Understanding Velocity Versus Time Graphs

There are two types of information that one normally wishes to find from a velocity versus time graph. The first and simplest objective is finding the acceleration. The second is to find a position versus time graph that produced the velocity versus time graph.


Acceleration

Acceleration is defined as the derivative, or rate of change, of the velocity with respect to time graph and written . Graphically, the acceleration at a particular time is defined as the slope of a line tangentto a velocity versus time graph at that particular time. In first-year physics we deal only with constant acceleration. Therefore the velocity versus time graph must have constant slope. In other words we deal only with straight line velocity versus time graphs. Such graphs can have positive, zero, or negative slopes. These three cases are shown in figure [1] below along with a tangent line showing the acceleration in each case.

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Figure 1. Acceleration for the various velocity versus time graphs.

Finding the correct position versus time graph

Finding the correct position versus time graph can be a little tricky because you are working backwards. Let’s start with the simplest case – zero acceleration and a flat velocity versus time graph. If the object is not accelerating we know that the corresponding position versus time graph must be a straight line. A straight line can have zero, positive, or negative slope. The slope of a position versus time graph indicates the direction of velocity. Consider the three possible velocity versus time graphs shown in figure [2] below.

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Figure 2. Constant velocity versus time graphs.

Figure [2a] is the simplest to understand. It has a constant zero velocity. It is not moving. Figure [2b] has positive velocity. Positive velocity means that the object is moving in the positive direction – upwards on a position versus time graph. The corresponding position versus time graph is a straight line of positive slope. Figure [2c] has negative velocity. Negative velocity means that the object is moving in the negative direction – downwards on a position versus time graph. The corresponding position versus time graph is a straight line of negative slope. Assuming that in each case the objects starts at the origin, the corresponding position versus time curves for figure [2] are shown in figure [3].

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Figure 3. Position versus time graphs demonstrating constant velocity.

Note that though the graphs in figure [2] are very similar to one another, the corresponding graphs of figure [3] are not. In figure [2] a flat line is merely higher or lower while the orientations of the lines in figure [3] are tilted. There are even greater variations when we consider non-zero acceleration.

Figure [4] shows positive acceleration but the straight line has again been raised or lowered on the vertical or velocity axis.


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Figure 4. Velocity versus time graphs with positive acceleration.

The fact that the acceleration is not zero indicates that the position versus time graphs will be curved. Examine the graph in [4a]. The velocity is always positive, so the object is always moving in the positive direction. The position versus time graph must also move upwards. Also in [4a] the velocity is getting faster. So the curvature of the position versus time graph must be getting steeper. The result is shown in figure [5a]. Considering the graph in [4b] note that the velocity is always negative, so the object is always moving in the negative direction. The position versus time graph must therefore move downwards. Also in [4b] the velocity is getting slower. So the curvature of the position versus time graph must be getting less steep. The result is shown in figure [5b]. Figure [4c] is interesting since it has a region of negative velocity at the start, passes through zero velocity, and then has a region of positive velocity. So a position versus time graph, as shown in [5c], must first move downwards, turn around, and move upwards like a yo-yo. The graph must have a trough, the base of the trough occurring when the velocity is zero. Note in figure [5] that we have assumed, as usual, that the object started from zero.

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Figure 5. Position versus time graphs demonstrating positive acceleration.

Figure [6] shows negative acceleration but the straight line has again been raised or lowered on the vertical or velocity axis.

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Figure 6. Velocity versus time graphs with negative acceleration.

Again, the fact that the acceleration is not zero indicates that the position versus time graphs will be curved. Examine the graph in [6a]. The velocity is always positive, so the object is always moving in the positive direction. The position versus time graph must also move upwards. Also in [6a] the velocity is getting smaller, the object is slowing. So the curvature of the position versus time graph must be getting less steep. The result is shown in figure [7a]. Considering the graph in [6b] note that the velocity is always negative, so the object is always moving in the negative direction. The position versus time graph must therefore move downwards. Also in [6b] the velocity is getting faster. So the curvature of the position versus time graph must be getting steeper. The result is shown in figure [7b]. Figure [6c] is interesting since it has a region of positive velocity at the start, passes through zero velocity, and then has a region of negative velocity. So a position versus time graph, as shown in [7c], must first move upwards, turn around, and move downwards like a ball thrown up into the air. The graph must have a peak, the base of the peak occurring when the velocity is zero. Note in figure [7] that we have again assumed that the object started from zero.

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Figure 7. Position versus time graphs demonstrating negative acceleration.


Alternate Approach

There is an alternate approach to finding a position versus time graph given the velocity versus time graph. It is based on the concept that displacement of an object can be found from the area under its velocity versus time graph. Mathematically this is written as an integral . The method works by considering small equal time intervals. In each interval we note whether the area is positive, negative, or zero indicating whether the object moved in the positive direction, negative direction, or returned to the same position. Further we compare the size of the area in each interval to see if the object have moved more or less than in the previous interval. Let’s try a few examples.

In example 1, shown in figure [8a] below, we have an object with constant positive velocity. In [8b] we have considered three equal width intervals. The area in each interval is positive indicating that the object moves in the positive direction. The fact that the areas are the same indicates that the same displacement in each time interval. So each dot in figure [8c] is the same distance away from the previous dot. The result is a straight line of positive slope where we assume, as always, that the object starts at zero.

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Figure 8. Example 1 – using area to find the position versus time graph.

In example 2, shown in figure [9a] below, we have an object with increasing positive velocity. In [9b] we have considered three equal width intervals. The area in each interval is positive indicating that the object moves in the positive direction. The area in each interval is also getting bigger indicating that object moves a greater distance in each time interval. So each dot in figure [9c] is a greater distance away from the previous dot. The result is an upward curve where we assume, as always, that the object starts at zero.

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Figure 9. Example 2 – using area to find the position versus time graph.

In example 3, shown in figure [10a] below, we have an object that starts with negative velocity but that turns around (since v = 0) and finishes with increasing positive velocity. In [10b] we have considered three equal width intervals. The area in the first interval is negative indicating that the object moves in the negative direction – downwards on the position versus time graph. The area in the next interval is positive indicating that in this interval the object moved in the positive direction or upwards on the position versus time graph. The area of the second interval is equal in magnitude to the interval in the first interval, so the object must have returned to its starting position which was assumed to be zero. The last interval is a bigger positive area so the object has travelled farther than in the second interval. Each dot in figure [10c] is placed accordingly. Note that the trough occurs when v = 0.

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Figure 10. Example 3 – using area to find the position versus time graph.