Unit 1: Kinematics

Displacement, Velocity, and Acceleration

• Displacement is a vector quantity representing the change in position of an object. Unlike distance, which is scalar, displacement includes both magnitude and direction. It is defined mathematically as \(\Delta \vec{r} = \vec{r}_f - \vec{r}_i\), where \(\vec{r}_f\) and \(\vec{r}_i\) are the final and initial position vectors.
• Velocity is the rate of change of displacement with respect to time. In calculus form, instantaneous velocity is given by \[ \vec{v}(t) = \frac{d\vec{r}}{dt} \] which gives both magnitude and direction at a particular instant. Average velocity over an interval is \(\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t}\).
• Acceleration is the rate of change of velocity with respect to time: \[ \vec{a}(t) = \frac{d\vec{v}}{dt} \] It is a vector, meaning it can change an object’s speed, direction, or both. Constant acceleration problems often lead to kinematic equations, but in AP Physics C, variable acceleration may require integration.

Calculus in Kinematics

• Differentiation allows us to move down the chain from position to velocity to acceleration. For example, if position is known as a function of time, taking its derivative yields velocity, and differentiating again yields acceleration.
• Integration allows us to move up the chain. If acceleration is known, integrating with respect to time gives velocity, and a second integration gives position. Initial conditions (constants of integration) are determined using known values at a specific time.
• This use of calculus distinguishes AP Physics C from AP Physics 1, where constant-acceleration formulas were often sufficient. Here, you must be able to handle variable acceleration problems using both derivatives and integrals.

Kinematic Equations for Constant Acceleration

• When acceleration is constant, we can derive the familiar kinematic equations using calculus. Velocity as a function of time is: \[ v(t) = v_0 + at \] where \(v_0\) is initial velocity. This comes from integrating constant acceleration.
• Position as a function of time is: \[ x(t) = x_0 + v_0t + \frac{1}{2}at^2 \] where \(x_0\) is the initial position. This equation combines the effects of initial velocity and constant acceleration.
• By eliminating time, we get: \[ v^2 = v_0^2 + 2a(x - x_0) \] which is useful when time is not known. These equations form the backbone of motion analysis when acceleration is constant.

Two-Dimensional Motion

• In two dimensions, position is represented as a vector: \[ \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} \] where \(x(t)\) and \(y(t)\) describe the horizontal and vertical components of motion. The velocity vector is obtained by differentiating position: \[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} \] and acceleration is the derivative of velocity.
• Projectile motion is a classic example. If an object is launched with initial velocity components \(v_{0x}\) and \(v_{0y}\), its position is: \[ x(t) = v_{0x}t, \quad y(t) = v_{0y}t - \frac{1}{2}gt^2 \] Calculus confirms that velocity in the x-direction is constant (since \(a_x = 0\)) while velocity in the y-direction decreases linearly due to gravity.
• Vector calculus makes it easier to handle more complex motions. For example, acceleration that changes with time (such as air resistance approximations) requires integration: \[ v(t) = \int a(t) dt \] With initial conditions applied, this yields a complete velocity function.

Relative Velocity

• Relative velocity describes the velocity of one object as observed from another reference frame. If object A has velocity \(\vec{v}_A\) and object B has velocity \(\vec{v}_B\), then the velocity of A relative to B is: \[ \vec{v}_{A/B} = \vec{v}_A - \vec{v}_B \] This vector subtraction accounts for both magnitude and direction.
• Calculus can be applied when one or both objects have velocities that change with time. In this case, acceleration functions may need to be integrated to find position functions before relative displacement can be calculated.
• Relative velocity is particularly important in analyzing problems involving boats crossing rivers or airplanes flying in wind. These problems often require separating motion into perpendicular components and using vector addition or subtraction to find the resultant motion.

Polar Coordinates and Radial Motion

• Some two-dimensional problems are easier to analyze in polar coordinates \((r, \theta)\) rather than Cartesian coordinates \((x, y)\). Position is expressed as: \[ \vec{r}(t) = r(t)\hat{r} \] where \(\hat{r}\) is the unit vector pointing radially outward from the origin.
• Velocity in polar coordinates includes both radial and angular components: \[ \vec{v}(t) = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta} \] The first term represents the rate of change of radius, while the second represents tangential velocity due to angular motion.
• Acceleration in polar coordinates is more complex: \[ \vec{a}(t) = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta} \] These terms show radial acceleration (toward or away from the center) and angular acceleration (changing speed around a circle). Such expressions are critical when analyzing planetary motion or objects in circular tracks.

