Interpreting Velocity Graphs

Describe the motion corresponding to the following graph and plot the position vs time and acceleration vs time assuming the initial position is \(\text{0}\):

Step 1: Determine what is required

The velocity vs time graph corresponding to the motion of an object is given, using the graph we need to describe the actual motion. Notice that the graph has discontinuities, this naturally breaks the graph into phases. We should process each phase separately and then try to infer (figure out) what probably happened to change from one phase to the next.

The first thing that you normally do in a vertical projectile motion problem is choose a positive direction. In this case you can't do that because it has already been decided. We need to use the information given to figure out which direction was chosen as positive.

Step 2: Determine the direction chosen as positive

We are given a velocity vs time graph. We know that the slope of the velocity time graph is the acceleration and we know that in these problems we are only dealing with gravitational acceleration. We can use what we know about gravitational acceleration and the slope of the graph to determine which direction was chosen as positive.

Gravitational acceleration is always directed towards the centre of the Earth. If we choose:

• upwards as the positive direction then the slope of the velocity vs time graph will be negative
• downwards as the positive direction then the slope of the velocity vs time graph will be positive

In this problem the slope of the velocity vs time graph is negative and so we know that upwards was chosen as the positive direction.

Step : First phase

In the plot below we highlight only the first phase of interest.

The initial velocity is positive and then decreases linearly, passes through zero and reaches the same magnitude in the negative direction. The positive starting value means the object was going upwards, the linear decreases means a constant acceleration in the negative direction, passing through zero means that the object reaches a maximum height before beginning to fall in the negative direction. The fact that the magnitude of the initial and final velocities is the same means that there is time symmetry.

This is just the plot of an object being thrown/shot/projected upwards, peaking and falling back to the same level as it was thrown from.

Step 3: Second phase

In the plot below we highlight only the second phase.

Just as in the first phase, the initial velocity is positive and then decreases linearly, passes through zero and reaches the same magnitude in the negative direction. The positive starting value means the object was going upwards, the linear decreases means a constant acceleration in the negative direction, passing through zero means that the object reaches a maximum height before beginning to fall in the negative direction. The fact that the magnitude of the initial and final velocities is the same means that there is time symmetry.

This is just the plot of an object being thrown upwards, peaking and falling back to the same level as it was thrown from.

Step 4: Combining phases

We know that the two phases look the same, the object starts off moving upwards, peaks, falls down to the initial position and then repeats the process. This could be a description of a ball bouncing as an example.

Step 5: Position vs time

To draw an accurate position graph we know that we are dealing with the Case 1 situation from earlier. We need to determine the height that the object rises to and then we can plot the position vs time. We know the timing information from the velocity vs time plot.

We can verify that the acceleration is in fact from gravity by using \(\vec{v}_f=\vec{v}_i+\vec{g}t\):

\begin{align*} \vec{v}_f & =\vec{v}_i+\vec{g}t \\ (-\text{40}) & = (\text{40}) +\vec{g}(\text{8,16}) \\ (-\text{80}) & = \vec{g}(\text{8,16}) \\ \vec{g} & = \frac{-\text{80}}{\text{8,16}} \\ \vec{g} & = -\text{9,8}\text{ m·s $^{-2}$} \end{align*}

We can work out the height using the equation for displacement:

\begin{align*} \vec{v}_f^2 & =\vec{v}_i^2+2\vec{g}\Delta \vec{x} \\ 0 & =(\text{40})^2+2(-\text{9,8})\Delta \vec{x} \\ \Delta \vec{x} & =\frac{-(\text{40})^2}{(-\text{19,6})} \\ \Delta \vec{x} & = \text{81,63}\text{ m} \end{align*}

So we can now draw the displacement vs time plot:

Step 6: Acceleration vs time

The acceleration is due to gravity except at one point, when the object bounces. At that point the acceleration must be some other, very large, value in the positive direction but we don't have enough information to determine what the value must be. We explicitly note that at that point the acceleration is NOT -\(\text{9.8}\) \(\text{m·s $^{-2}$}\).