Erika Giroux, Traci Klotz, Kelly Krater Velocity Graph:
Acceleration Graph:
The displacement, velocity, and acceleration graphs for the segment of Traci’s swing before the ball was hit are shown above.
The important segment of the displacement curve for the first part of the swing is highlighted on the graph. This highlighted area is a good representation of the bat’s motion as it is brought through the heart of the first part of the swing.
The slope of a position graph is the object's velocity, which is calculated through dividing change in distance by change in time. In this case, distance is approximately equal to 0.37 m, and time is approximately equal to 0.014 s.
V = Δd = 0.37 m_ = 26.43 m/s, which is very close to the actual 26.49 m/s. Δt 0.014 s
A velocity of 26.43 m/s is reflected by the velocity graph of the bat. The velocity of the bat in the aforementioned highlighted region before contact, according to this graph, ranges from a final velocity of 27 m/s to an initial velocity of 23.4 m/s. The data points do not all fall exactly within this best fit curve because of point placement in the Tracker program.
The displacement of the bat should be equal to the area under the velocity graph. The displacement charted on the position graph is 0.37 m. Correspondingly, the area under the velocity graph for the same set of data points is also equal to 0.37 m.
Since the bat's velocity is increasing during this segment of the swing, this means that the bat is accelerating at a constant rate, indicating that the acceleration graph will contain a horizontal line extending into the quadrant where "x" is positive (in this case, the quadrant on the right), at some value above zero. The best fit line of the acceleration graph essentially matches this prediction.
The bat's acceleration (its rate of change in velocity) can also be determined from the velocity graph. Acceleration is equal to change in velocity divided by change in time. According to the slope of the velocity graph, acceleration is 258.12 m/s^{2}. This can be calculated by dividing the values given in the graph for velocity and time.
A = Δv = 3.6 m/s = 257.14 m/s^{2}, which is very close to actual value 258.12 m/s^{2} Δt 0.014 s
An acceleration of 258.12 m/s^{2} is reflected by the acceleration graph of the bat before contact. The acceleration ranges from 900 m/s^{2} to 250 m/s^{2} on the graph, and all the data points average out to be around 250 m/s^{2}. These variations in the data are due to point placement in the Tracker program.
The area under the acceleration graph should be equal to the object’s change in velocity. The change in velocity was determined to be 3.6 m/s. The area under the acceleration graph is approximately equal to 3.7 m/s, which is nearly exact.
Swing After Hit: Position Graph: Velocity Graph:
Acceleration Graph:
The position, velocity, and acceleration graphs for Traci’s bat after it made contact with the ball are shown above.
After the ball was hit, the bat’s velocity greatly decreased. The slope of the position graph lessened greatly into only a gently sloping line traveling upwards to the right at a diagonal. A new segment of the bat’s path was then chosen to perform these next sets of calculations. During this segment of the swing, the velocity was 1.7 m/s, as calculated by:
V = Δd = 0.02 m_ = 1.67 m/s, which is very close to the actual 1.7 m/s. Δt 0.012 s
This change in velocity is reflected by the shift in positioning of data points on the velocity graph. The majority of the data from after contact is located around 2.3 m/s, with a scattering of points below that number to balance out to around 1.7 m/s.
The displacement for this segment of the graph is 0.02 m, which is also shown in the area under the velocity graph, equal to approximately the same 0.02 m.
The acceleration in this segment differs from that of the first segment. It is much less, since the bat was vastly slowed by the impact of the ball and is now gradually regaining its speed. This is shown by the calculations:
A = Δv = 1.7 m/s = 133.33 m/s^{2}, which is very close to actual value 122.4 m/s^{2}. Δt 0.012 s
The slope of the acceleration graph reflects this change in velocity. The best fit line is again nearly a horizontal line, indicating constant acceleration. The points on the graph range from approximately 500 m/s^{2} to 300 m/s^{2}, which even out to be around the actual 133 m/s^{2}. Again, all variation in data point location is due to point placement in the Tracker program.
The area under the acceleration graph should be equal to the object’s change in velocity; the area is around 1.6 m/s, and the actual velocity is nearly that, at 1.7 m/s.
Ball Before Hit: Position Graph: Velocity Graph:
Acceleration Graph:
The position, velocity, and acceleration graphs for the ball before it was hit are shown above.
