CIE IGCSE Physics (0625) Motion Study Notes - New Syllabus
CIE IGCSE Physics (0625) Topic 1.2 Motion Study Notes
Key Concepts:
Core
- Define speed as distance travelled per unit time; recall and use the equation v = s/t
- Define velocity as speed in a given direction
- Recall and use the equation average speed = total distance travelled / total time taken
- Sketch, plot and interpret distance–time and speed–time graphs
- Determine, qualitatively, from given data or the shape of a distance–time graph or speed–time graph when an object is: (a) at rest (b) moving with constant speed (c) accelerating (d) decelerating
- Calculate speed from the gradient of a straight-line section of a distance–time graph
- Calculate the area under a speed–time graph to determine the distance travelled for motion with constant speed or constant acceleration
- State that the acceleration of free fall g for an object near to the surface of the Earth is approximately constant and is approximately 9.8 m/s²
Supplement
- Define acceleration as change in velocity per unit time; recall and use the equation a = Δv/Δt
- Determine from given data or the shape of a speed–time graph when an object is moving with: (a) constant acceleration (b) changing acceleration
- Calculate acceleration from the gradient of a speed–time graph
- Know that a deceleration is a negative acceleration and use this in calculations
- Describe the motion of objects falling in a uniform gravitational field with and without air/liquid resistance, including reference to terminal velocity
Speed
Speed
Speed is the distance travelled per unit time.
- It tells us how fast an object is moving, but not the direction (so it is a scalar quantity).
- SI unit: metres per second (m/s).
Equation for Speed
\( \text{Speed} = d\frac{\text{Distance travelled}}{\text{Time taken}} \)
- Distance must be in metres (m).
- Time must be in seconds (s).
- Other units are sometimes used (e.g. km/h, cm/s), but must be converted into SI units for calculations.
Example:
A car travels 240 m in 20 s. Find its speed.
▶️ Answer/Explanation
Speed = Distance ÷ Time
= \( 240 \div 20 \)
= \(\boxed{12~\text{m/s}}\)
Example:
A cyclist covers 6 km in 20 minutes. Calculate the speed in m/s.
▶️ Answer/Explanation
Step 1: Convert units
Distance = 6 km = \( 6000~\text{m} \)
Time = 20 min = \( 20 \times 60 = 1200~\text{s} \)
Step 2: Use formula
Speed = \( 6000 \div 1200 = \boxed{5~\text{m/s}} \)
Velocity
Velocity
Velocity is the speed of an object in a given direction.
- It is a vector quantity because it has both:
- Magnitude → the speed (how fast the object is moving)
- Direction → the line along which the object moves
- Two objects can have the same speed but different velocities if they are moving in different directions.
- The SI unit of velocity is the same as speed: \( \text{m/s} \).
Example: A car moving at \( 20~\text{m/s} \) towards the north has a velocity of \( 20~\text{m/s north} \). Another car moving at \( 20~\text{m/s} \) towards the east has the same speed, but its velocity is different because the direction is different.
Comparison: Speed vs Velocity
| Aspect | Speed | Velocity |
|---|---|---|
| Definition | Distance travelled per unit time | Speed of an object in a given direction |
| Quantity Type | Scalar (magnitude only) | Vector (magnitude + direction) |
| Can be Negative? | No, always positive | Yes, depends on direction |
| Example | A car moves at \(30~\text{m/s}\) | A car moves at \(30~\text{m/s east}\) |
| SI Unit | \( \text{m/s} \) | \( \text{m/s} \) (with direction) |
Example:
A boat sails at \( 15~\text{m/s} \) east. What is its velocity?
▶️ Answer/Explanation
Speed of the boat = \( 15~\text{m/s} \)
Direction = East
Therefore, velocity = \( \mathbf{15~\text{m/s east}} \)
Example:
A car travels at \( 20~\text{m/s} \) due north for 30 seconds. Then it turns and continues at the same speed of \( 20~\text{m/s} \) due east. Compare its speed and velocity before and after the turn.
▶️ Answer/Explanation
Before the turn:
Speed = \( 20~\text{m/s} \), Direction = north → Velocity = \( 20~\text{m/s north} \).
After the turn:
Speed = \( 20~\text{m/s} \), Direction = east → Velocity = \( 20~\text{m/s east} \).
Conclusion: The speed of the car remains the same, but its velocity changes because the direction has changed.
Average Speed
Average Speed
Average speed is the total distance travelled divided by the total time taken.
- It is useful when the speed is not constant throughout a journey.
- SI unit: metres per second (m/s).
Equation for Average Speed
\( \text{Average Speed} = \dfrac{\text{Total Distance}}{\text{Total Time}} \)
- Distance must be in metres (m).
- Time must be in seconds (s).
- Other units (e.g. km/h) can be used, but must be consistent.
