Constant acceleration equations Knowledge-Based Study

Study notes

1. What is Constant Acceleration?

   In one-dimensional kinematics, an object undergoes uniformly accelerated motion when its velocity changes by equal amounts in equal intervals of time. This means the acceleration $a$ is constant in both magnitude and direction.

   Core Kinematic Variables

   To analyze this motion along a straight line, we track five fundamental quantities:

   • Displacement ($\Delta x$): The net change in position, measured in meters (m).
   • Initial velocity ($v_0$): The velocity at time $t = 0$, measured in meters per second (m/s).
   • Final velocity ($v$): The velocity at time $t$, measured in meters per second (m/s).
   • Acceleration ($a$): The constant rate of change of velocity, measured in meters per second squared ($\text{m/s}^2$).
   • Time interval ($t$): The elapsed time during the motion, measured in seconds (s).

   Deriving the Big Four Equations

   When acceleration is constant, the relationship between position, velocity, acceleration, and time can be expressed using four core equations:

   1. Velocity-Time Relation:
      $v = v_0 + at$
      (Relates initial velocity, final velocity, acceleration, and time; does not involve displacement.)
   2. Displacement-Time Relation:
      $\Delta x = v_0 t + \frac{1}{2}at^2$
      (Determines position over time; does not involve final velocity.)
   3. Time-Independent Relation:
      $v^2 = v_0^2 + 2a\Delta x$
      (Useful when elapsed time is unknown or not required.)
   4. Average Velocity Relation:
      $\Delta x = \left(\frac{v_0 + v}{2}\right)t$
      (Valid exclusively when acceleration is constant; does not require acceleration explicitly.)

   Problem-Solving Strategy

   Each of the four kinematic equations links exactly four of the five variables ($\Delta x$, $v_0$, $v$, $a$, $t$), leaving out precisely one variable.

   Variable Selection Matrix

   • If the problem neither provides nor asks for $\Delta x$, use: $v = v_0 + at$
   • If the problem neither provides nor asks for $v$, use: $\Delta x = v_0 t + \frac{1}{2}at^2$
   • If the problem neither provides nor asks for $t$, use: $v^2 = v_0^2 + 2a\Delta x$
   • If the problem neither provides nor asks for $a$, use: $\Delta x = \left(\frac{v_0 + v}{2}\right)t$

   By identifying the given values, the desired unknown, and the omitted variable, you can immediately select the single equation needed to solve the problem.

   Vector Nature of 1D Motion

   Displacement, velocity, and acceleration are vector quantities. In one dimension, direction is established solely by algebraic signs (+ or -):

   • Choose a reference frame at the start (commonly, rightward or upward is chosen as positive).
   • An object moving in the negative direction has a negative velocity ($v < 0$).
   • An acceleration vector opposing the velocity vector causes the object to slow down (deceleration).
   • An acceleration vector in the same direction as velocity causes the object to speed up.

   Note: A negative acceleration does not necessarily mean an object is slowing down; if velocity is also negative, the object is speeding up in the negative direction.

   Worked Example

   Scenario: A car travelling at an initial velocity of 15 m/s accelerates uniformly at $2.0\text{ m/s}^2$ for a duration of 6.0 s. Find its final velocity and total displacement.

   1. Identify Knowns and Unknowns

   • Given: $v_0 = 15\text{ m/s}$, $a = 2.0\text{ m/s}^2$, $t = 6.0\text{ s}$
   • Find: $v$ and $\Delta x$

   2. Calculate Final Velocity

   Use equation (1) since it directly yields $v$:

   $v = v_0 + at = 15 + (2.0)(6.0) = 15 + 12 = 27\text{ m/s}$

   3. Calculate Displacement

   Use equation (2) to find $\Delta x$:

   $\Delta x = v_0 t + \frac{1}{2}at^2 = (15)(6.0) + \frac{1}{2}(2.0)(6.0)^2$

   $\Delta x = 90 + 36 = 126\text{ m}$