Uniformly Accelerated Motion: Comprehensive Principles, Equations, and Worked Examples, and Derivations

Overview of Uniformly Accelerated Motion (UAM)

• Uniformly Accelerated Motion refers to the movement of an object along a straight path where its velocity changes at a constant, steady rate.

• Under UAM conditions, the acceleration of the object is constant, meaning the velocity changes by identical amounts within equal intervals of time.

• Key conceptual situations include:

  • When the magnitude of acceleration is zero ($|a| = 0$), the object maintains a constant velocity.

  • When the magnitude of acceleration is greater than zero (|a| > 0), the object is in UAM.

  • Determining if an object speeds up or slows down depends on the relative directions of the velocity and acceleration vectors.

Fundamental Kinematics and UAM Equations

There are four primary equations used to solve problems involving Uniformly Accelerated Motion. Each equation connects specific variables and is chosen based on which variable is missing from the problem context.

• Equation 1: $v_f = v_i + a \times \text{t}$

  • Variables solved for: Final velocity ($v_f$), initial velocity ($v_i$), acceleration ($a$), or time interval ($t$).

  • Missing variable: Displacement ($d$).

• Equation 2: $d = \frac{v_f + v_i}{2} \times t$

  • Variables solved for: Displacement ($d$), final velocity ($v_f$), initial velocity ($v_i$), or time interval ($t$).

  • Missing variable: Acceleration ($a$).

• Equation 3: $d = v_i \times t + \frac{1}{2} \times a \times t^2$

  • Variables solved for: Displacement ($d$), initial velocity ($v_i$), acceleration ($a$), or time interval ($t$).

  • Missing variable: Final velocity ($v_f$).

• Equation 4: $v_f^2 = v_i^2 + 2 \times a \times d$

  • Variables solved for: Final velocity ($v_f$), initial velocity ($v_i$), acceleration ($a$), or displacement ($d$).

  • Missing variable: Time interval ($t$).

Application Example 1: Solving for Displacement

In a road safety campaign experiment, a car accelerates uniformly from an initial velocity of $12.0\,m/s$ to a final velocity of $28.0\,m/s$ over a period of $8.0\,s$ while traveling in a straight line.

• Given:

  • Initial velocity ($v_i$) = $12.0\,m/s$

  • Final velocity ($v_f$) = $28.0\,m/s$

  • Time ($t$) = $8.0\,s$

• Required to Find (RTF): Displacement ($d$) = ?

• Formula Selection: Equation 2 is most appropriate because acceleration is not provided.

• Solution:

  • $d = \frac{28.0\,m/s + 12.0\,m/s}{2} \times 8.0\,s$

  • $d = \frac{40.0\,m/s}{2} \times 8.0\,s$

  • $d = 20.0\,m/s \times 8.0\,s$

• Final Answer: $d = 160\,m$

Application Example 2: Solving for Final Velocity

During a runway test, an airplane starts from rest and accelerates uniformly at a rate of $3.0\,m/s^2$ along a runway of $550\,m$.

• Given:

  • Initial velocity ($v_i$) = $0\,m/s$ (at rest)

  • Acceleration ($a$) = $3.0\,m/s^2$

  • Displacement ($d$) = $550\,m$

• Required to Find (RTF): Final velocity ($v_f$) = ?

• Formula Selection: Equation 4 is most appropriate because time is not provided.

• Solution:

  • $v_f^2 = (0\,m/s)^2 + 2 \times (3.0\,m/s^2) \times (550\,m)$

  • $v_f^2 = 0 + 3300\,m^2/s^2$

  • $\sqrt{v_f^2} = \sqrt{3300\,m^2/s^2}$

• Final Answer: $v_f = 57\,m/s$

Practice Exercise: Train Displacement

A train leaves a station with an initial speed of $5.0\,m/s$ and accelerates at a rate of $1.8\,m/s^2$ for a duration of $12.0\,s$.

• Given:

  • Initial velocity ($v_i$) = $5.0\,m/s$

  • Acceleration ($a$) = $1.8\,m/s^2$

  • Time ($t$) = $12.0\,s$

• Required to Find (RTF): Displacement ($d$) = ?

• Formula Selection: Equation 3 is most appropriate as final velocity is not provided.

• Solution:

  • $d = (5.0\,m/s) \times (12.0\,s) + 0.5 \times (1.8\,m/s^2) \times (12.0\,s)^2$

  • Step 1: Compute initial motion: $(5.0 \times 12.0) = 60\,m$

  • Step 2: Compute acceleration component: $0.5 \times 1.8 \times 144 = 130\,m$

  • Combine results: $60\,m + 130\,m = 190\,m$

  • Direct computation value: $189.6\,m$

• Final Answer: $d = 190\,m$

Equation Derivation: Solving for Acceleration from Equation 3

To find acceleration when given displacement, initial velocity, and time, Equation 3 must be rearranged.

• Original Form: $d = v_i t + \frac{1}{2} a t^2$

• Derivation Steps:

  1. Transpose the initial motion term to the other side: $d - v_i t = \frac{1}{2} a t^2$

  2. Multiply both sides by 2 to remove the fraction: $2(d - v_i t) = a t^2$

  3. Divide both sides by $t^2$ to isolate acceleration: $a = \frac{2(d - v_i t)}{t^2}$

• Variable Case Example: A car starts from rest ($v_i = 0\,m/s$), covering $110\,m$ in $5.21\,s$.

  • $a = \frac{2(110\,m - (0\,m/s \times 5.21\,s))}{(5.21\,s)^2}$

  • $a = \frac{220\,m}{27.1441\,s^2}$

  • Final Result: $a = 8.1\,m/s^2$ (or $8.1049\,m/s^2$)

Equation Derivation: Solving for Time from Equation 2

To calculate time based on displacement and velocity values, Equation 2 is rearranged.

• Original Form: $d = \frac{v_f + v_i}{2} \times t$

• Derivation Steps:

  1. Multiply both sides by 2: $2d = (v_f + v_i)t$

  2. Divide both sides by the sum of velocities: $t = \frac{2d}{v_f + v_i}$

• Variable Case Example: A bike accelerates from rest ($v_i = 0\,m/s$) to $7.10\,m/s$ over $35.4\,m$.

  • $t = \frac{2 \times 35.4\,m}{7.10\,m/s + 0\,m/s}$

  • $t = \frac{70.8\,m}{7.10\,m/s}$

  • Final Result: $t = 9.97\,s$ (or $9.9718\,s$)

Equation Derivation: Solving for Displacement from Equation 4

When designing infrastructure like airport runways, calculating minimum length ($d$) based on known speed and acceleration uses Equation 4.

• Original Form: $v_f^2 = v_i^2 + 2ad$

• Derivation Steps:

  1. Transpose the initial velocity squared: $v_f^2 - v_i^2 = 2ad$

  2. Divide both sides by $2a$ to isolate displacement: $d = \frac{v_f^2 - v_i^2}{2a}$

• Variable Case Example: An airplane with a minimum acceleration of $3\,m/s^2$ must reach a takeoff speed of $65\,m/s$ from rest.

  • $d = \frac{(65\,m/s)^2 - (0\,m/s)^2}{2 \times 3\,m/s^2}$

  • $d = \frac{4225\,m^2/s^2}{6\,m/s^2}$

  • Final Result: $d = 700\,m$ (rounded) or $704.1667\,m$