Kinematic Graphing and Motion Analysis Study Notes

Fundamental Principles of Kinematic Graphs

• The slope of a position–time ($P-T$) graph is equivalent to the velocity of the object.

• The slope of a velocity–time ($V-T$) graph is equivalent to the acceleration of the object.

Constant Positive Velocity (Rightward Motion)

• An object moving with a constant velocity in the positive (rightward) direction displays specific characteristics across different representations:

  • Dot Diagram: Each consecutive dot is located the same distance apart, indicating that the velocity is not changing.

  • Position–Time Graph: The graph exhibits a slope that is both constant and positive. This represents a constant velocity and a positive direction of motion.

  • Velocity–Time Graph: The graph shows a horizontal line at a constant value, meaning the slope is zero and thus the acceleration is zero ($0\,\text{m/s}^2$). The line is located in the positive region of the graph, corresponding to the positive velocity.

  • Acceleration–Time Graph: The graph displays a horizontal line exactly at the zero mark ($0\,\text{m/s}^2$), confirming zero acceleration.

Constant Negative Velocity (Leftward Motion)

• An object moving with a constant velocity in the negative (leftward) direction displays the following characteristics:

  • Dot Diagram: Consecutive dots are spaced equally, indicating a constant velocity.

  • Position–Time Graph: The slope is constant and negative, indicating motion in the negative direction at an unchanging rate.

  • Velocity–Time Graph: The graph shows a horizontal line located in the negative region. Because the line is horizontal, the slope is zero, signifying zero acceleration.

  • Acceleration–Time Graph: The graph shows a horizontal line at the zero mark, representing zero acceleration.

Positive Velocity and Positive Acceleration

• This scenario describes an object moving in the positive direction while speeding up (rightward velocity with a rightward acceleration):

  • Direction and Speed: Moving in the positive direction results in a positive velocity. If the object is speeding up, the acceleration vector points in the same direction as the motion (positive acceleration).

  • Dot Diagram: Consecutive dots are not the same distance apart; the distance between them increases, representing a changing and increasing velocity.

  • Position–Time Graph: The graph shows a changing slope (curved line) that remains positive. This signifies changing velocity and positive velocity.

  • Velocity–Time Graph: The graph shows a line with a positive (upward) slope. This indicated positive acceleration. The line is located in the positive region, corresponding to positive velocity.

  • Acceleration–Time Graph: The graph shows a horizontal line in the positive region of the graph, representing a constant positive acceleration.

Positive Velocity and Negative Acceleration

• This characterizes an object moving in the positive direction while slowing down (rightward velocity with a leftward acceleration):

  • Direction and Speed: The object has a positive velocity because it moves in the positive direction. Because it is slowing down, the acceleration vector is directed opposite to the motion, resulting in negative acceleration.

  • Dot Diagram: Consecutive dots are not the same distance apart, indicating changing velocity.

  • Position–Time Graph: The slope is changing (curved) and positive, indicating changing positive velocity.

  • Velocity–Time Graph: The graph shows a line with a negative (downward) slope, situated in the positive region of the graph. This indicates positive velocity with negative acceleration.

  • Acceleration–Time Graph: The graph shows a horizontal line in the negative region, indicating a constant negative acceleration.

Negative Velocity and Negative Acceleration

• This describes an object moving in the negative direction while speeding up (leftward velocity with a leftward acceleration):

  • Direction and Speed: Motion in the negative direction yields a negative velocity. Since the object is speeding up, the acceleration vector must be in the same direction as motion, which is negative.

  • Dot Diagram: The spacing between dots is not uniform, representing a changing velocity.

  • Position–Time Graph: The slope is changing and negative.

  • Velocity–Time Graph: The graph features a line with a negative (downward) slope located entirely in the negative region. This represents negative velocity and negative acceleration.

  • Acceleration–Time Graph: The graph features a horizontal line in the negative region, indicating constant negative acceleration.

Negative Velocity and Positive Acceleration

• This describes an object moving in the negative direction while slowing down (leftward velocity with a rightward acceleration):

  • Direction and Speed: The direction of motion is negative, resulting in a negative velocity. The object is slowing down, so the acceleration vector is in the opposite direction (positive acceleration).

  • Dot Diagram: Consecutive dots are not equally spaced, indicating a changing velocity.

  • Position–Time Graph: The slope is changing and negative.

  • Velocity–Time Graph: The line has a positive (upward) slope but is located in the negative region of the graph. This corresponds to a negative velocity and a positive acceleration.

  • Acceleration–Time Graph: The graph shows a horizontal line in the positive region, indicating constant positive acceleration.

