P16: Motion and Acceleration
Understanding distance, displacement, speed, velocity, acceleration, equations of motion, distance-time and velocity-time graphs, area under a graph, and terminal velocity.
Distance and Displacement
Distance is a scalar quantity that measures how far an object has travelled. Displacement is a vector quantity that measures the distance from the starting point in a straight line, including the direction.
Worked Example
A runner runs 400 m around a track and returns to the start. The distance travelled is 400 m. The displacement is 0 m because the runner ends at the same point they started.
Distance is always positive and depends on the route taken. Displacement can be positive or negative and only depends on the start and end positions.
Speed and Velocity
Speed = distance / time
v = d / t
v = speed (m/s), d = distance (m), t = time (s)
Speed is a scalar (magnitude only). Velocity is a vector (magnitude and direction). Two objects can have the same speed but different velocities if they travel in different directions.
| Typical Speeds | Value (m/s) |
|---|---|
| Walking | 1.5 |
| Running | 3.0 |
| Cycling | 6.0 |
| Car on main road | 13 |
| Train | 30-50 |
| Aeroplane | 250 |
The speed of sound in air is approximately 330 m/s. The speed of light in a vacuum is 3 × 10⁸ m/s.
Worked Example
A car travels 120 km in 1.5 hours. Calculate its average speed.
Speed = 120 / 1.5 = 80 km/h
Convert to m/s: 80 × 1000 / 3600 = 22.2 m/s
Acceleration
Acceleration = change in velocity / time taken
a = (v - u) / t
a = acceleration (m/s²), v = final velocity (m/s), u = initial velocity (m/s), t = time (s)
Acceleration is the rate of change of velocity. An object accelerates when it speeds up, slows down (deceleration - negative acceleration), or changes direction.
Worked Example
A car accelerates from 12 m/s to 28 m/s in 8 seconds. Calculate its acceleration.
a = (v - u) / t = (28 - 12) / 8 = 16 / 8 = 2 m/s²
Worked Example - Deceleration
A bus slows from 20 m/s to 8 m/s in 4 seconds. Calculate its acceleration.
a = (8 - 20) / 4 = -12 / 4 = -3 m/s² (a negative acceleration means deceleration)
Equations of Motion
v = u + at
v = final velocity, u = initial velocity, a = acceleration, t = time
s = (u + v) / 2 × t
s = distance or displacement, u = initial velocity, v = final velocity, t = time
v² = u² + 2as
v = final velocity, u = initial velocity, a = acceleration, s = displacement
s = ut + ½at²
s = displacement, u = initial velocity, a = acceleration, t = time
Worked Example
A cyclist travelling at 5 m/s accelerates at 0.5 m/s² for 10 seconds. Calculate the final velocity and distance travelled.
v = u + at = 5 + 0.5 × 10 = 5 + 5 = 10 m/s
s = (u + v) / 2 × t = (5 + 10) / 2 × 10 = 7.5 × 10 = 75 m
Worked Example
A ball is dropped from rest and accelerates at 9.8 m/s². Calculate its velocity after falling 20 m.
v² = u² + 2as = 0 + 2 × 9.8 × 20 = 392
v = √392 = 19.8 m/s
Always check which equation to use by identifying the quantities you know and the quantity you need to find. List u, v, a, s, t and cross out the ones you do not know to select the correct equation.
Distance-Time Graphs
In a distance-time graph, the gradient represents speed. A steeper gradient means a higher speed. A horizontal line means the object is stationary (at rest).
- A straight line with a positive gradient = constant speed
- A curved line with increasing gradient = accelerating
- A curved line with decreasing gradient = decelerating
- A horizontal line = stationary (speed = 0)
- A straight line through the origin = constant speed from rest
Graph Interpretation
On a distance-time graph, a section where the line curves upwards with increasing steepness means the object is accelerating - it covers more distance per unit time as time goes on. To find the speed at any point, draw a tangent to the curve at that point and calculate its gradient.
Speed = gradient of distance-time graph
Speed = Δdistance / Δtime
For a curved distance-time graph, the instantaneous speed at a point is found by drawing a tangent at that point and calculating the gradient of the tangent.
