Calculate acceleration from the gradient of a speed-time graph

1.2 Motion – Calculating Acceleration from the Gradient of a Speed‑Time Graph

Learning Objectives

• Recall the basic physical quantities involved in motion (distance, speed, velocity, acceleration) and the role of measurement uncertainties.
• Define velocity as a vector and acceleration as the rate of change of velocity.
• Use the formula \[ a = \frac{\Delta v}{\Delta t} \] correctly, including sign.
• Determine the numerical value of acceleration from the gradient (slope) of a speed‑time graph.
• Identify constant‑speed, accelerating and decelerating motion from a graph.
• Calculate the distance travelled from the area under a speed‑time graph (AO2).
• Link the result to the concepts of resultant force, work and momentum (syllabus integration).

Context within the Cambridge IGCSE Physics Syllabus (0625)

Syllabus Item (Core)	Present in these notes?	Action Required
1.1 Physical quantities & measurement	✗	Brief reminder of reading values from a graph and the effect of uncertainties.
1.2 Motion – speed, velocity, acceleration, graphs, gradient, area	✓ (expanded)	Explicit definition of velocity (vector) and its relation to speed.
1.5 Forces – resultant force, equilibrium	✗	Link‑on box: “Non‑zero acceleration ⇒ resultant force (F = ma).”
1.6 Momentum (supplementary)	✗	Extension note connecting constant‑acceleration motion to change in momentum.
1.7 Energy, work, power	✗	Real‑world example converting acceleration to work/force.

Key Concepts

• Speed – scalar magnitude of motion (m s⁻¹).
• Velocity – vector quantity; in 1‑D problems the direction is fixed, so the speed‑time graph gives the magnitude of velocity.
• Acceleration – rate of change of velocity: \[ a = \frac{\Delta v}{\Delta t} \] where \(\Delta v\) is the change in velocity (m s⁻¹) and \(\Delta t\) the change in time (s).
• On a speed‑time graph the gradient (rise ÷ run) equals the numerical value of acceleration (m s⁻²).
  • Positive gradient → speeding up (positive acceleration).
  • Negative gradient → slowing down (negative acceleration, i.e. deceleration).
  • Zero gradient → constant speed (zero acceleration).
• The area under a speed‑time graph gives the distance travelled: \[ \text{distance} = \int v\,dt \] For straight‑line segments the area can be calculated using simple geometric shapes (rectangles, triangles, trapezia).

Measurement Reminder

When reading a speed‑time graph, the values of speed and time are obtained from calibrated instruments (e.g. motion sensors). Record the uncertainties (±) for each reading; they propagate into \(\Delta v\) and \(\Delta t\) and consequently into the calculated acceleration.

Understanding the Gradient

For any straight‑line portion of a speed‑time graph:

\[ \text{Gradient} = \frac{\text{vertical change }(\Delta v)}{\text{horizontal change }(\Delta t)} = \frac{\Delta v}{\Delta t}=a \]

If the line is horizontal, \(\Delta v = 0\) and the acceleration is zero.

Step‑by‑Step Procedure for Finding Acceleration

Step	Action	What to Record
1	Identify the straight‑line segment of interest.	Start and end points \((t_1, v_1)\) and \((t_2, v_2)\).
2	Read the speeds (or velocities) at the two points.	\(\Delta v = v_2 - v_1\) (include uncertainty).
3	Read the corresponding times.	\(\Delta t = t_2 - t_1\) (include uncertainty).
4	Calculate the gradient using \(a = \dfrac{\Delta v}{\Delta t}\).	Acceleration (m s⁻²) with sign.
5	Interpret the sign.	Positive = speeding up; Negative = slowing down.

Calculating Distance from the Area Under the Graph

• Rectangle (constant speed): \(s = v \times \Delta t\).
• Triangle (uniform acceleration from rest or to rest): \(s = \tfrac{1}{2} v_{\max} \times \Delta t\).
• Trapezium (uniform acceleration between two non‑zero speeds): \[ s = \frac{(v_1+v_2)}{2}\,\Delta t \]
• For complex shapes, split the area into a combination of the above and sum the results.

