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Is your relationship worth repairing? Get "Build Before You Promise" → Relationship Repair When Love Is Still There but the Relationship Is Falling Apart Sometimes the relationship does not need another argument. It needs two people willing to stop defending themselves long enough to understand what is actually breaking. By Edwin Ogie • Author of The Relationship Repair Handbook , Build Before You Promise , Emotional Mastery , and The Value of You Not every broken relationship is a relationship without love. Sometimes two people still love each other deeply, but they have become exhausted by unresolved arguments, poor communication, resentment, disappointment and repeated misunderstandings. They may still care. They may still want the relationship. They may e...
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Dynamics 2

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Dynamics: Equations of Motion

Dynamics: Equations of Motion

Learn the fundamentals of motion with derivations and worked examples

Derivation of Equations of Motion

The three primary equations of motion relate the displacement, velocity, acceleration, and time of a moving object. They are derived using the assumption of uniform acceleration:

1. First Equation of Motion: v = u + at

From the definition of acceleration:

Acceleration (a) = Change in Velocity / Time a = (v - u) / t Rearranging gives: v = u + at

2. Second Equation of Motion: s = ut + (1/2)at²

Using the definition of average velocity:

Average Velocity = (Initial Velocity + Final Velocity) / 2 v_avg = (u + v) / 2 Displacement (s) = Average Velocity × Time s = [(u + v) / 2] × t

Substitute v = u + at from the first equation:

s = [u + (u + at)] × t / 2 s = ut + (1/2)at²

3. Third Equation of Motion: v² = u² + 2as

Starting from the first equation:

v = u + at Multiply through by v + u: v² - u² = 2as Rearrange to: v² = u² + 2as

Worked Examples

Example 1

Problem: A car travels a distance of 120 m in 15 seconds. Calculate the speed of the car.

Solution:

Using Speed = Distance / Time:

Speed = 120 / 15 = 8 m/s

Example 2

Problem: A cyclist starts from rest and accelerates uniformly at 2 m/s². Find the velocity after 10 seconds.

Solution:

Using v = u + at:

v = 0 + (2 × 10) = 20 m/s

Example 3

Problem: A ball is dropped from a height of 45 m. How long does it take to hit the ground? (Take g = 10 m/s²)

Solution:

Using s = ut + 1/2gt²:

45 = 0 + (1/2)(10)t² 45 = 5t² t² = 9 t = 3 s

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