Velocity-Time Graphs
In a velocity-time graph, the gradient represents acceleration and the area under the graph represents displacement (distance travelled in a given direction).
- A straight line with a positive gradient = constant acceleration
- A straight line with a negative gradient = constant deceleration
- A horizontal line = constant velocity (zero acceleration)
- A line below the x-axis = velocity in the opposite direction
Acceleration = gradient of velocity-time graph
a = Δvelocity / Δtime
Displacement = area under velocity-time graph
For a rectangle: area = base × height
For a triangle: area = ½ × base × height
For a trapezium: area = ½(a + b) × h
Worked Example
An object accelerates from rest to 20 m/s in 5 s, then travels at constant velocity for 10 s, then decelerates to rest in 4 s. Calculate the total distance travelled.
Phase 1 (acceleration): area = ½ × 5 × 20 = 50 m
Phase 2 (constant velocity): area = 10 × 20 = 200 m
Phase 3 (deceleration): area = ½ × 4 × 20 = 40 m
Total distance = 50 + 200 + 40 = 290 m
If the velocity-time graph goes below the x-axis, the area below counts as negative displacement. The total displacement is the positive area minus the negative area. The total distance is the sum of all areas (ignoring signs).
Terminal Velocity
Terminal velocity is the maximum velocity reached by an object when the driving force equals the total resistive force, so the resultant force is zero and acceleration is zero.
For a falling object, three stages occur:
- Initially: the object accelerates downwards due to gravity (weight is much greater than air resistance).
- As speed increases: air resistance increases. The resultant downward force decreases, so acceleration decreases.
- Terminal velocity: air resistance equals weight. The resultant force is zero, so the object falls at constant velocity.
Objects with a larger surface area reach a lower terminal velocity because air resistance builds up more quickly. A skydiver with an open parachute has a much lower terminal velocity than in freefall.
Worked Example
A skydiver of mass 80 kg falls from a plane. At terminal velocity, explain the forces acting.
Weight = 80 × 9.8 = 784 N downwards. Air resistance = 784 N upwards. Resultant force = 0 N, so acceleration = 0 m/s² and the skydiver falls at constant speed (terminal velocity).
You may be asked to sketch a velocity-time graph for a falling object reaching terminal velocity. The graph curves and then becomes horizontal - the curve shows decreasing acceleration and the horizontal section shows constant velocity.
Motion in a Circle (Higher)
Objects moving in a circle at constant speed are accelerating because their direction is constantly changing. The acceleration is directed towards the centre of the circle and is called centripetal acceleration.
Centripetal acceleration requires a centripetal force, which always acts towards the centre of the circle. Examples include:
- Friction between car tyres and the road for a car turning a corner
- Tension in a string for a ball swung on a string
- Gravitational force for a planet orbiting the Sun
Increasing speed or decreasing the radius of the circle increases the centripetal force required. If the centripetal force is not large enough, the object moves off in a straight line (tangent to the circle).
Practice Questions
1. A train accelerates from 10 m/s to 30 m/s in 20 seconds. Calculate its acceleration and the distance it travels.
a = (30 - 10) / 20 = 1 m/s². s = (10 + 30) / 2 × 20 = 400 m.
2. Describe what a horizontal line on a distance-time graph represents and what a horizontal line on a velocity-time graph represents.
A horizontal line on a distance-time graph means the object is stationary (not moving). A horizontal line on a velocity-time graph means the object is moving at constant velocity (zero acceleration).
3. A car decelerates from 25 m/s to rest in 5 seconds. Calculate the deceleration and the distance travelled.
a = (0 - 25) / 5 = -5 m/s². s = (25 + 0) / 2 × 5 = 62.5 m.
4. On a velocity-time graph, the area under the graph between two times is 150 m. What does this value represent?
The area under a velocity-time graph represents the displacement (distance travelled in the direction of motion).
5. Explain why a feather and a bowling ball dropped from the same height in a vacuum hit the ground at the same time, but in air the bowling ball hits first.
