AP Physics C Mechanics- 1.2 Displacement, Velocity, and Acceleration- Study Notes- New Syllabus
AP Physics C Mechanics- 1.2 Displacement, Velocity, and Acceleration – Study Notes
AP Physics C Mechanics- 1.2 Displacement, Velocity, and Acceleration – Study Notes – per latest Syllabus.
Key Concepts:
- Displacement and Change in Position
- Velocity and Average and Instantaneous Acceleration
- Instantaneous Position, Velocity, and Acceleration as a Function of Time
Displacement and Change in Position
In mechanics, the motion of an object can be described by the change in its position with respect to time. When analyzing motion, it is often convenient to treat an extended object as a point mass an object with size, shape, and internal configuration ignored. This simplifies analysis while preserving essential physical properties such as mass and charge.
Point Object Model:
- In this model, the object is considered to be concentrated at a single point in space.
- Useful when the object’s size is negligible compared to the distance it moves.
- Allows motion to be represented mathematically using coordinates and vectors.
Displacement:
Displacement represents the change in an object’s position from an initial point to a final point. It is a vector quantity, meaning it has both magnitude and direction.
If an object moves from point \( \mathrm{P_1(x_1, y_1, z_1)} \) to point \( \mathrm{P_2(x_2, y_2, z_2)} \), the displacement vector is:
\( \mathrm{\vec{d} = \vec{r}_2 – \vec{r}_1} \)
Where:
- \( \mathrm{\vec{r}_1 = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k}} \) → initial position vector
- \( \mathrm{\vec{r}_2 = x_2 \hat{i} + y_2 \hat{j} + z_2 \hat{k}} \) → final position vector
The magnitude of displacement is given by:
\( \mathrm{|\vec{d}| = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}} \)
Relevant Equation:
\( \mathrm{\vec{d} = \vec{r}_2 – \vec{r}_1} \)
Key Idea: Displacement depends only on the initial and final positions not the path taken between them.
Example:
An object moves from point \( \mathrm{A(2, 3)} \) to point \( \mathrm{B(7, 8)} \) in a plane. Find the displacement vector and its magnitude.
▶️ Answer / Explanation
Step 1: Write position vectors.
- \( \mathrm{\vec{r}_1 = 2\hat{i} + 3\hat{j}} \)
- \( \mathrm{\vec{r}_2 = 7\hat{i} + 8\hat{j}} \)
Step 2: Find displacement vector.
\( \mathrm{\vec{d} = \vec{r}_2 – \vec{r}_1 = (7 – 2)\hat{i} + (8 – 3)\hat{j} = 5\hat{i} + 5\hat{j}} \)
Step 3: Find magnitude of displacement.
\( \mathrm{|\vec{d}| = \sqrt{(5)^2 + (5)^2} = \sqrt{50} = 7.07 \, \text{units}} \)
Final Answer: \( \mathrm{\vec{d} = 5\hat{i} + 5\hat{j}}, \quad |\vec{d}| = 7.07 \, \text{units}} \)
Velocity and Acceleration: Average and Instantaneous
Motion can be described not only by how far an object moves (displacement) but also by how fast and in what direction it moves this is velocity. The rate at which velocity changes with time defines acceleration. Both quantities are vectors because they possess magnitude and direction.
Velocity:
Velocity is the rate of change of displacement with respect to time.
\( \mathrm{\vec{v} = \dfrac{d\vec{r}}{dt}} \)
- Direction of \( \mathrm{\vec{v}} \) is the same as the direction of motion.
- Velocity changes if either magnitude (speed) or direction of motion changes.
Average Velocity:
It is defined as the total displacement divided by the total time interval.
\( \mathrm{\vec{v}_{avg} = \dfrac{\Delta \vec{r}}{\Delta t} = \dfrac{\vec{r}_2 – \vec{r}_1}{t_2 – t_1}} \)
- Depends on initial and final positions only (path-independent).
- Gives the overall rate of change of position during the time interval.
Acceleration:
Acceleration is the rate of change of velocity with respect to time.
\( \mathrm{\vec{a} = \dfrac{d\vec{v}}{dt}} \)
- It measures how quickly and in what direction the velocity changes.
- Can be due to change in speed, change in direction, or both.
Average Acceleration:
It is defined as the change in velocity divided by the corresponding time interval.
\( \mathrm{\vec{a}_{avg} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{\vec{v}_2 – \vec{v}_1}{t_2 – t_1}} \)
Instantaneous Acceleration:
It is the rate of change of velocity at a particular instant the limit of average acceleration as time interval approaches zero.
\( \mathrm{\vec{a} = \lim_{\Delta t \to 0} \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{d\vec{v}}{dt} = \dfrac{d^2\vec{r}}{dt^2}} \)
Key Idea: Velocity tells how position changes with time; acceleration tells how velocity changes with time.
Example:
An object moves in a straight line. At \( \mathrm{t_1 = 0 \, s} \), its position is \( \mathrm{x_1 = 2 \, m} \). At \( \mathrm{t_2 = 4 \, s} \), its position is \( \mathrm{x_2 = 18 \, m} \). The velocity of the object then increases uniformly from \( \mathrm{3 \, m/s} \) at \( \mathrm{t = 4 \, s} \) to \( \mathrm{9 \, m/s} \) at \( \mathrm{t = 8 \, s} \). Calculate:
- (a) The average velocity between \( \mathrm{t_1} \) and \( \mathrm{t_2} \).
- (b) The average acceleration between \( \mathrm{t = 4 \, s} \) and \( \mathrm{t = 8 \, s} \).