The full path of the ball ends up being almost a Vshape. The displacement before it was hit is the left side of the V. Its motion originated at the right side of the coordinate plane and traveled toward the origin. Since its movement was to the left, from the perspective of the coordinate plane, its movement was also in the negative direction. A representative segment of the ball’s motion was chosen for the following calculations. Its displacement is calculated as such:
V = Δd = –0.75 m = –8.33 m/s, which is very close to the actual –8.064 m/s. Δt 0.009 s
The velocity indicated on the velocity graph should be relatively constant and should reflect this calculated 8.3 m/s. Most of the data points on the graph indicating velocity before contact with the bat are located very near 8 m/s, with a few above and a few below. The outlying points balance out to about 8 m/s.
The area under the velocity graph should be equal to the ball’s total displacement. The displacement derived from the position graph is 0.75 m, and the area under the velocity graph is 0.7 m. These measurements are nearly exact.
The slope of the velocity graph is equal to the ball’s acceleration. This slope is calculated by:
A = Δv = –0.08 m/s = –8.89 m/s^{2}, which is very close to actual value –8.9 m/s^{2}. Δt 0.009 s
This value of 8.9 m/s^{2} approximately matches the acceleration graph. The line of the acceleration graph should be horizontal and located slightly above zero. The best fit line of the graph does basically follow this path. The points above and below cancel out to about equal 9.
The area under the acceleration graph should be the same as total velocity. The ball’s velocity is –8 m/s, and the area under the acceleration graph is about –6, so the two numbers essentially match.
Ball After Hit: Position Graph: Acceleration Graph:
The position, velocity, and acceleration graphs for the ball after it was hit are shown above. Again, a representative segment of the ball’s motion was selected for these calculations.
Once the ball was hit, the right side of the “V” sloped upward at a steeper angle than the left side had sloped downward. This increased slope means that the velocity increased also because the ball moved away from the bat more quickly than it approached it. The velocity increased to 15.84 m/s.
V = Δd = 0.125 m = 15.63 m/s, which is very close to the actual 15.84 m/s. Δt 0.008 s
The velocity indicated on the velocity graph changed drastically after the ball was hit, primarily because the ball now traveled in the positive direction, which made the signs on the numbers positive also. The velocity graph had to alter its direction from a low negative number to a higher positive number in very little time, which led to a sharp upward slope of the line. This line balanced out around 16 m/s.
The area under the velocity graph should be equal to the ball’s total displacement. The displacement from the position graph is 0.125 m, and the area under the velocity graph is 0.125 m; the numbers are identical.
The acceleration also changes after the ball is hit. It changes to a positive acceleration since the ball is speeding up while moving in the positive direction. Its acceleration is around 183.8 m/s^{2}.
A = Δv = 1.4 m/s = 156 m/s^{2}, which is close to actual value 183.8 m/s^{2}. Δt 0.009 s
The points on the acceleration graph roughly balance out to equal this 183.8 m/s^{2}. The data range from 500 m/s^{2} to –450 m/s^{2}, and the points average out to around 183 m/s^{2}.
The area under the acceleration graph should equal the ball’s velocity. Its velocity is 15.84 m/s, and the area under the graph is about the same. Bat During Hit: Displacement Graph: Velocity Graph:
Acceleration Graph:
During the portion of the video where the ball is hit, both the bat and the ball undergo drastic changes in their motion. First, we will examine the bat.
For the duration of its contact with the ball, the bat experiences a drastic acceleration. Its average velocity during this period is 20.65 m/s, as demonstrated by the slope of the displacement graph.
v = Δd = 0.16 m = 20 m/s, which is very close to the actual 20.65 m/s. Δ t 0.008 s
However, the average velocity does not tell the entire story of the incident. The velocity graph tracks the bat’s velocity at different instants of time. It shows that the bat’s velocity dropped drastically during the hit, from nearly 22 m/s all the way to 4 m/s, in a span of time approximately 0.008 s in duration.
This change in velocity occurred over a very small displacement, indicated by the total area under the velocity graph. This displacement is equal to about 0.16 m, or 16 cm.