In a vacuum there is no air resistance, so both objects accelerate at g = 9.8 m/s². In air, air resistance has a much greater effect on the feather (due to its large surface area relative to its weight), so it reaches a much lower terminal velocity and falls more slowly.
Maths Skills
Equations of Motion
The four equations of motion (suvat) are used when acceleration is constant:
- v = u + at
- s = ½(u + v)t
- v² = u² + 2as
- s = ut + ½at²
Identify the three known quantities and the one unknown, then select the equation that contains all four.
Worked Example
A car accelerates from 8 m/s with constant acceleration of 1.5 m/s² for 12 seconds. Calculate the distance travelled.
s = ut + ½at² = (8 × 12) + ½ × 1.5 × 12² = 96 + 108 = 204 m
Interpreting Distance-Time and Velocity-Time Graphs
On a distance-time graph, the gradient = speed. On a velocity-time graph, the gradient = acceleration and the area under the graph = displacement. For a curved distance-time graph, draw a tangent at a point to find the instantaneous speed (gradient of tangent). For a curved velocity-time graph, estimate the area by counting squares or approximating with triangles and rectangles.
Tangent Method for Instantaneous Speed
When a distance-time graph is curved, the instantaneous speed at a specific point is found by: (1) drawing a tangent to the curve at that point, (2) extending the tangent to form a right-angled triangle, (3) calculating the gradient = Δy / Δx. A steeper tangent means a higher instantaneous speed.
Common Misconceptions
Distance vs Displacement
Distance and displacement are the same thing. Distance is a scalar — it measures the total path length travelled, regardless of direction. Displacement is a vector — it measures the straight-line distance from start to finish, including direction. A runner completing one lap of a 400 m track has a distance of 400 m but a displacement of 0 m (back at the start).
Constant Speed and Acceleration
If an object travels at constant speed, its acceleration is zero. Constant speed in a straight line means zero acceleration. However, an object moving in a circle at constant speed is accelerating because its direction is constantly changing. Acceleration is any change in velocity — and velocity includes direction. Changing direction at constant speed is still acceleration.
6-Mark Extended Question
Describe how the motion of an object falling through a fluid changes from the moment it is released until it reaches terminal velocity. Explain the forces involved. [6 marks]
When the object is first released, the only force acting is its weight (gravity) downwards, so it accelerates downwards (1). As it accelerates, its speed increases and so the drag (air resistance or fluid resistance) acting upwards increases (1). The resultant downward force is weight minus drag, so it decreases, meaning the acceleration decreases but the object still speeds up (1). Eventually, the drag force increases until it equals the weight of the object (1). At this point, the resultant force is zero and the acceleration becomes zero (1). The object now falls at a constant maximum speed called terminal velocity, where weight and drag are balanced and there is no further change in speed (1).
AO3: Analyse and Evaluate
A velocity-time graph shows the following motion of a cyclist:
| Time (s) | Velocity (m/s) |
|---|---|
| 0 | 0 |
| 5 | 10 |
| 15 | 10 |
| 19 | 0 |
Between 0–5 s the line is straight and increasing. Between 5–15 s the line is horizontal. Between 15–19 s the line is straight and decreasing.
(a) Calculate the acceleration during the first 5 seconds. (b) Calculate the total distance travelled. (c) Compare the magnitude of the acceleration and deceleration and explain which phase involves a larger force on the cyclist.
Evaluation
(a) a = Δv/t = (10 − 0)/5 = 2.0 m/s².
(b) Phase 1 area (triangle) = ½ × 5 × 10 = 25 m. Phase 2 area (rectangle) = 10 × 10 = 100 m. Phase 3 area (triangle) = ½ × 4 × 10 = 20 m. Total = 25 + 100 + 20 = 145 m.
(c) Deceleration = (0 − 10)/4 = 2.5 m/s², which is greater than the acceleration of 2.0 m/s². Since F = ma and the mass is constant, a larger deceleration requires a larger braking force. This means the braking force during deceleration is larger than the driving force during acceleration.
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