Motion
Velocity is a vector quantity, meaning it has both magnitude and direction. Building on the concept of speed, velocity is defined as speed in a given direction. This means that if an object moves at a constant speed but changes its direction (e.g., moving in a circle), its velocity is changing because the direction component is changing.
| Aspect | Value |
|---|---|
| Type | Vector (magnitude + direction) |
| Definition | Displacement per unit time (or speed in a given direction) |
| Example | 5 m/s North |
The equation for speed is:
v = \frac{s}{t}
Where:
v = speed in metres per second (m/s)
s = distance travelled in metres (m)
t = time taken in seconds (s)
Average speed is the total distance travelled divided by the total time taken. This is used when an object does not move at a constant speed throughout its journey.
The equation for average speed is:
\text{average speed} = \frac{s_{total}}{t_{total}}
Where:
- s_{total} = total distance travelled in metres (m)
- t_{total} = total time taken in seconds (s)
Solution:
- Identify total distance: s_{total} = 100 \text{ m} + 50 \text{ m} = 150 \text{ m}
- Identify total time: t_{total} = 20 \text{ s} + 10 \text{ s} + 10 \text{ s} = 40 \text{ s}
- Apply the formula: \text{average speed} = \frac{150}{40} = 3.75 \text{ m/s}
Note: Do not simply average the speeds (5 \text{ m/s} and 5 \text{ m/s}) or ignore the stationary time. The definition requires total distance over total time.
Correct Understanding: While the speed is constant, the velocity is changing because the direction of motion is continuously changing. Velocity requires both magnitude and direction to be constant.
Acceleration is defined as the change in velocity per unit time. It describes how quickly an object's velocity is changing.
The equation for acceleration is:
a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}
Where:
a = acceleration in metres per second squared (m/s²)
- \Delta v = change in velocity in metres per second (m/s)
v = final velocity in m/s
u = initial velocity in m/s
t = time interval for the change in seconds (s)
In calculations:
If an object is slowing down, the final velocity v is less than the initial velocity u.
Therefore, (v - u) will be negative, resulting in a negative value for a.
When asked for the 'deceleration', you may need to state the magnitude (positive value) of this negative acceleration.
g is approximately constant.
Value: g \approx 9.8 \text{ m/s}^2.
Steeper gradient = faster speed.
Speed = \frac{\text{change in y}}{\text{change in x}}. Curved line: The object is accelerating or decelerating. The gradient is changing.
Gradient increasing (curve getting steeper) = acceleration.
Gradient decreasing (curve flattening) = deceleration.
Acceleration = \frac{\text{change in y}}{\text{change in x}}. Straight diagonal line (downwards): The object is moving with constant deceleration (negative acceleration). Curved line: The object is moving with changing acceleration. The gradient at any specific point gives the instantaneous acceleration.
Gradient increasing = acceleration is increasing.
Gradient decreasing = acceleration is decreasing. Area under the graph: The area between the graph line and the time axis represents the distance travelled.
For constant speed (rectangle): Area = speed \times time.
For constant acceleration (triangle/trapezium): Use geometric area formulas.
Question: A car accelerates uniformly from rest to 20 m/s over 10 seconds, then travels at this constant speed for 5 seconds. Calculate the total distance travelled.
Solution:
- Section 1 (Acceleration): The graph is a triangle with base 10 s and height 20 m/s.
\text{Distance}_1 = \frac{1}{2} \times 10 \times 20 = 100 \text{ m} - Section 2 (Constant Speed): The graph is a rectangle with width 5 s and height 20 m/s.
\text{Distance}_2 = 5 \times 20 = 100 \text{ m} - Total Distance: 100 + 100 = 200 \text{ m}.
Error: Thinking the area under a distance-time graph represents speed, or that the gradient of a speed-time graph represents distance.
Correct Understanding:
- Gradient of distance-time = Speed. Area under speed-time = Distance. Gradient of speed-time = Acceleration.
Free Fall (No Air Resistance):
An object falling in a vacuum accelerates downwards at a constant rate of g (9.8\ \text{m/s}^2). Its speed increases linearly with time.Falling with Air/Liquid Resistance:
When an object falls through a fluid (air or liquid), it experiences a resistive force (drag) that increases with speed. Initial Drop: Speed is low, so drag is small. Weight > Drag. Net force is downwards. The object accelerates downwards. Increasing Speed: As speed increases, drag increases. The net downward force decreases, so acceleration decreases. Terminal Velocity: Eventually, the upward drag force equals the downward weight. Net force becomes zero. Acceleration becomes zero. The object continues to fall at a constant maximum speed called terminal velocity.
On a speed-time graph for an object reaching terminal velocity:
- Starts with a steep gradient (high acceleration).
- Gradient gradually decreases (curve flattens) as acceleration reduces.
- Becomes a horizontal line when terminal velocity is reached.
Correct Usage Example:
- Incorrect: 'The car speeds up.'
- Correct: 'The car moves with constant acceleration because the speed-time graph is a straight line with a positive gradient.'
- Incorrect: 'It stops accelerating.'
- Correct: 'The acceleration becomes zero and the object moves at terminal velocity (constant speed) because the gradient of the speed-time graph becomes zero.'
Why Examiners Accept This: Examiners look for precise terminology. Saying 'the speed increases' is better than 'it goes faster'. Saying 'constant acceleration' is required instead of just 'accelerating' if the line is straight.
Correct Usage Example: Draw a large right-angled triangle on the graph line. Ensure the vertices lie exactly on grid intersections if possible.
State: \text{Gradient} = \frac{\Delta y}{\Delta x}.
Substitute values from your triangle, not just two points on the line that are close together.
Include units in your final answer (e.g., m/s or m/s²).
Why Examiners Accept This: Marks are awarded for the method (using a large triangle) and the calculation. A small triangle leads to larger reading errors.
Notes built from years of past papers
- The exact keywords examiners look for
- The most common questions, year after year
- The common mistakes that cost marks
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