AJ

Constant Acceleration and SUVAT: Lecture Notes

Key Concepts

  • Position as a function of time: x(t). The slope of the x–t curve is velocity v(t).
  • Derivative and integral in motion:
    • Velocity: v = \frac{dx}{dt}
    • Acceleration: a = \frac{dv}{dt}
    • Integration is the inverse operation to differentiation; in motion, integrating velocity over time gives position up to a constant.
  • Constant acceleration implications:
    • Velocity curves are straight lines (linear in time): v(t) = v_0 + a t
    • Position curves are quadratic in time: x(t) = x0 + v0 t + \frac{1}{2} a t^2
  • Core equations for constant acceleration (SUVAT set):
    • Velocity as a function of time: v = v_0 + a t
    • Position as a function of time: x = x0 + v0 t + \frac{1}{2} a t^2
    • Velocity squared as a function of position: v^2 = v0^2 + 2 a (x - x0)
  • Notation:
    • x_0 = initial position
    • v_0 = initial velocity
    • t = time
    • a = (constant) acceleration
  • Dimensional considerations:
    • In many problems, initial position x0 = 0, but problems may give x0 \neq 0; the v^2 relation remains valid with the nonzero initial position.
  • Velocity at rest vs acceleration zero:
    • "At rest" means the velocity is zero at that instant: v = 0 at that moment.
    • Acceleration zero ( a = 0 ) means velocity is constant (could be nonzero); velocity being zero is not implied by zero acceleration.
  • Constant acceleration assumption:
    • We assume a is constant across the interval of interest to use the SUVAT equations.
  • Practical note on problem setup:
    • When solving for a final velocity and time t is unknown, you can:
    • Solve for t from v = v0 + a t: t = \frac{v - v0}{a}
    • Substitute into x = x0 + v0 t + \frac{1}{2} a t^2 to eliminate t and obtain an equation relating v and x:
      2 a (x - x0) = (v - v0)^2 + 2 v0 (v - v0)
      which simplifies to
      v^2 = v0^2 + 2 a (x - x0)

Equations and Derivations

  • Relationship between derivatives and motion variables:
    • v = \dfrac{dx}{dt}
    • a = \dfrac{dv}{dt}
  • From constant acceleration, integrate velocity to get position:
    • Start with v(t) = v_0 + a t
    • Then integrate to find position:
    • x(t) = x0 + \int0^t v(s) \; ds = x0 + \int0^t (v0 + a s) \; ds = x0 + v_0 t + \frac{1}{2} a t^2
  • Derivation of the v^2–position relation (eliminate time):
    • Given v = v0 + a t, solve for time: t = \frac{v - v0}{a}
    • Substitute into x = x0 + v0 t + \frac{1}{2} a t^2:
    • x - x0 = v0 \frac{v - v0}{a} + \frac{1}{2} a \left(\frac{v - v0}{a}\right)^2
    • Multiply by 2a: 2a(x - x0) = 2 v0 (v - v0) + (v - v0)^2
    • Expand and simplify:
    • (v - v0)^2 + 2 v0 (v - v0) = v^2 - v0^2
    • Therefore: 2 a (x - x0) = v^2 - v0^2
    • Rearranged: v^2 = v0^2 + 2 a (x - x0)
  • Important takeaway: The equation v^2 = v0^2 + 2 a (x - x0) is a direct consequence of the two primary equations and is often used when time is not known.

Initial conditions and special cases

  • Initial conditions:
    • Initial position x0, initial velocity v0, and constant acceleration a determine the motion via the SUVAT equations.
  • Common special cases:
    • If a = 0 (no acceleration):
    • v = v_0 (constant velocity)
    • x = x0 + v0 t
    • If the object is at rest at a given instant: v = 0 at that instant; this does not imply that the acceleration is zero unless specified.
  • Dimensional caveat:
    • Even when problems specify an initial position x0, it is common to take x0 = 0 for simplicity unless the problem gives a nonzero value.

Problem-Solving Strategy (practical approach)

  • Identify what is known and what is unknown in the problem.
  • If you need the final velocity and the time is unknown:
    1) Use v = v0 + a t to express time as t = \frac{v - v0}{a}.
    2) Substitute this expression for t into the position equation x = x0 + v0 t + \frac{1}{2} a t^2.
    3) After substitution and simplification, you obtain an equation relating velocity and position, typically leading to the relation v^2 = v0^2 + 2 a (x - x0).
  • If time is known, compute velocity directly from v = v_0 + a t.
  • If you need to find displacement given velocity data, you can integrate velocity over time or use the appropriate SUVAT form depending on knowns.
  • Note on algebraic workflow observed in lecture:
    • The instructor emphasized solving for time from velocity and then plugging into the position equation to derive a single-equation relation between velocity and position.
    • The algebraic trick of multiplying the position equation by 2 (to clear the 1/2) is a common step to simplify derivations.

Quick reference: summary of formulas (condensed)

  • Velocity as a function of time:
    v = v_0 + a t
  • Position as a function of time:
    x = x0 + v0 t + \frac{1}{2} a t^2
  • Velocity–position relation (no time):
    v^2 = v0^2 + 2 a (x - x0)
  • Derivative relations:
    • v = \dfrac{dx}{dt}
    • a = \dfrac{dv}{dt}
  • Integral relation for position from velocity:
    • x(t) = x0 + \int0^t v(s) \; ds
  • Special cases:
    • If a = 0: v = v0,\quad x = x0 + v_0 t
    • If the object is at rest: v = 0 at that moment; not necessarily forever.

Real-world relevance and connections

  • These equations model everyday motions: car acceleration, free-fall vertical motion under gravity (with a = -g), projectile motion on a level plane (with constant horizontal acceleration zero, vertical acceleration constant), etc.
  • They connect foundational calculus (derivatives and integrals) to kinematics through the constant-acceleration assumption, illustrating how calculus underpins simple motion laws.
  • They illustrate how to derive one equation from others, reinforcing understanding of interdependence among the equations and when it is advantageous to eliminate time from the equations.

Note on transcript specifics

  • The speaker mentions a potential confusion between v-dot (acceleration) and rest: v-dot being zero implies no acceleration; rest means velocity zero at that moment. Emphasize the distinction: a = 0 implies constant velocity; v = 0 at an instant does not imply a = 0.
  • The instructor also mentions that while there are three core SUVAT equations, in practice you may derive one from the others depending on which variables are known; the derived form v^2 = v0^2 + 2 a (x - x0) is a common and useful alternative when time is not given.
  • Initial position x0 is sometimes zero in problems, but not always; formulas are valid for any x0.
  • The derivation example discussed in the transcript shows solving for t using v = v_0 + a t, then substituting into the x-equation and simplifying to obtain the v^2 relation.