Example:
A bus travels 60 km in 1.5 hours. Find its average speed in km/h.
▶️ Answer/Explanation
Total distance = \( 60~\text{km} \)
Total time = \( 1.5~\text{h} \)
Average speed = \( \dfrac{60}{1.5} = \boxed{40~\text{km/h}} \)
Example:
A runner completes 400 m in 80 s, then 200 m in 40 s. Find the average speed.
▶️ Answer/Explanation
Total distance = \( 400 + 200 = 600~\text{m} \)
Total time = \( 80 + 40 = 120~\text{s} \)
Average speed = \( \dfrac{600}{120} = \boxed{5~\text{m/s}} \)
Acceleration
Acceleration
Acceleration is the rate of change of velocity per unit time.
- It tells us how quickly an object is speeding up or slowing down.
- SI Unit: \( \text{m/s}^2 \)
Equation for Acceleration:
$a = \dfrac{\Delta v}{t} = \dfrac{v – u}{t}$
- \( a \) = acceleration (\( \text{m/s}^2 \))
- \( v \) = final velocity (\( \text{m/s} \))
- \( u \) = initial velocity (\( \text{m/s} \))
- \( t \) = time taken (s)
Example:
A car increases its velocity from \( 10~\text{m/s} \) to \( 25~\text{m/s} \) in \( 5~\text{s} \). Find its acceleration.
▶️ Answer/Explanation
Initial velocity, \( u = 10~\text{m/s} \)
Final velocity, \( v = 25~\text{m/s} \)
Time taken, \( t = 5~\text{s} \)
\( a = \dfrac{v – u}{t} = \dfrac{25 – 10}{5} = \dfrac{15}{5} = \boxed{3~\text{m/s}^2} \)
Example:
A motorcycle is moving at \( 20~\text{m/s} \). The rider applies brakes and the speed reduces uniformly to \( 8~\text{m/s} \) in \( 4~\text{s} \). Find the acceleration of the motorcycle. State whether it is acceleration or deceleration.
▶️ Answer/Explanation
Initial velocity, \( u = 20~\text{m/s} \)
Final velocity, \( v = 8~\text{m/s} \)
Time taken, \( t = 4~\text{s} \)
Using formula: \( a = \dfrac{v – u}{t} \)
\( a = \dfrac{8 – 20}{4} = \dfrac{-12}{4} = -3~\text{m/s}^2 \)
Therefore: The motorcycle has a deceleration of \( \boxed{3~\text{m/s}^2} \).
Distance–Time & Speed–Time Graphs
Distance–Time Graphs
A distance–time graph shows how far an object has travelled over time.
- Gradient (slope) = speed
- Horizontal line → object is at rest
- Straight sloped line → constant speed
- Curved line → changing speed (acceleration/deceleration)
Speed–Time Graphs
A speed–time graph shows how an object’s speed changes over time.
- Gradient (slope) = acceleration
- Area under the graph = distance travelled
- Horizontal line → constant speed
- Upward slope → acceleration
- Downward slope → deceleration
- Curved lines → changing acceleration
Interpretation from a Distance–Time Graph
| Graph Shape | Interpretation |
|---|---|
| Flat horizontal line | Object is at rest |
| Straight upward line | Moving with constant speed |
| Curve getting steeper | Accelerating |
| Curve getting flatter | Decelerating |
Interpretation from a Speed–Time Graph
| Graph Shape | Interpretation |
|---|---|
| Horizontal line | Constant speed |
| Straight upward slope | Constant acceleration |
| Straight downward slope | Constant deceleration |
| Curved slope | Changing acceleration |
| Line on time axis | Object is at rest (speed = 0) |
Acceleration Types from Speed–Time Graphs
| Graph Type | Line Shape | Acceleration |
|---|---|---|
| Speed–Time | Straight slope | Constant acceleration |
| Speed–Time | Curved slope | Changing acceleration |
Example:
A distance–time graph shows a flat line from 0 to 10 seconds at 15 m. What does this mean?
▶️ Answer/Explanation
The distance does not change over time. This means the object is not moving.
Conclusion: The object is at rest for 10 seconds.
Example:
A cyclist travels 100 m in 20 seconds in a straight line. Draw and describe the distance–time graph.
▶️ Answer/Explanation
Speed = \( \dfrac{100}{20} = 5~\text{m/s} \)
The graph will be a straight diagonal line from (0, 0) to (20, 100).
Conclusion: The object moves with constant speed.
Example:
A car accelerates uniformly from 0 to 30 m/s in 6 seconds. Calculate the acceleration and distance travelled.
