Motion with Constant Acceleration

1. Kinematic Equations

• Four equations for constant acceleration:
  • $\displaystyle v = v_0 + a t$
  • $\displaystyle x = x_0 + v_0 t + \frac{1}{2} a t^2$
  • $\displaystyle v^2 = v_0^2 + 2a(x - x_0)$
  • $\displaystyle x = x_0 + \frac{(v_0 + v)}{2} t$
  • Variables:
    • $x_0$ = initial position, $x$ = final position
    • $v_0$ = initial velocity, $v$ = final velocity
    • $a$ = constant acceleration, $t$ = time interval
• Variables in another perspective:
  • Initial condition: $x_0$ = initial position, $v_0$ = initial velocity
  • Quantity that characterizes the motion: $a$ = constant acceleration
  • Final state: $t$ = time interval, $x$ = final position, $v$ = final velocity

2. When and how to use Kinematic equations:

• When to use these equations:
  • Constant acceleration, meaning the direction and magnitude of the acceleration is constant
  • These equations are used to calculate displacement, velocity, which are vector quantities. One may or may not directly use these equations to calculate the distance, which is a scalar quantities. More discussed below.
• How to use these equations:
  • Set up coordinate system: which way is positive?
  • Among all variables discussed above, which are mentioned and which are not, and pick the one that matches the problem's description:
    • Example: If the problem gives the final speed and total distance traveled, ask about the acceleration, then $\displaystyle v^2 = v_0^2 + 2a(x - x_0)$ is clearly the best choice, because neither the problem nor this equation talks about time $t$.

3. Critical confusions

• Myth: "Kinematic equations apply to any motion."

  • Reality: Only valid for constant acceleration. For non-constant acceleration (e.g., air resistance on a falling object, variable engine force), these equations fail.
  • Example: A rocket with thrust decreasing over time requires calculus-based integration, not kinematic equations.

• Sign convention errors:

  • Example: A ball thrown upward with $v_0 = +15 \, \text{m/s}$ and $a = -9.8 \, \text{m/s}^2$ (up positive). If $a$ is mistakenly taken as $+9.8 \, \text{m/s}^2$, max height calculation gives:
    $$ \text{Incorrect: } v^2 = 15^2 + 2(9.8)(h) \implies h = \frac{-225}{19.6} \quad (\text{negative height!}) $$
    Correct: $v^2 = 15^2 + 2(-9.8)h \implies h = \frac{225}{19.6} \approx 11.5 \, \text{m}$.

• Misusing the average velocity formula:

  • $\frac{v_0 + v}{2}$ is only valid for constant acceleration. For non-uniform acceleration, this gives incorrect displacement.
  • Example: A car accelerates non-uniformly from $v_0 = 0$ to $v = 10 \, \text{m/s}$ in $5 \, \text{s}$. Using $\frac{0+10}{2} \times 5 = 25 \, \text{m}$ displacement is invalid if acceleration isn’t constant (actual displacement could be $<25 \, \text{m}$ or $>25 \, \text{m}$).

• Confusing displacement and distance:

  • Kinematic equations compute displacement ($\Delta x = x - x_0$), not total distance traveled.
  • Example: A ball thrown upward $10 \, \text{m}$ then falling back $5 \, \text{m}$ to a point $5 \, \text{m}$ above start:
    • Displacement = $+5 \, \text{m}$ (correctly calculated by kinematic equations),
    • Total distance = $15 \, \text{m}$ (requires summing path segments).

• Assuming $v_0 = 0$ when it’s not:

  • Example: A car moving at $10 \, \text{m/s}$ then accelerates at $2 \, \text{m/s}^2$ for $3 \, \text{s}$.
    • Incorrect: $x = 0 + 0 \cdot t + \frac{1}{2}(2)(3)^2 = 9 \, \text{m}$
    • Correct: $x = 10 \cdot 3 + \frac{1}{2}(2)(3)^2 = 30 + 9 = 39 \, \text{m}$.

• Ignoring time intervals:

  • Kinematic equations relate quantities over a specific time interval starting at $t=0$. Using $t$ from a different reference point causes errors. The example below may be silly, but this can happen when the problem involves more than one steps in the process, and/or involves more than one moving object.
  • Example: A ball thrown upward with $v_0 = 20 \, \text{m/s}$ ($a = -9.8 \, \text{m/s}^2$). To find velocity at $t=2 \, \text{s}$:
    • Correct: $v = 20 + (-9.8)(2) = 0.4 \, \text{m/s}$ (still moving upward),
    • Incorrect: Using $t=2 \, \text{s}$ after reaching peak (when $v=0$ at $t \approx 2.04 \, \text{s}$) → wrong result.

• Quadratic equation pitfalls:

  • Solving $x = x_0 + v_0 t + \frac{1}{2} a t^2$ for $t$ may yield two solutions.
  • Example: Ball thrown upward from ground ($x_0=0$) with $v_0 = 10 \, \text{m/s}$, $a = -9.8 \, \text{m/s}^2$:
    $$ 0 = 10t - 4.9t^2 \implies t(10 - 4.9t) = 0 \implies t=0 \text{ (launch)} \text{ or } t \approx 2.04 \, \text{s} \text{ (return)} $$
    Both solutions are physically valid.

• Coordinate system inconsistencies:

  • Choosing inconsistent directions for velocity and acceleration.
  • Example: A car moves left ($v = -5 \, \text{m/s}$) with rightward acceleration ($a = +2 \, \text{m/s}^2$).
    • Correct: Acceleration opposes velocity → slowing down (velocity becomes less negative).
    • Incorrect: Using $a = -2 \, \text{m/s}^2$ (wrong sign) → incorrectly predicts speeding up.