Equation 1: First Equation of Motion

Use when: You don't know or need displacement

Application: Finding final velocity after accelerating for a given time[web:121][web:127]

Equation 2: Second Equation of Motion

Use when: You don't know or need final velocity

Application: Finding displacement when an object comes to rest[web:121][web:127]

Equation 3: Third Equation of Motion

Use when: You don't know or need time

Application: Finding stopping distance or final velocity without time information[web:121][web:127]

Equation 4: Average Velocity Equation

Use when: You don't know or need acceleration

Application: Finding displacement using average velocity[web:122][web:125][web:127]

Deriving \( v = u + at \)

By definition, acceleration is the rate of change of velocity[web:121][web:122]:

\( a = \frac{v - u}{t} \)

Multiply both sides by \( t \):

\( at = v - u \)

Rearrange to get:

\( v = u + at \) ✓

Deriving \( s = ut + \frac{1}{2}at^2 \)

Displacement equals average velocity times time[web:122][web:125]:

\( s = v_{avg} \times t = \frac{u + v}{2} \times t \)

Substitute \( v = u + at \):

\( s = \frac{u + (u + at)}{2} \times t = \frac{2u + at}{2} \times t \)

\( s = \frac{2ut + at^2}{2} = ut + \frac{1}{2}at^2 \)

\( s = ut + \frac{1}{2}at^2 \) ✓

Deriving \( v^2 = u^2 + 2as \)

Start with equations 1 and 4[web:122][web:125]:

From \( v = u + at \), we get \( t = \frac{v - u}{a} \)

Substitute into \( s = \frac{u + v}{2} \times t \):

\( s = \frac{u + v}{2} \times \frac{v - u}{a} = \frac{v^2 - u^2}{2a} \)

Multiply both sides by \( 2a \):

\( v^2 = u^2 + 2as \) ✓

Example 1: Finding Final Velocity

Problem: A car accelerates from rest at 3.5 m/s² for 5 seconds. Find the final velocity[web:129].

Solution:

Known: \( u = 0 \) m/s (starts from rest), \( a = 3.5 \) m/s², \( t = 5 \) s

Unknown: \( v = ? \)

Missing: Displacement (\( s \))

Equation: \( v = u + at \)

\( v = 0 + (3.5)(5) = 17.5 \) m/s

Answer: The final velocity is 17.5 m/s

Example 2: Finding Distance Traveled

Problem: A car moving from rest with acceleration 6.5 m/s² travels for 15 seconds. Find the distance covered[web:129].

Solution:

Known: \( u = 0 \) m/s, \( a = 6.5 \) m/s², \( t = 15 \) s

Unknown: \( s = ? \)

Missing: Final velocity (\( v \))

Equation: \( s = ut + \frac{1}{2}at^2 \)

\( s = 0(15) + \frac{1}{2}(6.5)(15)^2 = 0 + \frac{1}{2}(6.5)(225) = 731.25 \) m

Answer: The distance traveled is 731.25 meters

Example 3: Finding Deceleration

Problem: A car traveling at 14 m/s applies brakes and stops after covering 45 meters. Find the deceleration[web:129].

Solution:

Known: \( u = 14 \) m/s, \( v = 0 \) m/s (stops), \( s = 45 \) m

Unknown: \( a = ? \)

Missing: Time (\( t \))

Equation: \( v^2 = u^2 + 2as \)

\( 0^2 = 14^2 + 2a(45) \)

\( 0 = 196 + 90a \)

\( a = -\frac{196}{90} = -2.18 \) m/s²

Answer: The deceleration is 2.18 m/s² (negative indicates slowing down)

Example 4: Free Fall

Problem: A ball is dropped from rest. How far does it fall in 3 seconds? (Use \( g = 9.8 \) m/s²)[web:126][web:138]

Solution:

Known: \( u = 0 \) m/s (dropped from rest), \( a = 9.8 \) m/s² (gravity), \( t = 3 \) s

Unknown: \( s = ? \)

Equation: \( s = ut + \frac{1}{2}at^2 \)

\( s = 0(3) + \frac{1}{2}(9.8)(3)^2 = \frac{1}{2}(9.8)(9) = 44.1 \) m

Answer: The ball falls 44.1 meters

💡 Constant Acceleration Only

Kinematic equations ONLY work when acceleration is constant. They cannot be used for changing acceleration[web:122][web:125][web:126].

💡 Sign Conventions

Choose a positive direction (usually right or upward) and stick to it. Opposite direction values are negative[web:126][web:130].

💡 "Starts from Rest"

This phrase means initial velocity \( u = 0 \). Similarly, "comes to rest" means final velocity \( v = 0 \)[web:126][web:127].

💡 Free Fall

For objects falling under gravity (no air resistance), \( a = g = 9.8 \) m/s² or 10 m/s² (approximation)[web:126][web:138].

💡 Units Matter

Always convert to consistent SI units: meters (m), seconds (s), meters per second (m/s), and meters per second squared (m/s²).

💡 Scalar vs Vector

Distance and speed are scalars (no direction). Displacement, velocity, and acceleration are vectors (have direction).

💡 Graph Connections

Kinematic equations can be derived from velocity-time graphs using areas and slopes[web:122][web:125].

💡 Curriculum Coverage

Kinematic equations appear in IB Physics, AP Physics 1 & 2, A-Level Physics, GCSE/IGCSE Physics, SAT Physics, and all introductory physics courses.

❌ Using Wrong Signs

Forgetting to use negative values for quantities in the opposite direction to your chosen positive direction.

❌ Applying to Variable Acceleration

Using kinematic equations when acceleration is changing (requires calculus instead).

❌ Mixing Up Variables

Confusing initial velocity (\( u \)) with final velocity (\( v \)), or distance with displacement.

❌ Unit Inconsistencies

Mixing units like km/h with m/s without proper conversion.

❌ Forgetting the Square

In \( v^2 = u^2 + 2as \), forgetting to square the velocities or take the square root of the final answer.

Adam

Co-Founder @RevisionTown

Math Expert in various curricula including IB, AP, GCSE, IGCSE, and more.