Initial velocity and average extension in constant-acceleration motion

Overview

• The transcript fragment references:

  • Sec is your initial velocity, indicating a discussion of initial velocity denoted by how velocity is described in the problem.

  • Introduction of another set of equations, noting that the first set is the same as the one written elsewhere, implying two equivalent formulations of the same physics (likely SUVAT/kinematics equations for motion with constant acceleration).

  • A lead-in to the concept of average extension, suggesting we’ll compute or discuss the average extension over a certain interval.

• Goal of this section appears to be linking two equivalent equation sets and then defining/using average extension in the context of motion with initial velocity.

Key notation and concepts

• Initial velocity: usually denoted by $u$ or $v_0$ (the transcript uses a term that is described as the initial velocity, e.g., “Sec is your initial velocity”).

• Extension or displacement: often denoted by $s$ or $x(t)$, representing how far the object has moved from its initial position over time.

• Acceleration: denoted by $a$, assumed constant in the standard SUVAT framework.

• Time interval: denoted by $t$ or $T$ when discussing a duration.

• Final velocity: denoted by $v$ or $v<em>f$, related to the initial velocity and acceleration by $v = u + a t$ (or with $T$ as the time interval, $v</em>f = u + aT$).

• Two equivalent sets of equations: likely the standard SUVAT set of equations for constant acceleration, where different equations are used depending on which quantities are known.

SUVAT / equations of motion under constant acceleration

• Position as a function of time:

  • $x(t) = x_0 + u t + frac{1}{2} a t^2$

  • Here, $x_0$ is the initial position (or initial extension if working with displacement from equilibrium).

• Velocity as a function of time:

  • $v(t) = u + a t$

  • Final velocity at time $T$: $v_f = u + a T$

• Displacement from initial state after time $T$:

  • $s = u T + frac{1}{2} a T^2$

  • If using endpoints: $s = frac{u + v_f}{2} T$

• Relationship between velocity and displacement without time:

  • $v_f^2 = u^2 + 2 a s$

• Equivalence of the two forms: the first time-based form (with $t$) and the endpoint form (with $T$) are interchangeable representations of the same motion under constant acceleration.

Derivations and connections

• From velocity-time relation, integrating v = u + a t yields the position equation:

  • $x(t) = x_0 + ",u t" + frac{1}{2} a t^2$

• The average velocity over an interval [0, T] is:

  • $\bar{v} = rac{x(T) - x<em>0}{T} = rac{u + v</em>f}{2}$

• The displacement over the interval can also be written as:

  • $s = u T + frac{1}{2} a T^2$

• The v^2 relation connects velocity and displacement without time:

  • $v_f^2 = u^2 + 2 a s$

Average extension (displacement) over a time interval

• If extension is the displacement from the initial position, the average extension over a time interval $[0, T]$ is:

  • $\bar{x} = \frac{1}{T} \int<em>{0}^{T} x(t) \, dt = x</em>0 + \frac{u T}{2} + \frac{a T^2}{6}$

• Important distinction:

  • The average of the endpoints is not generally equal to the time-average for nonzero acceleration:

  • Endpoints average: $\frac{x(0) + x(T)}{2} = x_0 + \frac{u T}{2} + \frac{a T^2}{4}$

  • Time-average (integral) yields: $\bar{x} = x_0 + \frac{u T}{2} + \frac{a T^2}{6}$

• Practical interpretation:

  • The integral-based average gives a true mean extension over the interval, accounting for how the object speeds up or slows down during the interval.

• Example calculation (to illustrate the concept): assume

  • Initial velocity $u = 5\ \text{m/s}$, acceleration $a = -2\ \text{m/s}^2$, initial position $x_0 = 0$, and time interval $T = 3\ \text{s}$.

  • Final velocity: $v_f = u + aT = 5 + (-2)(3) = -1\ \text{m/s}$

  • Displacement after time T: $s = u T + \tfrac{1}{2} a T^2 = 5(3) + \tfrac{1}{2}(-2)(9) = 15 - 9 = 6\ \text{m}$

  • Time-average extension over [0,3]:

  • $\bar{x} = x_0 + \frac{u T}{2} + \frac{a T^2}{6} = 0 + \frac{5\cdot 3}{2} + \frac{-2\cdot 9}{6} = \frac{15}{2} - 3 = 7.5 - 3 = 4.5\ \text{m}$

  • Endpoint-average extension over [0,3]:

  • $\frac{x(0) + x(3)}{2} = \frac{0 + x(3)}{2} = \frac{0 + 6}{2} = 3\ \text{m}$

  • Note the two results differ, illustrating why the integral mean (time-average) is the correct average extension over the interval for varying velocity.

Examples and practice problems (conceptual)

• Example 1 (two-equation consistency):

  • Given $u = 4\ \text{m/s}, a = 2\ \text{m/s}^2, t = 3\ \text{s}$, find $x(3)$ and $v(3)$ assuming $x_0 = 0$.

  • Solutions:

  • $v(3) = u + a t = 4 + 2(3) = 10\ \text{m/s}$

  • $x(3) = x_0 + u t + \tfrac{1}{2} a t^2 = 0 + 4(3) + \tfrac{1}{2}(2)(9) = 12 + 9 = 21\ \text{m}$

• Example 2 (average extension):

  • Use the same numbers as above with interval [0, 3], compute time-average extension and endpoint-average extension to compare.

• Real-world relevance:

  • Car braking distances, projectile motion, elevator or elevator-cable systems, and any scenario with constant acceleration.

Connections to foundational principles

• Newton's laws underpin constant-acceleration motion: a is constant due to a constant net force, leading to linear velocity change and quadratic displacement.

• Work-energy perspective ties into the kinematic equations via energy change and constant acceleration (F = ma) and potential energy considerations for conservative forces.

• Dimensional analysis sanity checks: all terms in $x(t)$ have dimension of length, velocities in m/s, accelerations in m/s^2, etc.

Common pitfalls and tips

• Remember the difference between time-average (integral) and endpoint-average; they are generally not the same unless acceleration is zero.

• Use the correct form of the equation depending on the known quantities; if you know initial and final velocities and time, you can use $s = \tfrac{(u + v)}{2} t$; if you know acceleration, use $s = ut + \tfrac{1}{2} a t^2$.

• Keep track of reference points: x_0 or x(0) can be used interchangeably as long as you are consistent.

• When solving problems, check limits: if $a = 0$, the formulas reduce to linear motion with constant velocity, and $x = x_0 + u t$ and $v = u$.

Summary

• There are two equivalent representations of motion under constant acceleration; one emphasizes time-based development (x(t) = x_0 + u t + 1/2 a t^2, v(t) = u + a t), and the other emphasizes relationships between endpoints and velocities (s = (u + v) t / 2, v^2 = u^2 + 2 a s).

• The concept of average extension over a time interval is best captured by the time-average integral formula $\bar{x} = \dfrac{1}{T} \int<em>{0}^{T} x(t) \, dt = x</em>0 + \dfrac{u T}{2} + \dfrac{a T^2}{6}$, and not merely the average of endpoints.

• Practical applications involve solving for position, velocity, and displacement in problems with constant acceleration, which is common in physics and engineering contexts.