Let us learn how to find displacement on a velocity-time graph. This is one of the core graphical tools used to describe how objects move and change position over time, and it builds directly on the vector quantity described in our detailed comparison of distance and displacement.
Finding displacement on a velocity-time graph is straightforward: the area under a velocity-time graph gives the displacement of the object under consideration.
The area enclosed between a velocity-time graph and the time axis, over a given time interval, is equal to the displacement of the object during that interval.
A velocity-time graph is a graph plotted between the time taken and the velocity acquired by a moving object during that time.
- Time taken is plotted along the x-axis.
- Velocity acquired is plotted along the y-axis.
Displacement is the shortest straight-line change in an object’s position, from its initial point to its final point. We know that displacement of an object is the product of velocity and time.
So,
$$d = v \times t$$
In this article, we will explore how displacement can be calculated using a velocity-time graph. Let us look at the following three cases.
Case 1: When Velocity Is Uniform (Constant)
Uniform velocity means the object covers equal distances in equal intervals of time, so its velocity stays constant. Consider a truck moving along a straight line with a uniform velocity of 50 km/h. The velocity-time graph for a truck moving with uniform velocity is shown below.
From the above figure, we can see that the graph for a truck moving with a uniform velocity of 50 km/h is a straight line parallel to the time axis.
Displacement covered by this truck between time $t_1 = 2\,h$ at $P$ and $t_2 = 8\,h$ at $Q$ is given by:
$$d = 50 \times (t_2 – t_1)$$
$$d = 50 \times (8 – 2) = 50 \times 6 = 300\ \text{km}$$
From the above graph, we can clearly see that:
- $PS = 50\ \text{km/h}$
- $PQ = t_2 – t_1 = 6\,h$
Hence, displacement $= PS \times PQ =$ area of rectangle $PQRS$, as shown in the shaded region.
Thus, when velocity is uniform, the velocity-time graph is a horizontal line, and displacement equals velocity multiplied by time: $d = v \times t$.
Case 2: When Acceleration Is Uniform (Constant)
Uniform acceleration means the velocity of the object increases (or decreases) by equal amounts in equal intervals of time. For this case, let us consider the table given below, which shows the change in velocity of a truck at regular intervals of time.
| Time (s) | Velocity of truck (m/s) |
| 0 | 0 |
| 10 | 4 |
| 20 | 8 |
| 30 | 12 |
| 40 | 16 |
| 50 | 20 |
| 60 | 24 |
The table above shows the velocity-time relationship of the truck. We get the following graph when we plot this information.
From the graph, we can see that it is a straight line, which shows velocity increasing by equal amounts in equal intervals of time. So, this motion represents motion with uniform acceleration. The area under this line $OC$ on a velocity-time graph is equal to the displacement of the truck (or object under consideration).
In this case, the displacement covered in any time interval can be found by drawing perpendiculars on the time axis (for example, $AB$ and $DC$ shown in the above figure) at the given times. The area between these perpendiculars, under the graph, gives the displacement.
Let us now find the displacement covered by the truck between the time interval $10\,s$ and $60\,s$.
$$\text{Displacement} = \text{Area of trapezium } ABCDA$$
$$= \frac{AB + DC}{2} \times AD$$
$$= \frac{4 + 24}{2}\ \text{m/s} \times (60 – 10)\ s$$
$$= 14\ \text{m/s} \times 50\ s = 700\ \text{m}$$
Alternatively, for finding displacement from this v-t graph, we can break the shape of the graph into simple geometric shapes. The total area under the line can be calculated by adding the areas of those shapes.
For example, to find the displacement of the truck between time interval $10\,s$ and $60\,s$, we can break this graph into two shapes:
- Triangle $BCE$
- Rectangle $ABED$
We can then find the areas of both these shapes and add them to get the net displacement between the given time interval.
Thus, when acceleration is uniform, the velocity-time graph is a straight sloped line, and displacement equals the average of the initial and final velocity multiplied by the time interval — the area of a trapezium.
Case 3: When Velocity and Acceleration Are Both Variable (Non-Uniform)
The figure below shows the velocity-time graph of a body moving with variable velocity.
For such cases, we divide the time interval between $t_1$ and $t_2$ into small intervals of time $\Delta t$. Over each such small interval, the change in velocity is negligible, so velocity can be taken as constant.
Therefore, for this small time interval, displacement $\Delta s = v \Delta t$, or the area of the red strip.
For the complete time interval between $t_1$ and $t_2$, displacement is equal to the area enclosed between the curve and the time axis between $t_1$ and $t_2$, or the area of the portion $PQRS$.
Thus, when motion is non-uniform, displacement is found by summing the areas of many small strips under the curve — this total is the area between the curve and the time axis.
Is the Area Under a Velocity-Time Graph Distance or Displacement?
Speed is a scalar quantity — the magnitude of velocity, with no direction. It is particularly important to note that:
- The area under a speed-time graph gives distance, not displacement.
- The area under a velocity-time graph gives displacement.
IMPORTANT NOTE:
- Velocity can be negative, so the area under a velocity-time graph can also be negative.
- A negative area represents displacement in the negative direction.
- If the truck in the examples above had reversed direction, its total distance travelled would exceed its net displacement — the two are equal here only because the motion is one-directional.
FAQs
What does the area under a velocity-time graph represent?
The area under a velocity-time graph, between the curve and the time axis, represents the displacement of the object over that time interval.
How do you find displacement from a velocity-time graph when velocity is constant?
When velocity is constant, the velocity-time graph is a horizontal line, and the area under it is a rectangle. Displacement equals velocity multiplied by the time interval: $d = v \times t$.
How do you find displacement when acceleration is uniform?
When acceleration is uniform, the graph is a straight sloped line, and the area under it is a trapezium. Displacement equals the average of the initial and final velocity, multiplied by the time interval.
What is the difference between the area under a speed-time graph and a velocity-time graph?
The area under a speed-time graph gives distance, a scalar quantity that is always positive. The area under a velocity-time graph gives displacement, a vector quantity that can be positive, negative, or zero.
Can displacement be negative on a velocity-time graph?
Yes. If velocity is negative over an interval, the corresponding area lies below the time axis and represents displacement in the negative direction.
How is displacement found for non-uniform motion on a velocity-time graph?
For non-uniform motion, the time interval is divided into very small strips $\Delta t$, over which velocity is nearly constant. The displacement in each strip is $v \Delta t$, and the total displacement is the sum of all strip areas — equivalent to the total area enclosed between the curve and the time axis.