▶️ Answer/Explanation
Step 1: Acceleration
\( a = \dfrac{\Delta v}{\Delta t} = \dfrac{30 – 0}{6} = \boxed{5~\text{m/s}^2} \)
Step 2: Distance travelled (Area under triangle)
Area = \( \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 6 \times 30 = \boxed{90~\text{m}} \)
Example:
The speed-time graph below shows the motion of a vehicle over a period of 30 seconds.
Use the graph to calculate the total distance travelled by the vehicle in 30 seconds.
Break the graph into the following intervals and calculate the distance for each:
- From 0 to 10 seconds (acceleration)
- From 10 to 25 seconds (constant speed)
- From 25 to 30 seconds (deceleration)
▶️ Answer/Explanation
From 0 s to 10 s (acceleration):
This is a triangle:
Distance = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 20 = \boxed{100~\text{m}} \)
From 10 s to 25 s (constant speed):
This is a rectangle:
Distance = \( \text{width} \times \text{height} = 15 \times 20 = \boxed{300~\text{m}} \)
From 25 s to 30 s (deceleration):
This is another triangle:
Distance = \( \frac{1}{2} \times 5 \times 20 = \boxed{50~\text{m}} \)
Total Distance Travelled:
\( 100 + 300 + 50 = \boxed{450~\text{m}} \)
Acceleration Due to Free Fall
Free Fall:
Free fall is the motion of an object under the influence of gravity alone.
- There is no air resistance involved during true free fall.
- The only force acting on the object is its weight (gravitational force).
Acceleration due to gravity (g):
- The acceleration an object experiences during free fall is due to Earth’s gravity.
- This acceleration is denoted by the symbol g.
- Its value is nearly constant near the surface of the Earth.
Standard value: \( g = 9.8~\text{m/s}^2 \)
Key points:
- This acceleration acts vertically downward.
- It does not depend on the mass of the object.
- All objects fall at the same rate in a vacuum, regardless of their weight or size.
- The value of \( g \) is approximately constant at Earth’s surface, but may vary slightly by location (altitude, equator vs poles).
Important Concept: The value of \( g \) is also the gravitational field strength, which means the force per unit mass acting on an object:
\( g = \dfrac{F}{m} \)
- Where \( F \) is the weight in newtons and \( m \) is mass in kilograms.
Units of acceleration:
- SI Unit of acceleration: $m/s^2$ (metres per second squared)
- This means the velocity of a free-falling object increases by 9.8 m/s every second.
Example:
A stone is dropped from a tower. What is its acceleration during free fall?
▶️ Answer/Explanation
Since no other force except gravity acts on the stone, it undergoes free fall.
Acceleration = \( \boxed{9.8~\text{m/s}^2} \), vertically downward.
Example:
If a ball falls freely from rest, what is its speed after 3 seconds?
▶️ Answer/Explanation
Use the equation: \( v = g \cdot t \)
Given: \( g = 9.8~\text{m/s}^2 \), \( t = 3~\text{s} \)
Then, \( v = 9.8 \times 3 = \boxed{29.4~\text{m/s}} \)
Falling in a Uniform Gravitational Field
Falling in a Uniform Gravitational Field
- In a uniform gravitational field (like near Earth’s surface), all objects experience the same downward acceleration due to gravity: \( g = 9.8~\text{m/s}^2 \).
- This gravitational pull acts on all objects equally regardless of their mass (ignoring air resistance).
Case 1: Falling Without Air or Liquid Resistance (Vacuum):
- No resistive force is acting — only gravity.
- All objects fall with the same acceleration: \( g = 9.8~\text{m/s}^2 \).
- Speed increases uniformly (i.e. object accelerates constantly).
Case 2: Falling With Air or Liquid Resistance:
- As the object falls, air or fluid pushes against it — this is a resistive force (also called drag).
- This resistance increases with speed.
- Eventually, the upward resistive force becomes equal to the object’s weight.
- At this point, net force = 0 and acceleration = 0.
This constant maximum speed is called: Terminal velocity
- The object continues falling at constant speed from this point onward.
Key forces involved:
- Weight (W) — acts downward and stays constant
- Air resistance (R) — acts upward and increases with speed
| Stage | Forces | Motion |
|---|---|---|
| Start of fall | Weight > Air resistance | Accelerates downward |
| Falling faster | Weight ≈ Air resistance | Acceleration decreases |
| Terminal velocity | Weight = Air resistance | Constant speed (no acceleration) |
Example:
A metal ball and a paper sheet are dropped in a vacuum. Which hits the ground first?
▶️ Answer/Explanation
There is no air resistance in a vacuum. Both objects fall with the same acceleration.
They hit the ground at the same time.
Example:
A skydiver jumps from a plane and eventually reaches a constant speed. What is happening?
▶️ Answer/Explanation
At first, the skydiver accelerates as weight > air resistance.
As speed increases, air resistance increases until it equals the weight.
No net force → No acceleration → Skydiver falls at terminal velocity.