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In this Physics tutorial, you will learn:
|3.10||Velocity v's Time and Speed v's Time Graph|
As discussed in the previous Physics tutorial "Position vs Time & Distance vs Time Graph", we obtain useful information from motion graphs. They represent another "language" by which Physics communicates with its readers besides the language of words. Therefore, motion graphs are very important when trying to find the relationship between physical quantities involved in Kinematics but not only.
In this tutorial, we will deal with Velocity vs Time and Speed vs Time graphs. The approach will be very gradual, so the reader will have the chance to understand in depth these kind of motion graphs.
Like in the Position vs Time graph, in the Velocity vs Time graph the horizontal axis contains the Time, t, while the velocity is shown at the vertical axis.
In uniform motion, the velocity is constant. This means the Velocity vs Time graph will be a horizontal line, which lies v⃗ units above or below zero depending on the sign of velocity. This line extends horizontally from zero up to the moving time t.
Let's explain this point by illustrating it with numbers. Suppose we are required to find the displacement and distance of the object whose motion is shown in the graph below.
From the graph we can see that the velocity during the first 10 s is constant and positive (v⃗1 = 12 m/s). This means during this interval the object is moving at constant velocity in the positive direction of the position axis.
During the next 15 seconds, the object has a negative constant velocity (v⃗2 = -4 m/s). This means it is moving at constant velocity towards the negative direction of position axis (although it may still be moving on the positive part of the position axis).
Therefore, applying the equation of uniform motion for both parts of the motion, we obtain
Hence, the total displacement ∆x⃗tot is
As we have explained in the tutorial "Speed and Velocity in One Direction", the speed cannot be negative (and time as well). Therefore, the Speed vs Time graph will contain only the upper-right part of a two dimensional graph. Let's explain this point by considering the numerical values of the previous example. Any negative part of the Velocity vs Time graph is flipped vertically and it will be considered as positive in the corresponding Speed vs Time graph as shown below.
The total Distance stot will therefore be:
The above results comply with the numerical values of the area enclosed by the graph and the horizontal axis. If the graph in the velocity vs time graph is in the negative part, the area is taken as negative. Therefore, we obtain a very important property of the Velocity vs Time and Speed vs Time graphs:
In the previous tutorials, we stated that in uniformly accelerated (decelerated) motion, i.e. in the motion with constant acceleration, the velocity increases or decreases at the same rate. This means the Velocity vs Time graph of this kind of motion will be a sloped straight line as shown below
In the first part of the graph, the acceleration is negative as the velocity decreases, while in the second part, the acceleration is positive as the velocity increases.
Also, here again we can find the displacement by calculating the area under the graph as stated in the previous paragraph. Thus, the area bordered by the graph and the horizontal axis gives the numerical value of the displacement. Let's explain this point by illustrating it with numbers.
The velocity vs time graph below represents the motion of an object in a linear direction.
From the graph, we can extract the following values for the first interval. We have
Thus, we obtain
The same procedure is used to calculate the acceleration during the second interval Δt2. We have
We have to calculate the area under the graph to calculate the displacement. Both parts of the graph represent trapeziums, so their area is calculated by the equation
In the first interval, we have:
The same procedure is used in the second interval as well. We have:
Therefore, the total displacement is
In the above problem, it was confirmed the veracity of the second formula of the uniformly accelerated (decelerated) motion
Like in the distance vs time graph, the speed vs time graph is similar to the velocity vs time graph when the acceleration (slope) is positive. On the other hand, in the part of the graph in which the slope is negative (where the acceleration is negative), the graph is obtained by flipping vertically the corresponding section of the velocity vs time graph, as shown below.
The second graph (the speed vs time graph) does not provide any information on the direction of motion; it only shows that the object first decelerates until it stops, then it accelerates. This may either occur when the object changes direction or when it only slows down, stops and accelerates again at the same direction. Therefore, the velocity vs time graph is more comprehensive than the corresponding speed vs time graph.
The velocity vs time graph for a moving object is shown in the figure below.