Vector Components and Vector Operations

• Displacement, velocity, and acceleration in two or three dimensions are vector quantities and must be broken into components to analyze motion accurately. Any vector \( \vec{A} \) in 2D can be expressed as \( \vec{A} = A_x\hat{i} + A_y\hat{j} \), where \( A_x = A\cos\theta \) and \( A_y = A\sin\theta \).
• Vector addition and subtraction follow component-wise arithmetic. This is essential for solving problems involving relative motion, two-dimensional projectiles, and inclined plane components in circular motion.
• In more advanced problems, dot products (\( \vec{A} \cdot \vec{B} = AB\cos\theta \)) are used to calculate scalar quantities like work, while cross products (\( \vec{A} \times \vec{B} = AB\sin\theta \hat{n} \)) are used in rotational kinematics and torque.

Instantaneous vs. Average Quantities

• Average velocity is the total displacement divided by the total time: \( \vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} \). It gives a broad overview of motion over an interval but ignores how the velocity changes during that time.
• Instantaneous velocity is the velocity at a specific moment in time and is found using calculus: \( \vec{v}(t) = \frac{d\vec{r}}{dt} \). This distinction is crucial when motion is not uniform, such as during acceleration.
• Likewise, instantaneous acceleration is \( \vec{a}(t) = \frac{d\vec{v}}{dt} \), while average acceleration is \( \vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} \). Understanding the difference helps prevent confusion when analyzing non-uniform motion or interpreting graphs.

Position, Velocity, and Acceleration Graphs

• Graphs are powerful tools in kinematics. The slope of a position-time graph gives the velocity: \( v(t) = \frac{dx}{dt} \). A steep slope indicates fast motion, and a flat slope indicates rest.
• The area under a velocity-time graph gives the displacement: \( \Delta x = \int v(t)dt \). Positive area indicates forward motion, and negative area indicates backward motion.
• The slope of a velocity-time graph represents acceleration: \( a(t) = \frac{dv}{dt} \), and the area under the acceleration-time graph gives change in velocity: \( \Delta v = \int a(t)dt \). These calculus relationships allow full reconstruction of motion from graphs alone.

Non-Constant Acceleration

• Unlike AP Physics 1, AP Physics C requires solving problems where acceleration is not constant. In such cases, velocity must be found by integrating the acceleration function: \[ v(t) = \int a(t)\, dt + v_0 \] where \( v_0 \) is the initial velocity. This makes calculus essential for solving real-world motion problems.
• If acceleration is a function of time, position can be found by integrating velocity: \[ x(t) = \int v(t)\, dt + x_0 \] These two integrals connect acceleration, velocity, and position without assuming constant acceleration.
• For example, if \( a(t) = kt \), integrating once gives \( v(t) = \frac{1}{2}kt^2 + v_0 \), and integrating again gives \( x(t) = \frac{1}{6}kt^3 + v_0t + x_0 \). This demonstrates how motion changes more rapidly when acceleration itself grows with time.

Relative Motion in Non-Inertial Frames

• Non-inertial frames are reference frames that accelerate relative to an inertial (non-accelerating) frame. In these frames, Newton’s laws require the addition of fictitious or “pseudo” forces to account for the acceleration of the frame itself.
• A classic example is an elevator accelerating upward. Inside the elevator, objects appear to weigh more because the floor exerts a greater normal force to accelerate them along with the elevator. This apparent increase in weight is explained by adding a fictitious force acting downward.
• Mathematically, if the frame accelerates with \( \vec{a}_f \), then the observed acceleration inside the frame is: \[ \vec{a}_{obs} = \vec{a}_{real} - \vec{a}_f \] This adjustment allows correct analysis of motion in accelerating vehicles, rotating systems, or other non-inertial situations.

