Velocity-time graphs

Velocity-time graphs are another useful tool of representing graphically the information regarding a certain motion. As we have mentioned during the study of Position-time graphs, the horizontal axis represents the time and vertical axis the velocity. As the velocity can be either positive or negative (the sign of velocity depends on the direction of motion - if the object is increasing its position, its velocity is positive, and on contrary, if the object is decreasing its position then its velocity is negative), the vertical axis (the velocity axis) can extend both above and below the origin (remember that the time axis can only extend to the right of origin as the time can only be positive).

Let's examine some situations represented by velocity-time graphs:  The above graph includes 2 types of motion: uniform motion and non-uniform motion with constant acceleration. As you see, there are 7 different situations as the graph is made up by 7 segments. All these segments represent different types of motion while within the same segment the motion has the same characteristics. Let's analyze them:

1. Uniform motion: This type of motion is shown in the part (3) and (7) of the graph where the velocity-time graph is horizontal. The vertical coordinate (here the velocity) doesn't change, therefore the object is moving at constant velocity (or steady speed). However, there are 2 important differences between these 2 motions:

a.) The magnitude of velocity is not the same. as you can easily see in the part (3), the object is moving faster than in the part (7), because the horizontal line of the part (3) graph is farther from the horizontal axis than the horizontal line of the part (7).

b.) The direction of motion in the part (3) is positive (as the graph lies in the positive region of the vertical axis (velocity)). Therefore, as the velocity is positive, the direction of motion is also positive. As for the part (7) of the graph, it lies in the negative region of the vertical axis (velocity), therefore the velocity is negative. It means that the object is moving to the negative direction of motion. It is either moving towards the origin (if the position is still positive) or is moving away from the origin and its position is becoming more negative (if the position has become negative already).

2. At rest (stopped, no motion at all). Such a situation occurs at part (5) of the graph. The graph lies on the horizontal axis (time) where the vertical coordinate (velocity) is zero. Therefore the velocity is zero for the entire segment and thus the object is not moving. It has stopped (is at rest).

3. Motion with constant acceleration. This type of motion is shown in the parts (1), (2), (4) and (6) of the graph. As the graph changes at the same rate within the same interval it is clear that the slope of the graph is the same in that interval. So, the slope has a very important meaning for the velocity-time graph. It represents the rate of velocity change (the acceleration). Hence, the slope of velocity time graph gives the acceleration

Slope of velocity-time graph = $$\frac{\Delta{v}}{\Delta{t}}=\vec{a}$$

Let's analyze the motion in the four abovementioned intervals.

a.) Part (2) of the graph has a smaller acceleration than part (1), as the slope at part (2) is smaller than the slope of part (1). However, both accelerations are positive and the velocity is increasing in both parts (1) and (2) of the graph (speeding up motion), because the velocity increased as the time passes. Another common point of these two motions is that they are made at the positive direction of motion (as the velocity is positive).

b.) Part (4) of the graph represents a slowing down (decelerated) motion. The slope is negative (velocity decreases as the time increases). Therefore, the acceleration is negative. But the object still moves at the positive direction (forward) despite the fact that the position is increasing at a slower rate than before until the object stops.

4. At the part (6) of the graph the object turns back and starts to move by increasing speed (accelerating) towards the negative direction of motion although the position will still be positive because there have been many intervals where the motion has been done at the positive direction and only 1 interval is not enough to make the object go to the negative direction of motion (for this example).

Very important! In the part (6) of the graph most people (wrongly) think  that the motion is slowing down as the acceleration is negative. A slowing down motion occurs when acceleration is negative but the initial and final velocities are both positive and only their difference is negative. Here we have both velocities negative only because the direction of motion is negative. It is true that the difference between final and initial velocities of the part (6) of the graph is negative (and therefore the acceleration is negative), but the motion is speeding up because the object is increasing more and more the rate of motion towards the negative direction.

Another useful information obtained by a velocity-time graph is the information regarding the displacement. Let's look at the following velocity-time graph of a uniform motion.

Example 1.

