P15: Motion and Acceleration
Distance, displacement, speed, velocity and acceleration
Motion And Acceleration
π Key Definitions
Distance: How far an object has travelled overall (scalar — magnitude only). Measured in metres (m).
Displacement: How far an object is from its starting point in a straight line, including direction (vector — magnitude and direction). Measured in metres (m).
Speed: How fast an object is travelling, regardless of direction (scalar). Measured in m/s.
Velocity: Speed in a given direction (vector). Measured in m/s.
Acceleration: The rate of change of velocity. Measured in m/s².
π Key Equations
v = s / t (speed = distance / time)
a = Δv / t (acceleration = change in velocity / time)
a = acceleration (m/s²), Δv = change in velocity (m/s), t = time (s)
v² − u² = 2as (final velocity² − initial velocity² = 2 × acceleration × distance)
s = ½(u + v)t (distance = average velocity × time)
s = ut + ½at² (distance = initial velocity × time + ½ × acceleration × time²)
Where: s = distance, u = initial velocity, v = final velocity, a = acceleration, t = time
Worked Example 1: Calculating speed
A cyclist travels 600 m in 40 s. Calculate the speed.
v = s / t = 600 / 40 = 15 m/s
Worked Example 2: Calculating acceleration
A car accelerates from 5 m/s to 25 m/s in 4 seconds. Calculate the acceleration.
a = Δv / t = (25 − 5) / 4 = 20 / 4 = 5 m/s²
Worked Example 3: Using v² = u² + 2as
A car accelerates at 3 m/s² from 4 m/s over a distance of 24 m. Calculate the final velocity.
v² = u² + 2as = 4² + 2 × 3 × 24 = 16 + 144 = 160
v = √160 = 12.6 m/s
Worked Example 4: Using s = ½(u + v)t
A train has an initial velocity of 10 m/s and a final velocity of 30 m/s. It accelerates for 8 seconds. Calculate the distance travelled.
s = ½(u + v)t = ½(10 + 30) × 8 = ½ × 40 × 8 = 20 × 8 = 160 m
Worked Example 5: Deceleration
A car travelling at 20 m/s brakes to a stop in 5 seconds. Calculate the deceleration.
a = Δv / t = (0 − 20) / 5 = −20 / 5 = −4 m/s²
The negative sign indicates deceleration (slowing down).
Worked Example 6: Distance using s = ut + ½at²
An object starts from rest (u = 0) and accelerates at 6 m/s² for 3 seconds. Calculate the distance.
s = ut + ½at² = 0 × 3 + ½ × 6 × 3² = 0 + 3 × 9 = 27 m
π Distance-Time Graphs
| Feature of graph | What it shows |
|---|---|
| Gradient (slope) | Speed — steeper gradient = higher speed |
| Flat (horizontal) section | Stationary (not moving, speed = 0) |
| Straight sloping line | Constant speed |
| Curved line (getting steeper) | Acceleration |
| Curved line (getting less steep) | Deceleration |
To find speed from a distance-time graph: calculate the gradient of the line. Speed = gradient = change in distance / change in time. For a curved line, draw a tangent at the point of interest and find its gradient for the instantaneous speed.
π Velocity-Time Graphs
| Feature of graph | What it shows |
|---|---|
| Gradient (slope) | Acceleration — steeper gradient = greater acceleration |
| Flat (horizontal) section | Constant velocity |
| Positive gradient | Acceleration |
| Negative gradient | Deceleration |
| Area under the graph | Displacement (distance travelled) |
To find acceleration from a velocity-time graph: calculate the gradient. Acceleration = gradient = change in velocity / change in time. To find displacement: calculate the area under the line (area of rectangle + triangle, etc.).
π§ͺ Practical: Measuring Speed and Acceleration
Light gates can be used to measure speed and acceleration accurately.
Measuring speed
- Set up two light gates a known distance apart
- Attach a card of known length to a trolley
- As the trolley passes through the first light gate, the timer starts when the card breaks the beam and stops when the card clears it
- Speed = length of card / time the beam was broken
Measuring acceleration
- Set up two light gates connected to a data logger
- The first gate measures initial velocity (u), the second measures final velocity (v)
- Measure the distance between the two gates (s)
- Use v² − u² = 2as to calculate acceleration
Why light gates are better than stopwatches: They eliminate human reaction time errors, providing more accurate and precise measurements.
β Practice Questions
Q1: Foundation A runner travels 100 m in 12.5 s. Calculate the average speed.
Q2: Foundation A car accelerates from rest to 30 m/s in 6 s. Calculate the acceleration.
Q3: Higher A car has initial velocity 8 m/s and accelerates at 2 m/s² for 5 s. Calculate the final velocity and the distance travelled.
Q4: Higher Describe what a flat horizontal section and a negative gradient represent on a velocity-time graph.
Q5: Foundation On a distance-time graph, what does the gradient represent and how can you tell if an object is accelerating?
Q6: Higher A velocity-time graph shows a car accelerating from 0 to 20 m/s in 4 s, then travelling at constant velocity for 6 s. Calculate the total distance travelled.
β Answers
- v = s / t = 100 / 12.5 = 8 m/s
- a = Δv / t = (30 − 0) / 6 = 30 / 6 = 5 m/s²
- v = u + at = 8 + 2 × 5 = 18 m/s. s = ½(u + v)t = ½(8 + 18) × 5 = ½ × 26 × 5 = 65 m
- Flat horizontal section = constant velocity (not accelerating). Negative gradient = deceleration (slowing down).
- Gradient represents speed. If the line curves upwards (getting steeper), the object is accelerating.
