Physics Explained Through a Video Game/Speed and Average Velocity

Topic 1.3 - Speed and Average Velocity

When using the concept of displacement, we can also describe the motion of an object over a period of time. The rate at which the displacement of an object is called its velocity. We can define the average velocity as the average rate at which the displacement is changing over a period of time. Since displacement is a vector, velocity is also a vector. Because velocity is a vector, it describes a object by (1) identifying the direction of motion and (2) finding the average rate at which an object is moving over some distance relative to time.

By using the described definition for average velocity, we can represent average velocity by using the formula in terms of displacement and time as shown below.

${\vec {v}}_{average}={\frac {\text{displacement}}{{\text{change}}\quad {\text{in}}\quad {\text{time}}}}={\frac {\Delta x}{\Delta t}}={\frac {(x_{final}-x_{initial})}{(t_{final}-t_{initial})}}$

To better understand velocity, consider the video on the right. In the map "DumbBell Brown" by GayfishDeluxe players are moving back and forth to fight against one another.

Because velocity is a vector, it varies depending in which direction an object is traveling in. For instance, in the clip, Hot Wheel (reddish-orange color) fires an arrow in both the leftward and rightward directions. By the end of this topic, we will explore the idea that even though the arrow may have been shot just as fast in both cases by Hot Wheel, the arrows will have had different velocities immediately at launch.

Block on Ice Activity

To apply how average velocity is calculated, an interactive demo is provided below. For this activity, we will be observing the motion of a block sliding across ice to gather information about its average velocity.

Task:

• Find any two different times on the on-screen timer where the front of the block is directly above a distance marker.
• Record these distances$(x_{\text{final}}\;and\;x_{\text{initial}})$and times $(t_{\text{final}}\;and\;t_{\text{initial}})$.
• Substitute your data into the equation for average velocity and estimate the block's ${\vec {v}}_{\text{average}}$.

Note:

1. In this experiment, the distance values are defined as being in meters ${\text{(m)}}$ and the time values are defined in seconds ${\text{(s)}}$.
2. For more accurate data collection, record times with the on-screen count-up timer.
3. The overhead, on-screen timer allows for us to gather data for when the block's right-hand side passes over a certain distance marker.

Task Solution:

Here's a table and graph of the collected positions of the block and their associated times. The graph allows for us to see that the position is linearly increasing with respect to time. We can include a curve fit to showcase this relationship.

Also, because ${\vec {v}}_{\text{average}}={\frac {\Delta x}{\Delta t}}$, $\Delta x$ is our vertical change and $\Delta t$ is our horizontal change on the graph. As such, ${\vec {v}}_{\text{average}}$ is the slope of the graph on the right.

Since the graph is linear, this means that the slope (${\vec {v}}_{\text{average}}$) will be the same regardless of what our final and initial (Time, Position) data points were. This explains why the task described above did not specify which two specific data points to collect; ${\vec {v}}_{\text{average}}$ should be similar regardless of the chosen values.

As an example of two selected data points, consider the (Time, Position) data points of (0.71, 0.0) and (3.28, 4.0). Since $t=3.28\;(s)$ occurs after $t=0.71\;(s)$, this means that the data point (3.28, 4.0) are the final values and (0.71, 0.0) are the initial values.

We can substitute these data points into the equation for ${\vec {v}}_{average}$ and simplify.

${\vec {v}}_{average}={\frac {(x_{final}-x_{initial})}{(t_{final}-t_{initial})}}\implies {\vec {v}}_{average}={\frac {(4.0-0)\quad {\text{meters}}}{(3.28-0.71)\quad {\text{seconds}}}}$

$\implies {\vec {v}}_{average}={\frac {(4.0)\quad {\textsf {meters}}}{(2.57)\quad {\textsf {seconds}}}}\implies {\vec {v}}_{average}=1.5564...\;{\textsf {meters/seconds}}$

$\mathbf {\implies {\vec {v}}_{average}=1.6} \;{\textsf {m/s}}\;[{\textsf {rounded}}\;{\textsf {to}}\;2\;{\textsf {sigfigs}}]$

To clarify, because our measured values for calculating ${\vec {v}}_{average}$, distance (m) and time (s), respectively have 2 and 3 significant figures, it's convention that our final answer should only have 2 significant figures. Also, note that answers may vary slightly depending on which exact data points were collected while observing the video.

The Difference Between Speed and Velocity

On the right is a video showcasing circles moving in different directions but having the same speed. Unlike the average velocity of a system, the speed of the system is only dependent on the magnitude of the rate at which an object is moving. Thus, there's a few notable properties of speed:

• The direction angle doesn't affect the speed of an object.
• Since speed is dependent on the magnitude of movement, speed is always a non-negative value (as in either being 0 or positive).
• To calculate for an average speed s in a direction x, use the equation: $s_{x}={\frac {d_{x}}{\Delta t}}$ where $d_{x}$ is the total distance traveled in the x direction and $\Delta t$ is the time elapsed.

The total distance traveled refers to the length of the path in which an object has taken. This is separate from the displacement, which only considers the start and end points of the path. Using the videos below, monkey butler (orange color) is able to jump over the barrier and fill in the Capture Zone in a single jump. In contrast, Other Guy (green color) tries several times in order to cross over the barrier until he is able to do so. Although both monkey butler and Other Guy have the same displacement between their spawning location and the Capture Zone, Other Guy moved a greater total distance because of him running back and forth.

Since speed is always a non-negative value, this means that average speed is not necessarily equal to the average velocity. We can see this through an example of a clip played in Football Mode as shown below.

This is a graph representing the position of the football with respect to time. Whenever the football crosses one of the marked meter lines, a data point is recorded.

This is a graph representing the average velocity of the football with respect to time. The data is extrapolated from the average velocity the football had between each data point in the Position vs. Time graph (above).

From this video, we are able to take a record of whenever the football passes over one of the grid lines and record its time. This allows for us to begin getting an idea of the horizontal motion of the football during the clip.

Through the previous graph of the position of the football relative to time, we can consider each two blue data points from the X-Value Passed vs. Time graph. From this, on the Average Velocity vs. Time graph, we can plot red data points representing the average velocity of the football between each pair of blue data points.

To note the differences between the average velocity and the average speed in this situation, because the average speed of the object can never be negative, whenever ${\vec {v}}_{x,avg}<0;{\vec {v}}_{x,avg}\neq {\text{horizontal}}\;{\text{average}}\;{\text{speed.}}$

Also, if the football was to change direction while between the grid lines, the average velocity cannot equal the average speed. As previously mentioned, average speed is dependent on the total distance of the path taken. This is dissimilar to using displacement to calculate average velocity. Therefore, if the football loops back and forth while traveling, the total distance traveled will be greater than the displacement of the football during the same time interval.

An example of this is between 0.90 and 1.20 seconds in the Football video above. Because the football bounces off the wall and back onto the 9.0 meter line, there's no displacement during this time interval. However, the ball travels a non-zero total distance towards the wall and back. Thus, although ${\vec {v}}_{x,avg}={\frac {x_{t=1.2}-x_{t=0.9}}{t_{1.2}-t_{0.9}}}={\frac {0}{0.3}}=0\;{\textsf {m/s}}$, the average speed over this time interval must be greater than 0 m/s.