Find the displacement of the object for the motion shown in the graph below by using the velocity-time graph.  During all 5 seconds of motion the object has maintained the same velocity (+10 m/s). Therefore, the displacement is:

$$\Delta{\vec{x}}=\vec{v}\times{t}=10\;m/s\times{5\;s}=50\;m$$

But the same result would be also obtained by calculating the area under the graph (here the area of the rectangle by length = 5 units and width = 10 units.)

$$Area\space{of}\space{rectangle}=length\times{width}=5\times{10}=50\space{units}^2$$

Thus, we obtain another important rule:

The area under the velocity-time graph (when velocity is positive) and over the velocity-time graph (when velocity is negative) is numerically equal to the displacement.

This rule helps a lot in finding the displacement, especially when the motion is not very regular. Let's take another example to illustrate this idea.

Example 2.

Calculate the total displacement for the motion given in the velocity-time graph below:  Solution:

There are 2 methods for solving this problem:

1. Analytic method (by using the equations of motion) and

2. Graphical method (by using the velocity-time graph to find the position.

Let's find it in both ways:

1. Analytical method: As there are 3 segments contained in the graph line, there are 3 types of motion: during the first 5 seconds the object has had a uniformly accelerated motion by starting from rest $$(\vec{v_0}=0)$$ and reaching a final velocity of 4 m/s. Thus, the acceleration for this part is:

$$\vec{a}=\frac{\vec{v}-\vec{v_0}}{t}=\frac{4\;m/s-0\;m/s}{5\;s}=0.8\;m/s^2$$

Thus, for the displacement of the first interval we obtain:

$$\vec{v}^2-\vec{v_0}^2 = 2a\vec{\Delta{x}_1}$$

$$4^2-0^2=2\times{0.8}\times{\Delta{x}}_1$$

$$16=1.6\times{\Delta{x_1}}$$

$$\Delta{x_1}=10\;m$$

The second interval (from 5 s to 12 s) represents a uniform motion. The velocity for the entire interval is the same $$(\vec{v}=4\;m/s)$$. Hence, the displacement for the second interval is:

$$\Delta{\vec{x_2}}=\vec{v}\times{\Delta{t_2}}=4\;m/s\times(12\;s-5\;s)=4\;m/s\times{7\;s}=28\;m$$

The third interval (from 12 s to 15 s) is done by uniformly decreasing velocity (from $$v=4\;m/s$$ to $$v_3=-2\;m/s$$). From here we can find the acceleration for this interval:

$$a=\frac{v_3- v}{t}=\frac{-2\;m/s-4\;m/s}{3\;s}=-2\;m/s$$

Hence, the displacement for the third interval will be:

$$\vec{v_3}^2-\vec{v}^2=2\vec{a}\Delta{\vec{x_3}}$$

$$(-2)^2-4^2=2\times{(-2)}\Delta{x_3}$$

$$4-16=-4\Delta{x_3}$$

$$-12=-4\Delta{x_3}$$

$$\Delta{x_3}=\frac{-12}{-4}=3m$$

The total displacement will be the sum of all these 3 displacements. Thus,

$$\Delta{\vec{x_t}}=\Delta{\vec{x_1}}+\Delta{\vec{x_2}}+\Delta{\vec{x_3}}=10\;m+28\;m=3\;m=51\;m$$

2. Graphical method: Here we must calculate 4 different areas as the motion is divided into 4 parts:

a. From 0 s to 5 s we must calculate the area of a triangle. Thus,

$$Area_1=(Base\times{height})/2=(5\times{4})/2=20/2=10\;m$$

b. From 5 s to 12 s we must calculate the area of a rectangle. Thus,

$$Area_2=length\times{width}=(12-5)\times4=28\;m$$

From 12 s to 14 s, we must calculate the area of a triangle. Thus,

$$Area_3=(base\times{height})/2=[(14-12)\times4]/2=(2\times{4})/2=4\;m$$

From 14 s to 15 s, we have a negative displacement. The area of the triangle will have a negative sign. Thus,

$$Area_4=(base\times{height})/2=[(15-14)\times(-2)]/2=(-2)/2=-1\;m$$

Hence, the total displacement will be again the sum of all these 4 displacements. By adding them we obtain

$$\Delta{x}_t=10\;m+28\;m+4\;m+(-1\;m)=51\;m$$

This result is the same as the result obtained through the analytical method but as you see, the graph method is much shorter and easier.