The slope of the velocity graph indicates the bat’s acceleration during the hit, which is –2682 m/s^{2}. The average acceleration can be found using the data points on the velocity graph, which, for this segment, begin at 4 m/s and end at 23 m/s, along with the time. This calculation is as follows:
a = Δv = 19 m/s = –2375 m/s^{2}, which is close to the actual –2682 m/s^{2}. Δ t 0.008 s
Once again, the average acceleration does not provide all of the pertinent information. The acceleration graph tracks the bat’s acceleration at different instants of time. In this case, it shows that the bat’s acceleration changed during the hit, from 250 m/s^{2} to –1200 m/s^{2}, in a span of time lasting approximately 0.008 s.
The change in velocity during this time span is indicated by the area under the acceleration graph, which totals –19 m/s.
Ball During Hit: Displacement Graph: Velocity Graph:
Acceleration Graph:
Now, let’s look at the changes in the ball’s motion.
For the duration of its contact with the bat, the ball experiences a much more drastic acceleration than even the bat. Its average velocity during this period is –4.032 m/s, as demonstrated by the slope of the displacement graph.
v = Δd = 0.25 m = –4.17 m/s, which is very close to the actual –4.032 m/s. s Δ t 0.006 s
However, the average velocity once again does not tell the entire story of the incident. The velocity graph tracks the ball’s velocity at different instants of time. It shows that the ball’s velocity changed during the hit, from nearly 10 m/s all the way to 15 m/s, in a span of time approximately 0.006 s in duration.
This change in velocity occurred over a very small displacement, indicated by the total area under the velocity graph. This displacement is equal to about 0.018 m, or 1.18 cm.
The slope of the velocity graph indicates the bat’s acceleration during the hit, which is 4619 m/s^{2}. The average acceleration can be found using the data points on the velocity graph, which, for this segment, begin at 10 m/s and end at 15 m/s, along with the time. This calculation is as follows:
a = Δv = 25 m/s = 4166.7 m/s^{2}, which is close to the actual 4619m/s^{2}. Δ t 0.006 s
Once again, the average acceleration does not provide all of the pertinent information. The acceleration graph tracks the bat’s acceleration at different instants of time. In this case, it shows that the bat’s acceleration changed during the hit, starting at 1500 m/s^{2} and spiking up to nearly 5000 m/s^{2}, in a span of time lasting approximately 0.006 s.
The change in velocity during this time span is indicated by the area under the acceleration graph, which totals approximately 22 m/s. Forces Bat Mass: 0.52 kg Ball Mass: 0.3 kg
Force During Hit: On Bat: On Ball: a = F_{net} a = F_{net} m m
–2682 m/s^{2} = F_{net__}_ 4619 m/s^{2} = F_{net__} 0.52 kg 0.3 kg
–1394.64 N = F_{net (xdirection) }1385.7 N = F_{net (xdirection)}
Newton’s First Law states, essentially, that an object at rest will remain at rest and an object moving at constant velocity will remain moving at constant velocity unless acted upon by a net force. This is also known as the “law of inertia,” since inertia is an object’s tendency to resist changes in its motion. A change in motion is caused by an unbalanced force (a “push” or “pull”) acting on the object, which is known as the net force. In our Physics of Sports demonstration, this law held true: the ball continued moving on a path toward the bat at constant velocity until the bat made contact with it. Then the force of the bat on the ball caused the ball to accelerate away from the bat in the direction from which it originated.
Newton’s Second Law states that acceleration of an object is equal to net force on the object divided by its mass. This equation can also be rearranged so that net force equals acceleration of an object multiplied by its mass. With this equation in mind, the Second Law also held true in our Physics of Sports demonstration.
Newton’s Third Law states, in oversimplied terms, that for every action, there is an equal and opposite reaction. A better definition would be that every force is part of an interaction pair, in which the force of A on B is equal in magnitude (size) but opposite in direction (sign) to the force of B on A. These interaction forces act upon different objects. To call them “actionreaction” forces would be a misnomer because it suggests that one causes the other, which is not true; they either exist together or not at all. This law held true in our Physics of Sports demonstration. The force of the bat on the ball (1387.5 N) was approximately equal in magnitude but opposite in direction to the force of the ball on the bat (1394.64 N). The slight discrepancy in the magnitudes is a result of point placement in the Tracker program which provided the acceleration magnitude used in the equation, but they are close enough to be considered the same.
