Kinematics: Constant Acceleration — Key Concepts and Worked Examples

Instantaneous Acceleration and Velocity

• Instantaneous acceleration is defined as the limit of the average rate of change of velocity as the time interval approaches zero:

  • a = rac{dv}{dt} = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}.

• An alternative, equivalent description (when a is constant) is to use the time derivative of velocity:

  • a = \frac{dv}{dt} (constant acceleration implies a linear change of velocity with time).

Types of Motion in One Dimension

• Uniform motion (Uniform Rectilinear Motion): both speed and direction are constant; velocity is constant over time.

  • Graphical implications:

  • Velocity vs time: a horizontal line (constant velocity).

  • Position vs time: a straight line with slope equal to the constant velocity (positive if moving in +x, negative for -x).

• Uniformly Accelerated Rectilinear Motion (UARM): velocity changes at a constant rate (constant acceleration).

  • Graphical implications:

  • Position vs time: a quadratic (parabolic) curve.

  • Velocity vs time: a straight line whose slope equals the constant acceleration a.

  • If velocity and acceleration point in the same direction, the object speeds up; if opposite directions, it slows down.

Velocity, Position, and Acceleration: Key Relationships

• For constant acceleration a, the basic kinematic equations are derived from the definition of a and from average velocity concepts:

  • Final velocity:

  • vf = vi + a\,t

  • Displacement during time t (using average velocity):

  • \Delta x = v_{\text{avg}}\,t

  • For constant acceleration, the average velocity is the mean of the initial and final velocities:

  • v{\text{avg}} = \frac{vi + v_f}{2}

  • Substituting into the displacement relation yields:

  • \Delta x = v_i\,t + \frac{1}{2} a t^2

  • Position after time t (from x_i):

  • xf = xi + v_i\,t + \frac{1}{2} a t^2

  • Velocity–displacement relation (without explicit time):

  • Starting from vf = vi + a t and eliminating t via t = (vf - vi)/a, one obtains:

  • vf^2 = vi^2 + 2 a \Delta x

Derivation Sketch (from the transcript)

• Assume constant acceleration a and set the initial time t = 0 for convenience.

• Use the definition a = dv/dt and consider limit as Δt → 0 to obtain the differential form, leading to v = v_i + a t.

• Define displacement using average velocity and derive Δx = v_i t + (1/2) a t^2.

• Combine the expressions to obtain the standard equations of motion:

  • Final velocity: vf = vi + a t

  • Displacement: \Delta x = v_i t + \frac{1}{2} a t^2

  • Position: xf = xi + v_i t + \frac{1}{2} a t^2

• Eliminating t yields the velocity–displacement relation: vf^2 = vi^2 + 2 a \Delta x

Problem-Solving Strategy for Kinematics (Constant a)

• Step 1: Identify given data and determine the type of motion (uniform vs uniformly accelerated). Check if the motion is one-dimensional (along x).

• Step 2: Determine what quantities are known and what is asked.

• Step 3: Choose the appropriate equation(s) based on the knowns and what needs to be found.

• Step 4: Solve algebraically, keeping track of initial conditions (xi, vi) and the acceleration a.

• Step 5: Check units and reasonableness of the answer (e.g., positive/negative signs correspond to direction).

Example 1: Police car vs. speeding car

• Problem setup:

  • A speeding car moves at a constant speed of v_c = 25\ \mathrm{m\,s^{-1}} (a = 0).

  • A police car starts from rest and accelerates at constant a_p = 5\ \mathrm{m\,s^{-2}}.

  • At time zero, both cars are at the same position (the speeder passes the police car).

• Known values:

  • Car: v{c,i} = 25\ \mathrm{m\,s^{-1}},\ ac = 0,\ x_{c,i} = 0

  • Police car: v{p,i} = 0,\ ap = 5\ \mathrm{m\,s^{-2}},\ x_{p,i} = 0

• Part (a): When does the police car catch the speeding car?

  • Position equations (taking x = 0 at the moment of passing):

  • Car: xc(t) = vc t = 25 t (since a = 0)

  • Police: xp(t) = x{p,i} + v{p,i} t + \tfrac{1}{2} ap t^2 = 0 + 0\cdot t + \tfrac{1}{2} (5) t^2 = \tfrac{5}{2} t^2

  • Catch occurs when xc(t) = xp(t):

  • 25 t = \frac{5}{2} t^2

  • Solve: t(25 - \frac{5}{2} t) = 0 \Rightarrow t = 0 \text{ or } t = \frac{50}{5} = 10\ \mathrm{s}

  • Time to catch: t = 10\ \mathrm{s} (ignore t = 0 because that’s the moment of passing).