Worked Example – Acceleration

A car’s speed‑time graph shows a straight line from \((2\;\text{s},\;4\;\text{m s}^{-1})\) to \((6\;\text{s},\;12\;\text{m s}^{-1})\).

1. \(\Delta v = 12 - 4 = 8\;\text{m s}^{-1}\)
2. \(\Delta t = 6 - 2 = 4\;\text{s}\)
3. \(a = \dfrac{8}{4} = 2\;\text{m s}^{-2}\)
4. Positive slope → acceleration is positive (speeding up).

Worked Example – Distance from Area

A cyclist travels at a constant speed of 5 m s⁻¹ for 8 s, then accelerates uniformly to 9 m s⁻¹ over the next 4 s. Find the total distance covered during the 12 s.

1. First segment (rectangle): \(s_1 = 5 \times 8 = 40\;\text{m}\).
2. Second segment (trapezium): average speed \(\displaystyle \frac{5+9}{2}=7\;\text{m s}^{-1}\); \(s_2 = 7 \times 4 = 28\;\text{m}\).
3. Total distance \(s = s_1 + s_2 = 68\;\text{m}\).

Link‑on: From Acceleration to Force, Work and Momentum

• Resultant Force: From Newton’s second law, a non‑zero acceleration implies a resultant force \(F = ma\). This connects the graph work to the forces strand (1.5).
• Work Done: If the force acts over the distance \(s\) calculated above, the work done is \(W = F \times s\) (assuming the force is in the direction of motion). This provides a bridge to the energy & work strand (1.7).
• Momentum Change: For constant mass \(m\), the change in momentum is \(\Delta p = m\Delta v = m a \Delta t\). This ties the topic to the supplementary momentum strand (1.6).

Common Mistakes to Avoid

• Using the total experiment time instead of the specific \(\Delta t\) for the chosen segment.
• Reading the wrong axis (e.g., distance instead of speed).
• Ignoring the sign of the gradient – a downward slope means negative acceleration.
• Mixing units – speed must be in m s⁻¹, time in s; consequently acceleration is in m s⁻².
• For distance, forgetting to include the area under the whole relevant portion of the graph.
• Neglecting measurement uncertainties, which affect the final answer’s precision.

Practice Questions

1. From a speed‑time graph a cyclist’s speed increases from 3 m s⁻¹ at \(t = 5\;\text{s}\) to 11 m s⁻¹ at \(t = 9\;\text{s}\). Calculate the acceleration.
2. A ball rolls down a slope and its speed‑time graph shows a straight line from \((0\;\text{s},\;0\;\text{m s}^{-1})\) to \((4\;\text{s},\;8\;\text{m s}^{-1})\). What is the acceleration? State whether the ball is speeding up or slowing down.
3. On a speed‑time graph a car moves at a constant speed of 15 m s⁻¹ for 10 s, then its speed decreases uniformly to 5 m s⁻¹ over the next 5 s. Determine the magnitude and direction of the acceleration during the deceleration phase.
4. Using the same graph as in question 3, calculate the total distance travelled by the car during the whole 15 s.
5. (Challenge) The acceleration you found in question 3 is \(-2\;\text{m s}^{-2}\). If the car has a mass of 1200 kg, calculate the resultant force acting on it and the work done by this force over the deceleration distance.

Summary

The gradient of a straight‑line segment on a speed‑time graph gives the object's acceleration during that interval:

\[ a = \frac{\Delta v}{\Delta t}\;\;(\text{m s}^{-2}) \]

Key points to remember:

• Measure \(\Delta v\) and \(\Delta t\) for the exact segment of interest, and include uncertainties.
• Keep units consistent (m s⁻¹ for speed, s for time).
• Interpret the sign of the gradient – positive = speeding up, negative = slowing down.
• Use the area under the speed‑time graph to find distance travelled.
• Link the result to force (F = ma), work (W = F s) and momentum change (\(\Delta p = m\Delta v\)) for a fuller physics picture.