- Phase 1 (acceleration): area = ½ × 4 × 20 = 40 m. Phase 2 (constant): area = 20 × 6 = 120 m. Total = 40 + 120 = 160 m
π― Exam Tips
- Distance-time graph: gradient = speed. Velocity-time graph: gradient = acceleration. Don't mix these up!
- On a velocity-time graph, the area under the line = displacement. Calculate it as rectangles and triangles
- When using SUVAT equations, write down the variables you know and the one you need, then pick the right equation
- Deceleration is just negative acceleration — use a negative value for a
- For “from rest”, u = 0. For “comes to a stop”, v = 0
- Always include units in your answer — speed in m/s, acceleration in m/s², distance in m
π¬ Required Practical
Required Practical: Investigating Acceleration (F = ma)
Aim: To investigate how the acceleration of an object depends on the force applied and on the mass of the object.
Method: 1. Set up a trolley on a runway with a pulley at the end β a string over the pulley connects a mass hanger to the trolley. 2. Use light gates and a data logger to measure acceleration: the first gate measures initial velocity, the second measures final velocity, and the time between them gives acceleration. 3. Vary the force by adding masses to the hanger (keeping total mass of the system constant by moving masses from the trolley to the hanger). 4. Repeat for different forces and plot a graph of acceleration against force. 5. Then keep the force constant and vary the mass of the trolley by adding masses to it. Plot a graph of acceleration against 1/mass.
Variables: IV: force (by changing mass on hanger) or mass of trolley, DV: acceleration (measured by light gates), Control: total mass of system (when varying force); force on hanger (when varying mass); same trolley, same runway surface.
π’ Maths Skills
Mathematical Skills
Use v = s/t, a = Ξv/t, vΒ² β uΒ² = 2as, s = Β½(u + v)t, and s = ut + Β½atΒ². On distance-time graphs, gradient = speed; on velocity-time graphs, gradient = acceleration and area under line = displacement. Calculate area as rectangles and triangles.
Maths Example
A car accelerates from 8 m/s to 20 m/s over 120 m. Find acceleration: vΒ² β uΒ² = 2as β 20Β² β 8Β² = 2 Γ a Γ 120 β 400 β 64 = 240a β 336 = 240a β a = 1.4 m/sΒ². On a velocity-time graph, a triangle of base 4 s and height 20 m/s gives area = Β½ Γ 4 Γ 20 = 40 m displacement.
β οΈ Common Misconceptions
Watch Out!
1. Wrong: A steeper line on a distance-time graph means the object is accelerating. Correct: A steeper straight line means a higher constant speed. Acceleration is shown by a CURVED line getting steeper, not by a straight line.
2. Wrong: A flat line on a velocity-time graph means the object is stationary. Correct: A flat line on a velocity-time graph means constant velocity (not accelerating). The object IS moving. A flat line at zero means stationary.
3. Wrong: Deceleration is a positive value of acceleration in the opposite direction. Correct: Deceleration means the acceleration value is negative β it is still acceleration, just in the opposite direction to the velocity.
βοΈ 6-Mark Question
Extended Answer
6 marks: Describe how to investigate the relationship between force and acceleration using a trolley, light gates and masses. Explain how the results confirm F = ma.
Set up a trolley on a sloped runway (to compensate for friction) with a pulley at the end. Attach a string over the pulley to a mass hanger. Set up two light gates connected to a data logger, a known distance apart. Attach a card of known length to the trolley so each light gate can measure velocity. Start with the trolley and a 1 N mass on the hanger. Release the trolley and record the acceleration from the data logger. Repeat with increasing force by adding masses to the hanger. To keep the total mass of the system constant, transfer masses from the trolley to the hanger each time. Plot a graph of acceleration (y-axis) against force (x-axis). The graph should be a straight line through the origin, showing that acceleration is directly proportional to force (confirming F = ma, or a = F/m). The gradient equals 1/m, confirming the inverse relationship with mass.
Mark scheme: 1 mark for correct apparatus (trolley, pulley, light gates); 1 mark for method of varying force (adding masses); 1 mark for measuring acceleration with light gates/data logger; 1 mark for keeping total mass constant (transferring masses); 1 mark for plotting graph and stating it is a straight line through origin; 1 mark for conclusion that a β F (confirming F = ma). (6 marks total)
π AO3: Analyse & Evaluate
Analysis and Evaluation
A student investigates F = ma using a trolley and light gates. Their results are shown below. The total mass of the system is 1.0 kg.
| Force (N) | Acceleration (m/sΒ²) | Expected acceleration (m/sΒ²) |
|---|---|---|
| 1.0 | 0.85 | 1.0 |
| 2.0 | 1.80 | 2.0 |
| 3.0 | 2.70 | 3.0 |
| 4.0 | 3.60 | 4.0 |
(a) Describe the pattern in the student's results compared to the expected values.
(b) Explain why all the measured accelerations are lower than expected.
(c) Suggest how the student could improve the experiment to get results closer to the expected values.
Answers: (a) The measured accelerations are consistently about 10β15% lower than expected, but they still show a linear relationship (acceleration is proportional to force). (b) Friction between the trolley wheels and the runway, and air resistance, oppose the motion. These forces reduce the resultant force on the trolley, so the actual acceleration is less than F/m predicts. (c) The student could compensate for friction by slightly tilting the runway until the trolley just moves at constant speed with no force applied. This balances the friction force, so when masses are added, the full applied force acts as the resultant force. Alternatively, they could calculate the friction force and subtract it from the applied force to find the true resultant force.
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