• Part (b): How far does the police car travel when it catches the speeder?

  • Using police displacement: \Delta xp = v{p,i} t + \frac{1}{2} a_p t^2 = 0 + \frac{1}{2} (5) (10)^2 = 250\ \mathrm{m}

  • Final position: xp = x{p,i} + \Delta x_p = 250\ \mathrm{m}

• Part (c): How fast is the police car traveling when it catches up?

  • Final velocity: vf = v{p,i} + a_p t = 0 + 5 \cdot 10 = 50\ \mathrm{m\,s^{-1}}

• Summary results: catch time = 10\ \mathrm{s}, distance traveled by police to catch = 250\ \mathrm{m}, speed of police at catch = 50\ \mathrm{m\,s^{-1}}.

Example 2: Two riders starting 10 m apart, accelerating in opposite directions

• Problem setup:

  • Two riders start from rest, 10 m apart along a straight track.

  • Both accelerate at constant a = 2\ \mathrm{m\,s^{-2}}, in opposite directions.

  • Time elapsed: t = 3\ \mathrm{s}.

• Coordinate choices:

  • Let rider A start at xA(0) = 0 with acceleration aA = -2 (to the left).

  • Let rider B start at xB(0) = 10 with acceleration aB = +2 (to the right).

  • Initial velocities: v{A,i} = v{B,i} = 0.

• Position calculations after 3 s:

  • Rider A: xA(t) = x{A,i} + v{A,i} t + frac{1}{2} aA t^2 = 0 + 0 + frac{1}{2}(-2)(3)^2 = -9\ \mathrm{m}

  • Rider B: xB(t) = x{B,i} + v{B,i} t + frac{1}{2} aB t^2 = 10 + 0 + frac{1}{2}(2)(3)^2 = 10 + 9 = 19\ \mathrm{m}

• Separation after 3 s:

  • Distance = xB - xA = 19 - (-9) = 28\ \mathrm{m}

• Result: After 3 seconds, the riders are 28 meters apart.

Additional Notes and Common Points from the Transcript

• When a is constant, you can derive multiple useful expressions and choose the one that fits the given data.

• The problem setup often includes choosing an origin x = 0 at t = 0 and tracking initial conditions (vi, xi).

• If one object has constant velocity (a = 0) and another has constant acceleration, you still solve by equating the position functions x(t) for the meet/catch condition.

• The four key equations of motion for constant acceleration cover most one-dimensional problems:

  • vf = vi + a t

  • \Delta x = v_i t + \tfrac{1}{2} a t^2

  • xf = xi + v_i t + \tfrac{1}{2} a t^2

  • vf^2 = vi^2 + 2 a \Delta x

• Practical tip: before applying a specific equation, confirm what quantities are known, what is unknown, and which variables correspond to initial conditions (subscripts i) and final conditions (subscripts f).

Quick Recap

• Instantaneous acceleration is the derivative of velocity with respect to time.

• In one dimension, constant acceleration yields four central equations that link velocity, displacement, position, and time.

• Uniform motion has constant velocity and straight-line position-time graph; UARM has a quadratic position-time graph and a linear velocity-time graph.

• Real problems often involve setting up x(t) for each moving object, applying the catch-up or separation condition, and solving for the desired quantity.

Practice Prompts (conceptual)

• If a car moves with constant velocity 15 m/s and then accelerates at 3 m/s^2 for 4 s, what is its final velocity? What displacement does it cover during the 4 s of acceleration?

• A particle starts from rest and accelerates uniformly at 2 m/s^2 for 5 s. What is its position after 5 s if it starts at x = 0?

• Two trains start at the same time from same position, one with v=20 m/s and the other from rest with a=2 m/s^2. After how long will the faster train be ahead by 100 m if the faster train keeps its velocity? (Assume motion along a straight track.)

(Note: The assignment problems referenced in the transcript (exercises 4 and 5) are said to appear in the posted assignment; practice these kinds of problems to reinforce the four main equations and the problem-solving approach.)