▶️ Answer / Explanation
Given Data:
- \( \mathrm{x_1 = 2 \, m, \, x_2 = 18 \, m, \, t_1 = 0 \, s, \, t_2 = 4 \, s} \)
- \( \mathrm{v_1 = 3 \, m/s, \, v_2 = 9 \, m/s, \, \Delta t = 8 – 4 = 4 \, s} \)
(a) Average Velocity:
Average velocity is given by the total displacement divided by total time:
\( \mathrm{v_{avg} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 – x_1}{t_2 – t_1}} \)
\( \mathrm{v_{avg} = \dfrac{18 – 2}{4 – 0} = \dfrac{16}{4} = 4 \, m/s} \)
→ The average velocity between \( \mathrm{t = 0 \, s} \) and \( \mathrm{t = 4 \, s} \) is \( \mathrm{4 \, m/s} \).
(b) Average Acceleration:
Average acceleration is the rate of change of velocity over a given time interval:
\( \mathrm{a_{avg} = \dfrac{\Delta v}{\Delta t} = \dfrac{v_2 – v_1}{t_2 – t_1}} \)
\( \mathrm{a_{avg} = \dfrac{9 – 3}{8 – 4} = \dfrac{6}{4} = 1.5 \, m/s^2} \)
→ The average acceleration between \( \mathrm{t = 4 \, s} \) and \( \mathrm{t = 8 \, s} \) is \( \mathrm{1.5 \, m/s^2} \).
Instantaneous Position, Velocity, and Acceleration as a Function of Time
The motion of an object can be completely described using functions of time that specify its position, velocity, and acceleration at any instant. These time-dependent relationships form the foundation of kinematics.
Instantaneous Position:
The position of an object at a given instant of time specifies its location relative to a chosen reference point (origin).
\( \mathrm{\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}} \)
- Here \( \mathrm{x(t), y(t), z(t)} \) are functions of time describing motion along each axis.
- The graph of position vs time gives insight into how an object’s location evolves.
Key Idea: Knowing \( \mathrm{\vec{r}(t)} \) allows us to determine both velocity and acceleration by differentiation.
Instantaneous Velocity:
The instantaneous velocity is the rate of change of position with respect to time.
\( \mathrm{\vec{v}(t) = \dfrac{d\vec{r}(t)}{dt}} \)
- Velocity describes both how fast and in what direction the position changes.
- It is tangent to the trajectory at each point.
- If \( \mathrm{x(t), y(t), z(t)} \) are known, then:
\( \mathrm{\vec{v}(t) = \dfrac{dx}{dt}\hat{i} + \dfrac{dy}{dt}\hat{j} + \dfrac{dz}{dt}\hat{k}} \)
Magnitude of Instantaneous Velocity (Speed):
\( \mathrm{|\vec{v}(t)| = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 + \left(\dfrac{dz}{dt}\right)^2}} \)
Instantaneous Acceleration:
The instantaneous acceleration is the rate of change of velocity with respect to time.
\( \mathrm{\vec{a}(t) = \dfrac{d\vec{v}(t)}{dt}} \)
Since velocity is itself the time derivative of position, acceleration can also be written as:
\( \mathrm{\vec{a}(t) = \dfrac{d^2\vec{r}(t)}{dt^2}} \)
- Acceleration tells how quickly velocity changes — in magnitude or direction.
- If \( \mathrm{x(t), y(t), z(t)} \) are known:
\( \mathrm{\vec{a}(t) = \dfrac{d^2x}{dt^2}\hat{i} + \dfrac{d^2y}{dt^2}\hat{j} + \dfrac{d^2z}{dt^2}\hat{k}} \)
Key Relationships (Time Derivatives):
- \( \mathrm{\vec{v}(t) = \dfrac{d\vec{r}(t)}{dt}} \)
- \( \mathrm{\vec{a}(t) = \dfrac{d\vec{v}(t)}{dt} = \dfrac{d^2\vec{r}(t)}{dt^2}} \)
Key Idea: Position, velocity, and acceleration are connected through differentiation. Given \( \mathrm{\vec{r}(t)} \), we can find \( \mathrm{\vec{v}(t)} \) and \( \mathrm{\vec{a}(t)} \); conversely, knowing \( \mathrm{\vec{a}(t)} \), we can obtain \( \mathrm{\vec{v}(t)} \) and \( \mathrm{\vec{r}(t)} \) through integration.
Example:
The position of a particle moving in a straight line is given by \( \mathrm{x(t) = 4t^3 – 3t^2 + 2t + 1} \) (in meters). Find its instantaneous velocity and acceleration as functions of time, and evaluate them at \( \mathrm{t = 2 \, s} \).
▶️ Answer / Explanation
Step 1: Differentiate \( \mathrm{x(t)} \) to find \( \mathrm{v(t)} \).
\( \mathrm{v(t) = \dfrac{dx}{dt} = 12t^2 – 6t + 2} \)
Step 2: Differentiate \( \mathrm{v(t)} \) to find \( \mathrm{a(t)} \).
\( \mathrm{a(t) = \dfrac{dv}{dt} = 24t – 6} \)
Step 3: Evaluate at \( \mathrm{t = 2 \, s} \).
- \( \mathrm{v(2) = 12(2)^2 – 6(2) + 2 = 48 – 12 + 2 = 38 \, m/s} \)
- \( \mathrm{a(2) = 24(2) – 6 = 42 \, m/s^2} \)
Final Answer:
\( \mathrm{v(t) = 12t^2 – 6t + 2, \quad a(t) = 24t – 6} \)
At \( \mathrm{t = 2 \, s: \, v = 38 \, m/s, \, a = 42 \, m/s^2} \)