AP Physics C Mechanics- 1.3 Representing Motion- Study Notes- New Syllabus
AP Physics C Mechanics- 1.3 Representing Motion – Study Notes
AP Physics C Mechanics- 1.3 Representing Motion – Study Notes – per latest Syllabus.
Key Concepts:
- Representations of Motion and Kinematic Equations for Constant Acceleration
- Acceleration Due to Gravity Near the Surface of Earth
- Graphs of Position, Velocity, and Acceleration as Functions of Time (Motion Graphs)
- Instantaneous Velocity, Instantaneous Acceleration, and Area Relationships Between Motion Graphs
Representations of Motion and Kinematic Equations for Constant Acceleration
Motion can be represented in several ways — each providing unique insight into how an object moves. Depending on the context, motion may be described through:
- Motion diagrams: show the position of an object at equal time intervals using dots or arrows.
- Figures: depict paths or trajectories of motion.
- Graphs: display relationships between position, velocity, acceleration, and time.
- Equations: provide mathematical relationships that describe motion quantitatively.
- Narrative descriptions: explain motion conceptually in words.
Motion with Constant Acceleration:
When an object moves with constant acceleration in one dimension, three key kinematic equations relate displacement, velocity, acceleration, and time. These equations describe instantaneous linear motion under uniform acceleration.
1. Velocity–Time Relation:
\( \mathrm{v_x = v_{0x} + a_x t} \)
- Relates velocity to time under constant acceleration.
- Graphically, it represents the slope of a straight line on a \( \mathrm{v_x \text{ vs } t} \) graph.
2. Displacement–Time Relation:
\( \mathrm{x = x_0 + v_{0x}t + \dfrac{1}{2}a_x t^2} \)
- Gives displacement as a function of time.
- Describes a parabolic position–time graph when acceleration is constant.
3. Velocity–Displacement Relation:
\( \mathrm{v_x^2 = v_{0x}^2 + 2a_x(x – x_0)} \)
- Relates velocity and displacement without involving time.
- Useful when time is not known or not required.
Important Notes:
- The subscript \( \mathrm{x} \) indicates motion in the x-direction. These equations can be similarly applied to motion in the y- or z-direction.
- All three equations are valid only under constant acceleration.
- Instantaneous velocity and position at any time can be obtained directly from these equations.
Key Idea: Each kinematic equation connects a subset of four fundamental motion variables — displacement, velocity, acceleration, and time. Knowing any three allows determination of the fourth.
Example:
A car starts from rest (\( \mathrm{v_{0x} = 0} \)) and accelerates uniformly at \( \mathrm{a_x = 3 \, m/s^2} \) for \( \mathrm{t = 6 \, s} \). Determine:
- (a) The final velocity
- (b) The total displacement
▶️ Answer / Explanation
Given: \( \mathrm{v_{0x} = 0, \, a_x = 3 \, m/s^2, \, t = 6 \, s} \)
(a) Final Velocity:
Using \( \mathrm{v_x = v_{0x} + a_x t} \):
\( \mathrm{v_x = 0 + (3)(6) = 18 \, m/s} \)
(b) Displacement:
Using \( \mathrm{x = x_0 + v_{0x}t + \dfrac{1}{2}a_x t^2} \):
\( \mathrm{x – x_0 = 0 + 0 + \dfrac{1}{2}(3)(6)^2 = 54 \, m} \)
Acceleration Due to Gravity Near the Surface of Earth
All freely falling objects near the surface of Earth experience a constant acceleration directed downward toward the center of Earth. This acceleration is produced by the force of gravity acting on the object’s mass.
Definition:
The acceleration due to gravity is denoted by \( \mathrm{g} \) and represents the rate at which the velocity of a freely falling body changes with time.
\( \mathrm{\vec{a} = \vec{g}} \)
Near the Earth’s surface, this acceleration is approximately constant in both magnitude and direction:
\( \mathrm{|\vec{g}| = g \approx 9.8 \, m/s^2 \, \text{(often rounded to } 10 \, m/s^2)} \)
- Direction of \( \mathrm{\vec{g}} \): vertically downward (toward Earth’s center).
- Magnitude is nearly constant for all objects close to Earth’s surface.
- Independent of the object’s mass (in the absence of air resistance).
