IB MYP 4-5 Physics- Speed, velocity, and acceleration- Study Notes - New Syllabus
IB MYP 4-5 Physics- Speed, velocity, and acceleration- Study Notes
Key Concepts
- Speed, velocity, and acceleration
Speed, velocity, and acceleration
Speed
Speed is a scalar quantity that refers to how fast an object is moving, regardless of its direction.
- Definition: Distance travelled per unit time.
- Formula:
\( \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \)
- SI Unit: \( \text{m/s} \) (metres per second)
- Other units: km/h, cm/s
- Type: Scalar (has only magnitude, no direction)
Velocity
Velocity is a vector quantity that refers to the rate of displacement i.e., speed in a specified direction.
- Definition: Displacement per unit time.
- Formula:
\( \text{Velocity} = \dfrac{\text{Displacement}}{\text{Time}} \)
- SI Unit: \( \text{m/s} \)
- Type: Vector (has both magnitude and direction)
- If direction changes, velocity changes even if speed stays constant.
Acceleration
Acceleration is the rate at which velocity changes with time.
- Definition: Change in velocity per unit time.
- Formula:
\( a = \dfrac{v – u}{t} \)
Where:
- \( a \): acceleration
- \( u \): initial velocity
- \( v \): final velocity
- \( t \): time
- SI Unit: \( \text{m/s}^2 \)
- Type: Vector (has both direction and magnitude)
- Positive acceleration: Speeding up in the positive direction
- Negative acceleration (deceleration): Slowing down
Average vs Instantaneous
Average values represent a quantity over a period of time or a range of values, while instantaneous values represent that quantity at a specific point in time
- Average speed: $\dfrac{\text{Total Distance}}{ \text{ total time}}$
- Instantaneous speed: Speed at a specific moment in time (measured by speedometer)
- Average velocity: $\dfrac{\text{Total displacement}}{ \text{ total time}}$
- Average acceleration: $\dfrac{\text{Change in velocity}}{ \text{ total time}}$
Example :
A car travels 120 km in 2 hours. What is its average speed in m/s?
▶️ Answer/Explanation
Step 1: Use the formula \( \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \).
Distance = 120 km = \( 120 \times 1000 = 120{,}000\,\text{m} \)
Time = 2 hours = \( 2 \times 3600 = 7200\,\text{s} \)
\( \text{Speed} = \dfrac{120{,}000}{7200} = 16.67\,\text{m/s} \)
Final Answer: \( \boxed{16.67\,\text{m/s}} \)
Example:
A person walks 3 km north, then 4 km south in 2 hours. What is the difference between their speed and velocity?
▶️ Answer/Explanation
Total distance travelled = \( 3 + 4 = 7\,\text{km} \)
Total displacement = \( 4 – 3 = 1\,\text{km south} \)
Time = 2 hours
- Average speed = \( \dfrac{7}{2} = 3.5\,\text{km/h} \)
- Average velocity = \( \dfrac{1}{2} = 0.5\,\text{km/h south} \)
This shows that speed depends on path, but velocity depends only on net displacement and direction.
Final Answer: \( \boxed{\text{Speed = 3.5 km/h, Velocity = 0.5 km/h south}} \)
Example:
Is it possible for an object to have zero velocity but non-zero acceleration? Give an example.
▶️ Answer/Explanation
Yes. A good example is a ball thrown vertically upward.
At the highest point, its velocity is zero momentarily, but it still experiences a downward acceleration due to gravity.
This shows that zero velocity does not mean zero acceleration.
Final Answer: \( \boxed{\text{Yes, e.g., at the top of vertical motion}} \)
Example:
A cyclist travels 300 m north in 1 minute, stops for 30 seconds, then travels 400 m south in 2 minutes. After the journey:
- (a) Calculate the total distance and total displacement.
- (b) Calculate average speed and average velocity.
- (c) Explain whether the motion involved acceleration, and why.
▶️ Answer/Explanation
(a) Total Distance and Displacement
Total distance = \( 300 + 400 = 700\,\text{m} \)
Total displacement = \( 400 – 300 = 100\,\text{m south} \)
(b) Average Speed and Velocity
Total time = \( 1 + 0.5 + 2 = 3.5\,\text{min} = 210\,\text{s} \)
Average speed = \( \dfrac{700}{210} = 3.33\,\text{m/s} \)
Average velocity = \( \dfrac{100}{210} = 0.476\,\text{m/s south} \)
(c) Was there acceleration?
Yes. The cyclist changed direction (north to south), which means a change in velocity vector, implying acceleration.
Also, the cyclist started from rest, moved, stopped, and then moved again — all involving changes in velocity.
Final Answers:
- (a) Distance = \( \boxed{700\,\text{m}} \), Displacement = \( \boxed{100\,\text{m south}} \)
- (b) Average speed = \( \boxed{3.33\,\text{m/s}} \), Average velocity = \( \boxed{0.476\,\text{m/s south}} \)
- (c) \( \boxed{\text{Yes, due to change in direction and speed}} \)