Interpreting Multi-Interval Velocity-Time Graphs

• Scenario A: An object moves in the positive ($+$) direction at a constant speed with zero acceleration during interval A. In interval B, it continues in the positive direction but slows down with a negative acceleration. In interval C, it moves at a constant speed in the positive direction, but at a slower speed than in interval A.

• Scenario B: In interval A, the object moves in the positive ($+$) direction while slowing down, which involves a negative acceleration. In interval B, the object remains at rest ($v = 0\,\text{m/s}$). In interval C, the object moves in the negative ($-$) direction while speeding up, which also involves a negative acceleration.

• Scenario C: In interval A, the object moves in the positive ($+$) direction with a constant velocity and zero acceleration. In interval B, it slows down while moving in the positive direction (negative acceleration) until it reaches zero velocity and stops. In interval C, the object moves in the negative ($-$) direction while speeding up, corresponding to a negative acceleration.

Displacement and the Area Under a Velocity–Time Graph

• For a velocity versus time graph, the displacement of an object is represented by the area bounded by the line and the axes.

• Rectangle Displacement: If the motion forms a rectangle between the line and the time-axis (e.g., from $0\,\text{s}$ to $6\,\text{s}$), the displacement is calculated as:

  • $\text{Area of a rectangle} = l \times w$

  • $A = s \times \text{m/s} = \text{meters}$

• Triangle Displacement: If the shape formed is a triangle (e.g., from $0\,\text{s}$ to $4\,\text{s}$ with a peak velocity of $40\,\text{m/s}$), the displacement is:

  • $A = \frac{1}{2} b \times h$

  • $A = \frac{1}{2} (4\,\text{s}) \times (40\,\text{m/s}) = 80\,\text{m}$

• Trapezoid Displacement: If the shape formed is a trapezoid (e.g., from $2\,\text{s}$ to $5\,\text{s}$ where the velocity changes from $20\,\text{m/s}$ to $50\,\text{m/s}$), the displacement is:

  • $A = \frac{1}{2} b \times (h_1 + h_2)$

  • $A = \frac{1}{2} (3\,\text{s}) \times (20\,\text{m/s} + 50\,\text{m/s}) = 105\,\text{m}$

• Alternative Trapezoid Method: A trapezoid can be broken down into a rectangle and a triangle. The total displacement is the sum of the area of the rectangle and the area of the triangle.

Questions & Discussion

• Consider a blue car moving at a constant speed of $10\,\text{m/s}$ and a red car at rest. At $t = 0$, both are at a stoplight when it turns green. The red car accelerates from rest at $4\,\text{m/s/s}$ for $3\,\text{s}$ and then maintains a constant speed. The blue car maintains $10\,\text{m/s}$ for $12\,\text{s}$.

• Question: What is the final velocity of the red car after three seconds of acceleration?

  • Answer: The final velocity is $12\,\text{m/s}$.

• Question: What is the displacement of each individual car after three seconds?

  • Answer: For the Red Car, the area is a triangle: $0.5 \times (3\,\text{s}) \times (12\,\text{m/s}) = 18\,\text{m}$. For the Blue Car, the area is a rectangle: $(3\,\text{s}) \times (10\,\text{m/s}) = 30\,\text{m}$.

• Question: What is the slope of the line for the red car for the first three seconds?

  • Answer: $\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{12\,\text{m/s} - 0\,\text{m/s}}{3\,\text{s}} = 4\,\text{m/s/s}$.

• Question: What is the displacement of each individual car after nine seconds?

  • Answer: For the Red Car, the displacement is the area of the triangle plus the area of the rectangle: $(0.5 \times 3\,\text{s} \times 12\,\text{m/s}) + (6\,\text{s} \times 12\,\text{m/s}) = 18\,\text{m} + 72\,\text{m} = 90\,\text{m}$. For the Blue Car, the displacement is the area of the rectangle: $(9\,\text{s}) \times (10\,\text{m/s}) = 90\,\text{m}$.

• Question: Does the red car pass the blue car at three seconds? If not, when does it pass?

  • Answer: No. At three seconds, the blue car ($30\,\text{m}$) is ahead of the red car ($18\,\text{m}$). The red car passes the blue car at $9\,\text{s}$ when both have reached a displacement of $90\,\text{m}$.

• Question: When lines on a velocity–time graph intersect, does it mean the cars are passing each other?

  • Answer: No. Intersecting lines on a velocity–time graph mean the cars have the same velocity at that moment. Cars are passing each other only when the lines intersect on a position–time graph.