Parametric Equations and 3D Motion

• Parametric equations describe motion in two or three dimensions by expressing each coordinate as a function of time. For example, projectile motion can be written as \[ x(t) = v_0\cos\theta \cdot t, \quad y(t) = v_0\sin\theta \cdot t - \tfrac{1}{2}gt^2 \] allowing analysis of horizontal and vertical motion simultaneously.
• In three dimensions, motion can be represented as \[ \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k} \] with velocity and acceleration found by differentiating each component. This approach extends seamlessly from 2D motion to full 3D analysis.
• Parametric analysis is especially useful in AP Physics C because it allows elimination of time to find trajectories. For example, eliminating \( t \) between \( x(t) \) and \( y(t) \) produces the path of a projectile, connecting algebraic and calculus-based approaches.

Example Problem 1: Non-Constant Acceleration in 1D

Context: A particle moves along a straight line with acceleration given by \( a(t) = 4t \) m/s². At time \( t = 0 \), the particle is at position \( x_0 = 2.0 \) m with initial velocity \( v_0 = 3.0 \) m/s.

• (a) Derive the expression for the velocity of the particle as a function of time.
  \[ v(t) = \int a(t)\, dt + v_0 = \int 4t\, dt + 3.0 = 2t^2 + 3.0 \]
• (b) Derive the position of the particle as a function of time.
  \[ x(t) = \int v(t)\, dt + x_0 = \int (2t^2 + 3)\, dt + 2.0 = \tfrac{2}{3}t^3 + 3t + 2.0 \]
• (c) Find the velocity and position of the particle at \( t = 2.0 \) s.
  \[ v(2) = 2(2^2) + 3 = 11 \,\text{m/s}, \quad x(2) = \tfrac{2}{3}(8) + 6 + 2 = 13.3\, \text{m} \]
• (d) Compute the average velocity of the particle from \( t = 0 \) to \( t = 2.0 \) s and compare it with the instantaneous velocity at \( t = 1.0 \) s.
  \[ v_{avg} = \frac{x(2)-x(0)}{2-0} = \frac{13.3-2}{2} = 5.65\, \text{m/s} \] \[ v(1) = 2(1)^2 + 3 = 5\, \text{m/s} \]
  The values are close, showing how instantaneous and average velocity can align when acceleration grows steadily.

Example Problem 2: Projectile Motion with Calculus

Context: A ball is launched from the ground with initial velocity \( v_0 = 20\, \text{m/s} \) at an angle of \( 60^\circ \). Neglect air resistance and use \( g = 9.8\, \text{m/s}^2 \).

• (a) Write parametric equations for the ball’s position as a function of time.
  \[ x(t) = v_0\cos\theta \cdot t = (20\cos60^\circ)t = 10t \] \[ y(t) = v_0\sin\theta \cdot t - \tfrac{1}{2}gt^2 = (20\sin60^\circ)t - 4.9t^2 \]
• (b) Find the time when the ball reaches its maximum height.
  At the top, vertical velocity = 0: \[ v_y(t) = v_0\sin\theta - gt = 20\sin60^\circ - 9.8t = 0 \] \[ t = \frac{20\sin60^\circ}{9.8} ≈ 1.77\, \text{s} \]
• (c) Determine the maximum height reached by the ball.
  \[ y(1.77) = (20\sin60^\circ)(1.77) - 4.9(1.77)^2 ≈ 15.3\, \text{m} \]
• (d) Calculate the total horizontal range of the projectile.
  Total flight time is when \( y(t) = 0 \) again: \[ t = \frac{2v_0\sin\theta}{g} = \frac{2(20)(\sqrt{3}/2)}{9.8} ≈ 3.53\, \text{s} \] \[ x(3.53) = 10(3.53) ≈ 35.3\, \text{m} \]

Example Problem 3: Motion in Polar Coordinates

Context: A particle moves in a plane such that its radial distance is given by \( r(t) = 2t \) and its angular position is \( \theta(t) = t^2 \), where \( r \) is in meters and \( t \) in seconds.