From the graph, we can extract the following values during the first 4 s:
Thus, we have
As you see, the acceleration is positive although the object is slowing down. This is because it is moving in the negative part of coordinates and slowing down in the negative part of coordinates has the same effect as speeding up in the positive one.
In the following 4 s (from 4s to 8s) we have:
In the third interval (from 8s to 14s) we have:
Hence, we have
The displacement in the first 4s is obtained by calculating the magnitude of the corresponding area of the figure, i.e.
The displacement in the next 4s (from 4s to 8s) is obtained by calculating the magnitude of the corresponding area of the figure, i.e.
The displacement in the last 6s (from 8s to 14s) is obtained by calculating the corresponding area of figure, i.e.
The total displacement therefore, is obtained by adding all the previous displacements
The speed vs time graph differs from the velocity vs time graph only in the negative part (the graph is flipped vertically). Thus,
How to plot the Position vs Time graph when the corresponding Velocity vs Time graph is given?
First, let's take a simple velocity vs time graph to explain this idea and then we will deal with more complicated graphs.
An object starts moving from x⃗0 = 12 m. Its velocity vs time graph is shown below.
Plot the corresponding Position vs Time graph for this motion.
The graph shows a uniformly accelerated motion where v⃗0 = 4 m/s, v⃗ = 16 m/s, t = 6s and x⃗0 = 12m. From these clues, we can calculate the acceleration a⃗ for this motion. Thus,
Thus giving that the equation of motion with constant acceleration is a quadratic equation of the type
we obtain after the substitutions
The graph of this quadratic function is a parabola. Therefore, we need to find several values in order to plot an accurate graph. These values are obtained by substituting t by 0, 1, 2, 3, 4, 5 and 6 respectively. The following table shows the results obtained.
|Time t (s)||0||1||2||3||4||5||6|
|Position x (m)||12||17||24||33||44||57||72|
Now, let's insert these points in the position vs time graph and connect them smoothly.
As you see, the graph is a positive (arms-up) parabola since the acceleration is positive (the object speeds up).
Now, let's consider a more complicated example in which multiple kinds of motion are involved.
Plot the position vs time graph for the motion described in the velocity vs time graph shown below if initially the object was at x⃗0 = 30 m.
From the graph we can see the motion is composed by three parts: the first part (from 0 to 12 s) represents an accelerated motion; in the second part (from 12 s to 22 s) the motion is uniform; and in the third part (from 22 s to 30 s) the motion is decelerated (by the end of which the object eventually stops because the final velocity becomes zero).
a. In the first interval the approach is similar to that of the previous example. We have v⃗0 = 5 m/s, v⃗1 = 8 m/s, t1 = 12s. Therefore, we have for the acceleration a⃗1:
Therefore, the equation of motion for the first interval is
Substituting the values, we obtain
We can find some points in the graph to make the graph more accurate. We have
|x⃗ = 30 + 5 × t + 0.25 × t2/2||46.125 m||64.5 m||85.125 m||108 m|
We will insert later these points in the final graph.
b. In the second interval, the velocity is constant (v⃗1 = v⃗2 = 8 m/s). Therefore, we have
This means by the end of the second interval the object will be at
c. In the third interval, the motion is decelerated. The approach is the same as in the first part. The only difference is that here the initial position is x⃗2 = 188 m.
The general equation of the third interval is
Substituting the known values, we obtain the specific equation of motion in the third interval,
Hence, at the end of the motion the object will be at
Again, we can use the table method to find some intermediate points in the graph to make its plotting more accurate. We have
|t (s)||2 (24 - 22)||4 (26 - 22)||6 (28 - 22)||8 (30 - 22)|
|x⃗ = 188 + 8 × t - 0.5 × t2||202 m||212 m||218 m||220 m|
Now, let's insert all the above values obtained during the solution of this exercise in the same Position vs Time graph
The Distance vs Time graph would be the same as the Velocity vs Time in this example as the entire movement has occurred only in the positive direction.
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