Key Concept:
For motion in the vertical direction, the acceleration is given by:
\( \mathrm{a_y = -g} \) (if upward is chosen as the positive direction)
Thus, motion equations for free fall or vertical projection become:
- \( \mathrm{v_y = v_{0y} – gt} \)
- \( \mathrm{y = y_0 + v_{0y}t – \dfrac{1}{2}gt^2} \)
- \( \mathrm{v_y^2 = v_{0y}^2 – 2g(y – y_0)} \)
Key Idea: Every object near Earth’s surface experiences the same downward acceleration \( \mathrm{g} \), regardless of its mass or composition — a fundamental result of Galileo’s experiments on falling bodies.
Example:
An object is dropped from rest from a height of \( \mathrm{45 \, m} \). Find:
- (a) The time it takes to reach the ground.
- (b) Its velocity just before impact.
- Take \( \mathrm{g = 9.8 \, m/s^2} \).
▶️ Answer / Explanation
Given: \( \mathrm{u = 0}, \, \mathrm{y – y_0 = -45 \, m}, \, \mathrm{g = 9.8 \, m/s^2} \)
(a) Time to reach the ground:
Using \( \mathrm{y = y_0 + ut – \dfrac{1}{2}gt^2} \):
\( \mathrm{-45 = -\dfrac{1}{2}(9.8)t^2} \)
\( \mathrm{t^2 = \dfrac{45 \times 2}{9.8} = 9.18} \Rightarrow \mathrm{t = 3.03 \, s} \)
(b) Velocity before impact:
Using \( \mathrm{v = u – gt} \):
\( \mathrm{v = 0 – (9.8)(3.03) = -29.7 \, m/s} \)
(Negative sign indicates downward direction.)
Graphs of Position, Velocity, and Acceleration as Functions of Time (Motion Graphs)
Graphs are powerful tools for visualizing motion. They provide a direct way to represent how an object’s position, velocity, and acceleration change with time and help reveal the mathematical relationships between these quantities.
Key Concept:
The slope and area of motion graphs are used to find how one quantity relates to another:
- The slope of a position–time (\( \mathrm{x\text{–}t} \)) graph gives velocity.
- The slope of a velocity–time (\( \mathrm{v\text{–}t} \)) graph gives acceleration.
- The area under a velocity–time graph gives displacement.
- The area under an acceleration–time graph gives change in velocity.
1. Position–Time Graph (\( \mathrm{x\text{–}t} \))
- The slope of the \( \mathrm{x\text{–}t} \) graph represents the object’s velocity:
\( \mathrm{v = \dfrac{dx}{dt}} \)
- A straight line indicates constant velocity.
- A curved line indicates changing velocity (i.e., acceleration).
- The steeper the slope, the greater the speed.
2. Velocity–Time Graph (\( \mathrm{v\text{–}t} \))
- The slope represents acceleration:
\( \mathrm{a = \dfrac{dv}{dt}} \)
- Constant acceleration → straight-line \( \mathrm{v\text{–}t} \) graph.
- The area under the graph gives displacement:
\( \mathrm{\Delta x = \displaystyle \int v \, dt} \)
- Positive slope → increasing velocity (acceleration).
- Negative slope → decreasing velocity (deceleration).
3. Acceleration–Time Graph (\( \mathrm{a\text{–}t} \))
- For constant acceleration, the graph is a horizontal line.
- The area under the graph gives change in velocity:
\( \mathrm{\Delta v = \displaystyle \int a \, dt} \)
- A positive value indicates acceleration in the positive direction; negative indicates deceleration.
Key Relationships Between the Graphs:
| Graph | Slope Represents | Area Under Curve Represents |
|---|---|---|
| Position–Time (\( \mathrm{x\text{–}t} \)) | Velocity (\( \mathrm{v = \dfrac{dx}{dt}} \)) | — |
| Velocity–Time (\( \mathrm{v\text{–}t} \)) | Acceleration (\( \mathrm{a = \dfrac{dv}{dt}} \)) | Displacement (\( \mathrm{\Delta x = \displaystyle \int v \, dt} \)) |
| Acceleration–Time (\( \mathrm{a\text{–}t} \)) | — | Change in velocity (\( \mathrm{\Delta v = \displaystyle \int a \, dt} \)) |
Key Idea: By examining the slopes and areas of these graphs, one can move between the three fundamental quantities of motion — position, velocity, and acceleration — both qualitatively and quantitatively.
Example:
A car moves with constant acceleration. Its velocity–time graph is a straight line starting from \( \mathrm{v = 0 \, m/s} \) at \( \mathrm{t = 0} \) and reaching \( \mathrm{v = 20 \, m/s} \) at \( \mathrm{t = 5 \, s} \). Using the graph, determine:
- (a) The acceleration of the car
- (b) The displacement during the 5 seconds
▶️ Answer / Explanation
Given: The velocity–time graph is a straight line from \( (0, 0) \) to \( (5, 20) \).