• (a) Find the velocity vector in polar coordinates.
  \[ \vec{v}(t) = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta} \] \[ \dot{r} = \frac{dr}{dt} = 2, \quad \dot{\theta} = \frac{d\theta}{dt} = 2t \] \[ \vec{v}(t) = 2\hat{r} + (2t)(2t)\hat{\theta} = 2\hat{r} + 4t^2\hat{\theta} \]
• (b) Find the acceleration vector in polar coordinates.
  \[ \vec{a}(t) = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta} \] \[ \ddot{r} = 0, \quad \ddot{\theta} = 2 \] \[ a_r = 0 - (2t)(2t)^2 = -8t^3, \quad a_\theta = (2t)(2) + 2(2)(2t) = 4t + 8t = 12t \] \[ \vec{a}(t) = (-8t^3)\hat{r} + (12t)\hat{\theta} \]
• (c) Evaluate velocity and acceleration at \( t = 1.0 \) s.
  \[ \vec{v}(1) = 2\hat{r} + 4(1)^2\hat{\theta} = 2\hat{r} + 4\hat{\theta} \] \[ \vec{a}(1) = (-8)(1)^3\hat{r} + (12)(1)\hat{\theta} = -8\hat{r} + 12\hat{\theta} \]

Tips & Problem-Solving Strategies for Kinematics

• Always start with calculus definitions. Remember that velocity is the derivative of position and acceleration is the derivative of velocity. If acceleration is given as a function of time, integrate it to find velocity, and integrate again to find position. This keeps your work consistent with the calculus framework of AP Physics C.
• Set up coordinate systems carefully. Choosing a clear positive direction makes vector analysis easier and prevents sign errors. For two-dimensional motion, break vectors into components using unit vectors (\(\hat{i}, \hat{j}, \hat{k}\)) or trig functions. This ensures each dimension is analyzed independently while still forming a complete vector solution.
• Use initial conditions to solve for constants of integration. When integrating acceleration to get velocity, don’t forget to add \(v_0\) as the constant of integration. Similarly, when integrating velocity to get position, add \(x_0\). Plugging in given conditions like "at \(t = 0\), \(x = 0\)" ensures your functions match the physical situation.
• Interpret graphs using calculus. The slope of a position-time graph is velocity, the slope of a velocity-time graph is acceleration, and the area under an acceleration-time graph gives change in velocity. Always check if you can solve a problem more easily through graphical interpretation instead of pure algebraic manipulation.
• Check the independence of motion components. In projectile and relative motion problems, solve horizontal and vertical motions separately, then combine results at the end. Remember they share the same time variable, so solving for time in one dimension often unlocks the other.
• Dimensional analysis is your friend. Always check that your final answer has the correct units. For example, if integrating acceleration (m/s²) over time (s), the result should be in m/s. This quick check catches many common mistakes without requiring you to redo the problem.
• Draw diagrams for clarity. Whether it’s a vector triangle for relative motion or a velocity-time graph for variable acceleration, visuals help track how quantities relate. In AP Physics C free-response questions, well-labeled diagrams can also earn you partial credit even if your calculations contain errors.

Common Misconceptions in Kinematics

• Students often confuse distance and displacement. Distance is the total path length traveled and is scalar, while displacement is the straight-line change in position and is vector. For example, running one lap around a track gives a large distance but zero displacement, which is essential in understanding velocity calculations.
• It is a common mistake to think that acceleration is zero whenever velocity is zero. At turning points in motion, such as the top of a projectile’s path, velocity is zero but acceleration due to gravity remains constant. This misconception leads to errors when analyzing projectile motion or interpreting velocity-time graphs.
• Many students misinterpret slopes and areas in motion graphs. The slope of a position-time graph is velocity, while the area under a velocity-time graph gives displacement. Forgetting these calculus-based connections often causes incorrect interpretations of motion data.
• Some students believe that kinematic equations always apply. In AP Physics C, they are valid only when acceleration is constant. For variable acceleration, students must use integration and differentiation, which reinforces why calculus is a core tool in this course.
• In two-dimensional motion, a frequent misconception is that horizontal and vertical motions affect each other. In reality, they are independent except for the shared time variable. Neglecting this independence leads to incorrect projectile calculations and misunderstandings of vector decomposition.
• When dealing with relative motion, students sometimes incorrectly add speeds as scalars instead of as vectors. Correct analysis requires vector subtraction or addition, which accounts for both magnitude and direction. This is especially important in problems involving planes in wind or boats in current.