Distance, Displacement, Speed, Velocity, Acceleration
CSEC Physics Essential Knowledge: Motion is described using quantities that define how far objects move and how fast they move. It is crucial to distinguish between scalars (magnitude only) like distance and speed, and vectors (magnitude and direction) like displacement, velocity, and acceleration. Understanding these differences is key to mastering mechanics and interpreting kinematics graphs.
Part 1: Distance vs. Displacement
Distance (Scalar)
Definition: The total length of the path traveled by an object, regardless of direction.
Symbol: \(d\) or \(s\)
SI Unit: Metres (m)
Key Property: Can never be negative; always increases or stays the same.
Displacement (Vector)
Definition: The shortest distance from the initial position to the final position in a specific direction.
Symbol: \(s\) or \(\Delta x\)
SI Unit: Metres (m)
Key Property: Can be positive, negative, or zero. Depends on direction.
A runner runs one complete lap around a 400m standard oval track.
Calculate:
(a) The distance traveled.
(b) The displacement.
Distance = 400 m
Displacement = 0 m
Part 2: Speed vs. Velocity
While distance and displacement describe “where” an object is, speed and velocity describe “how fast” it gets there.
Speed (Scalar)
Definition: The rate at which an object covers distance.
SI Unit: Metres per second (m/s) or km/h
Types: Average speed, Instantaneous speed.
Velocity (Vector)
Definition: The rate of change of displacement (speed in a given direction).
SI Unit: Metres per second (m/s)
Key Property: Velocity changes if speed changes OR if direction changes.
A student walks 1.5 km to school in 20 minutes. Calculate their average speed in m/s.
Distance: \(1.5 \text{ km} \times 1000 = 1500 \text{ m}\)
Time: \(20 \text{ mins} \times 60 = 1200 \text{ s}\)
Part 3: Acceleration
Acceleration describes how quickly an object’s velocity changes. Remember, since velocity is a vector, acceleration can be caused by a change in speed, a change in direction, or both.
Definition: Acceleration is the rate of change of velocity with respect to time.
Where:
- \(a\) = acceleration
- \(v\) = final velocity
- \(u\) = initial velocity
- \(t\) = time
SI Unit: Metres per second squared (m/s²)
Positive Acceleration
Object speeds up in the positive direction.
OR
Object slows down in the negative direction.
Negative Acceleration (Deceleration)
Object slows down in the positive direction.
OR
Object speeds up in the negative direction.
A car accelerates from rest (\(0 \text{ m/s}\)) to \(20 \text{ m/s}\) in 5 seconds. Calculate its acceleration.
Initial velocity (\(u\)) = 0 m/s
Final velocity (\(v\)) = 20 m/s
Time (\(t\)) = 5 s
Part 4: Kinematics Graphs
In CSEC Physics, you must be able to interpret Distance-Time and Speed-Time graphs. The gradient (slope) and area under the line provide physical quantities.
1. Distance-Time Graphs
Gradient (Slope) = Speed
- Horizontal line: Stationary (Speed = 0)
- Sloped straight line: Constant Speed
- Curved line: Changing Speed (Acceleration/Deceleration)
- Steeper slope: Higher Speed
2. Speed-Time Graphs
Gradient (Slope) = Acceleration
Area under line = Distance Traveled
- Horizontal line: Constant Speed (Acceleration = 0)
- Positive slope: Acceleration
- Negative slope: Deceleration
- Above x-axis: Moving in positive direction
Using the Speed-Time graph above (Fig 2), calculate the total distance traveled in the first 6 seconds.
\( \frac{1}{2} \times 2 \text{ s} \times 20 \text{ m/s} = 20 \text{ m} \)
\( (6 \text{ s} – 2 \text{ s}) \times 20 \text{ m/s} = 4 \times 20 = 80 \text{ m} \)
Part 5: Past Paper Questions
(a) Define displacement. [2 marks]
(b) A car travels 5 km East, then 5 km West, in a total time of 30 minutes.
(i) Calculate the distance traveled by the car. [1 mark]
(ii) Calculate the displacement of the car. [1 mark]
(iii) Calculate the average speed of the car in m/s. [3 marks]
(iv) Calculate the average velocity of the car in m/s. [2 marks]
(a) Displacement is the shortest distance between the starting point and the finishing point in a specific direction. (Alternatively: The change in position of an object in a specific direction).
(b)
(i) Distance = 5 km + 5 km = 10 km
(ii) Starting point and finishing point are the same. Displacement = 0 km
(iii) Speed = Distance / Time
Distance = 10 km = 10,000 m
Time = 30 mins = 1800 s
Speed = 10,000 / 1800 = 5.56 m/s
(iv) Velocity = Displacement / Time
Velocity = 0 / 1800 = 0 m/s
A cyclist starts from rest and reaches a speed of 8 m/s in 4 seconds. He then travels at this constant speed for 10 seconds before applying brakes and coming to rest in 2 seconds.
Draw the speed-time graph for this motion and calculate:
(a) The initial acceleration. [2 marks]
(b) The total distance traveled. [3 marks]
(a) \( u = 0, v = 8, t = 4 \)
\( a = \frac{v-u}{t} = \frac{8-0}{4} = \mathbf{2 \text{ m/s}^2} \)
(b) Total distance is the area under the graph.
Phase 1 (0-4s): Triangle area = \( 0.5 \times 4 \times 8 = 16 \text{ m} \)
Phase 2 (4-14s): Rectangle area = \( 10 \times 8 = 80 \text{ m} \)
Phase 3 (14-16s): Triangle area = \( 0.5 \times 2 \times 8 = 8 \text{ m} \)
Total Distance = 16 + 80 + 8 = 104 m
🎯 Motion Formulae Summary
\( v = \frac{d}{t} \)
\( v = \frac{\Delta s}{\Delta t} \)
\( a = \frac{v-u}{t} \)
- Distance-Time Graph: Slope = Speed
- Speed-Time Graph: Slope = Acceleration, Area = Distance
- Vector quantities: Displacement, Velocity, Acceleration
- Scalar quantities: Distance, Speed
Exam Day Checklist:
1. Did you answer the question asked? (Speed vs Velocity)
2. Did you convert units? (km/h to m/s, minutes to seconds)
3. Did you include the direction for vector answers?
4. For graph questions, did you calculate Area or Gradient? Identify which is needed!