Edexcel iGCSE Physics -1.6 Acceleration- Study Notes- New Syllabus
Edexcel iGCSE Physics -1.6 Acceleration- Study Notes- New syllabus
Edexcel iGCSE Physics -1.6 Acceleration- Study Notes -Edexcel iGCSE Physics – per latest Syllabus.
Key Concepts:
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Acceleration, Change in Velocity and Time
Acceleration describes how quickly the velocity of an object changes with time. An object accelerates if its speed increases, decreases, or if its direction of motion changes.
In many situations in mechanics, objects move with uniform acceleration, meaning the acceleration remains constant throughout the motion.
Key Relationship
The relationship between acceleration, change in velocity, and time taken is:![]()
\( \mathrm{a = \dfrac{v – u}{t}} \)
- \( \mathrm{a} \) = acceleration (m/s²)
- \( \mathrm{u} \) = initial velocity (m/s)
- \( \mathrm{v} \) = final velocity (m/s)
- \( \mathrm{t} \) = time taken (s)
Understanding the Equation
- \( \mathrm{v – u} \) represents the change in velocity.
- Dividing by time gives the rate at which velocity changes.
- A larger acceleration means the velocity changes more rapidly.
If the value of acceleration is negative, the object is decelerating (slowing down).
Rearranging the Equation
The equation can be rearranged to find velocity or time:
- Final velocity: \( \mathrm{v = u + at} \)
- Initial velocity: \( \mathrm{u = v – at} \)
- Time taken: \( \mathrm{t = \dfrac{v – u}{a}} \)
Link to Velocity–Time Graphs
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- Acceleration is the gradient of a velocity–time graph.
- A straight line on a velocity–time graph shows constant acceleration.
- A horizontal line shows zero acceleration (constant velocity).
Important Points to Remember
- All velocities must be in m/s and time in seconds.
- Acceleration is measured in m/s².
- Always identify initial and final velocities correctly.
Example
A car increases its velocity from \( \mathrm{10\ m/s} \) to \( \mathrm{30\ m/s} \) in \( \mathrm{5\ s} \). Calculate the acceleration of the car.
▶️ Answer / Explanation
Use: \( \mathrm{a = \dfrac{v – u}{t}} \)
\( \mathrm{a = \dfrac{30 – 10}{5}} \)
\( \mathrm{a = \dfrac{20}{5} = 4\ m/s^2} \)
Example
A cyclist slows down from \( \mathrm{12\ m/s} \) to \( \mathrm{4\ m/s} \) in \( \mathrm{4\ s} \). Calculate the acceleration of the cyclist.
▶️ Answer / Explanation
Use: \( \mathrm{a = \dfrac{v – u}{t}} \)
\( \mathrm{a = \dfrac{4 – 12}{4}} \)
\( \mathrm{a = \dfrac{-8}{4} = -2\ m/s^2} \)
The negative sign shows the cyclist is decelerating.
