IB MYP 4-5 Physics- Equations of motion- Study Notes - New Syllabus
IB MYP 4-5 Physics- Equations of motion- Study Notes
Key Concepts
- Equations of Motion
Equations of Motion
Equations of Motion
These equations describe the motion of objects under uniform acceleration. They relate displacement, velocity, time, and acceleration.
Symbols Used:
- \( u \) = Initial velocity (m/s)
- \( v \) = Final velocity (m/s)
- \( a \) = Acceleration (m/s²)
- \( s \) = Displacement (m)
- \( t \) = Time (s)
1. First Equation: Relates velocity, acceleration, and time
\( v = u + at \)
2. Second Equation: Relates displacement, initial velocity, time, and acceleration
\( s = ut + \dfrac{1}{2}at^2 \)
3. Third Equation: Relates velocity, acceleration, and displacement (no time)
\( v^2 = u^2 + 2as \)
Derivation of Equations of Motion
These equations apply when an object moves with uniform (constant) acceleration.
1. First Equation: \( v = u + at \)
Definition of acceleration:
\( a = \dfrac{v – u}{t} \Rightarrow v = u + at \)
2. Second Equation: \( s = ut + \dfrac{1}{2}at^2 \)
Displacement = $\text{Average velocity × Time}$
Average velocity \( = \dfrac{u + v}{2} \)
From First Equation: \( v = u + at \)
Substitute into average velocity:
\( s = \dfrac{u + (u + at)}{2} \cdot t = \dfrac{2u + at}{2} \cdot t = ut + \dfrac{1}{2}at^2 \)
3. Third Equation: \( v^2 = u^2 + 2as \)
Start from First Equation: \( v = u + at \)
Solve for \( t = \dfrac{v – u}{a} \)
Now use Second Equation:
\( s = ut + \dfrac{1}{2}at^2 \)
Substitute for \( t \):
\( s = u \cdot \dfrac{v – u}{a} + \dfrac{1}{2}a \cdot \left( \dfrac{v – u}{a} \right)^2 \)
Simplifying gives: \( v^2 = u^2 + 2as \)
Example:
A car accelerates from \( 10\,\text{m/s} \) to \( 30\,\text{m/s} \) in 4 seconds. Find the acceleration.
▶️ Answer/Explanation
Given: \( u = 10 \,\text{m/s}, v = 30 \,\text{m/s}, t = 4 \,\text{s} \)
Using \( v = u + at \):
\( 30 = 10 + a \times 4 \)
\( a = \dfrac{30 – 10}{4} = \dfrac{20}{4} = 5\,\text{m/s}^2 \)
Final Answer: \( \boxed{5\,\text{m/s}^2} \)
Example:
A body starts from rest and accelerates at \( 3\,\text{m/s}^2 \) for 6 seconds. Find the displacement.
▶️ Answer/Explanation
Given: \( u = 0 \,\text{m/s}, a = 3 \,\text{m/s}^2, t = 6 \,\text{s} \)
Using \( s = ut + \dfrac{1}{2}at^2 \):
\( s = 0 \times 6 + \dfrac{1}{2} \times 3 \times 6^2 \)
\( s = \dfrac{1}{2} \times 3 \times 36 = 54\,\text{m} \)
Final Answer: \( \boxed{54\,\text{m}} \)
Example:
A car starts from rest and accelerates at \( 2\,\text{m/s}^2 \) for 10 seconds. Find:
- Final velocity
- Displacement covered
▶️ Answer/Explanation
Given: \( u = 0 \,\text{m/s}, a = 2 \,\text{m/s}^2, t = 10\,\text{s} \)
1. Final Velocity: Use \( v = u + at \)
\( v = 0 + 2 \cdot 10 = \boxed{20\,\text{m/s}} \)
2. Displacement: Use \( s = ut + \dfrac{1}{2}at^2 \)
\( s = 0 + \dfrac{1}{2} \cdot 2 \cdot 10^2 = \dfrac{1}{2} \cdot 2 \cdot 100 = \boxed{100\,\text{m}} \)
Example:
An object moving at \( 15\,\text{m/s} \) comes to rest after covering a distance of \( 45\,\text{m} \). Find the acceleration.
▶️ Answer/Explanation
Given: \( v = 0, u = 15\,\text{m/s}, s = 45\,\text{m} \)
Use third equation: \( v^2 = u^2 + 2as \)
\( 0 = 15^2 + 2a(45) \Rightarrow 0 = 225 + 90a \)
\( a = \dfrac{-225}{90} = \boxed{-2.5\,\text{m/s}^2} \)
(Negative sign shows deceleration)