<𝑡<4, the curve slopes upward, so the gradient is positive.

We deduce that the gradient is positive in the time intervals 0<𝑡<1 and 3<𝑡<4. Therefore, these are the time intervals for which the velocity is positive, so the correct answer is E.

Let us finish by recapping some key concepts from this explainer.

• Distance is a scalar quantity. The distance that an object travels is the length of the path that the object takes from the starting point of its motion to the endpoint of its motion.
• Displacement is a vector quantity. The displacement of an object has a magnitude equal to the shortest distance between the starting point and endpoint of the motion of the object, and the direction of the motion must be specified.
• The average speed of an object is equal to the total distance traveled divided by the time taken: averagespeedtotaldistancetraveledtimetaken=. This is a scalar quantity measured as a distance per unit of time, for example, kilometres per hour (km/h) or metres per second (m/s).
• The average velocity of an object is equal to the displacement from its starting point divided by the time taken: averagevelocitydisplacementfromstartingpointtimetaken=. Since displacement is a vector quantity, average velocity is also a vector quantity and so may be positive or negative. Its magnitude, which is a scalar, is measured as a distance per unit of time and is therefore always positive.
• We can represent the motion of an object by using a displacement–time graph. These graphs always have the displacement from a given point represented on the vertical axis and time represented on the horizontal axis.
• On a displacement–time graph, the gradient represents the velocity:
  • Any horizontal sections (which have a gradient of zero) represent the time intervals for which the velocity is zero, and hence the given object is stationary.
  • Any upward- or downward-sloping straight-line sections represent the time intervals for which the given object is traveling at a constant velocity.
  • Any curved sections represent the time intervals for which the given object is accelerating or decelerating.