(a) Acceleration:
From the slope of the \( \mathrm{v\text{–}t} \) graph:
\( \mathrm{a = \dfrac{\Delta v}{\Delta t} = \dfrac{20 – 0}{5 – 0} = 4 \, m/s^2} \)
(b) Displacement:
The area under the \( \mathrm{v\text{–}t} \) graph (a triangle):
\( \mathrm{\Delta x = \dfrac{1}{2} \times base \times height = \dfrac{1}{2}(5)(20) = 50 \, m} \)
Final Answers:
- \( \mathrm{a = 4 \, m/s^2} \)
- \( \mathrm{\Delta x = 50 \, m} \)
Conceptual Note: The constant slope of the \( \mathrm{v\text{–}t} \) graph shows uniform acceleration, while the area under it gives total displacement during that interval.
Instantaneous Velocity, Instantaneous Acceleration, and Area Relationships Between Motion Graphs
Motion can be analyzed graphically by studying the slopes and areas of position–time, velocity–time, and acceleration–time graphs. The slope of a tangent line represents an instantaneous rate of change, while the area under a curve represents the accumulated change of one quantity with respect to another.
1. Instantaneous Velocity:
The instantaneous velocity of an object is the rate of change of its position with respect to time. On a position–time (\( \mathrm{x\text{–}t} \)) graph, it is equal to the slope of the line tangent to the curve at a specific point.
\( \mathrm{v_x = \dfrac{dx}{dt}} \)
- The direction of \( \mathrm{v_x} \) indicates whether the object is moving forward (positive slope) or backward (negative slope).
- A steeper slope indicates a higher magnitude of velocity.
- A horizontal tangent (\( \mathrm{v_x = 0} \)) indicates a momentary stop.
2. Instantaneous Acceleration:
The instantaneous acceleration is the rate of change of velocity with respect to time. On a velocity–time (\( \mathrm{v\text{–}t} \)) graph, it is equal to the slope of the line tangent to the curve at that instant.
\( \mathrm{a_x = \dfrac{dv_x}{dt}} \)
- A positive slope means the velocity is increasing (acceleration in the positive direction).
- A negative slope means the object is slowing down (acceleration opposite the motion).
- A zero slope (\( \mathrm{a_x = 0} \)) corresponds to constant velocity.
3. Area Interpretation — Displacement from Velocity Graph:
The displacement of an object during a time interval is equal to the area under the velocity–time curve between two time points \( \mathrm{t_1} \) and \( \mathrm{t_2} \).
\( \mathrm{\Delta x = \displaystyle \int_{t_1}^{t_2} v_x(t) \, dt} \)
- For constant velocity, the area is a rectangle: \( \mathrm{\Delta x = v_x (t_2 – t_1)} \).
- For variable velocity, the total displacement is the algebraic sum of all signed areas under the curve.
- Positive areas → motion in positive direction; negative areas → motion in negative direction.
4. Area Interpretation — Change in Velocity from Acceleration Graph:
The change in velocity of an object during a time interval is equal to the area under the acceleration–time curve between \( \mathrm{t_1} \) and \( \mathrm{t_2} \).
\( \mathrm{\Delta v_x = \displaystyle \int_{t_1}^{t_2} a_x(t) \, dt} \)
- For constant acceleration, \( \mathrm{\Delta v_x = a_x (t_2 – t_1)} \).
- The area above the time axis represents an increase in velocity, and below represents a decrease.
Example:
The velocity–time graph of an object is a straight line increasing uniformly from \( \mathrm{v_x = 0} \) at \( \mathrm{t = 0} \) to \( \mathrm{v_x = 12 \, m/s} \) at \( \mathrm{t = 4 \, s} \).
Find:
- (a) The instantaneous acceleration
- (b) The total displacement during the 4 s
▶️ Answer / Explanation
Given: The \( \mathrm{v\text{–}t} \) graph is a straight line from (0, 0) to (4, 12).
(a) Instantaneous Acceleration:
Since the graph is a straight line, acceleration is constant and equal to the slope:
\( \mathrm{a_x = \dfrac{\Delta v_x}{\Delta t} = \dfrac{12 – 0}{4 – 0} = 3 \, m/s^2} \)
(b) Displacement:
Displacement = area under the \( \mathrm{v\text{–}t} \) graph:
\( \mathrm{\Delta x = \dfrac{1}{2} \times base \times height = \dfrac{1}{2}(4)(12) = 24 \, m} \)