determine acceleration using the gradient of a velocity–time graph

Cambridge International AS & A Level Physics 9702 – Kinematics

Lesson Checklist (AS – Unit 2 Kinematics)

Learning Objectives

• Recall and use the SI units for displacement, velocity, acceleration and time.
• Derive the three equations of motion for constant acceleration and apply them in a range of contexts.
• Interpret displacement–time, velocity–time and acceleration–time graphs.
• Determine acceleration from the gradient of a straight‑line segment on a v‑t graph and recognise when the method is not applicable.
• Link the graphical method to Newton’s 2nd law and to the work‑energy principle.
• Identify common pitfalls and avoid them in exam questions.

1. Fundamental Quantities & Units

Reminder: In the gradient formula a = Δv / Δt use Δv in m s⁻¹ and Δt in s; the result is automatically in m s⁻².

2. Derivation of the Equations of Motion (Constant Acceleration)

Summary of the Three Core Equations

Worked Example – Using the Third Equation (No‑Time Case)

A car accelerates uniformly from 5 m s⁻¹ to 25 m s⁻¹ over a distance of 100 m. Find the acceleration.

1. Apply equation (3): \(\displaystyle a = \frac{v^{2}-u^{2}}{2s}\).
2. Insert the numbers: \[ a = \frac{(25)^{2}-(5)^{2}}{2\times100} = \frac{625-25}{200} = \frac{600}{200}=3.0\ \text{m s}^{-2} \]
3. Result: \(a = 3.0\ \text{m s}^{-2}\) (positive, i.e. in the direction of motion).

3. Graphical Interpretation of Motion

3.1 Displacement–Time (s‑t) Graphs

• Gradient = velocity (m s⁻¹).
• Curvature indicates changing velocity (non‑uniform acceleration).
• Area under the curve has no direct physical meaning.

3.2 Velocity–Time (v‑t) Graphs

• Gradient = acceleration (m s⁻²).
• Area under the curve = displacement (since \(\displaystyle s = \int v\,dt\)).
• A straight‑line segment → constant acceleration; a horizontal line → zero acceleration.

3.3 Acceleration–Time (a‑t) Graphs

• Gradient = rate of change of acceleration (jerk), rarely needed at AS level.
• Area under the curve = change in velocity (\(\displaystyle \Delta v = \int a\,dt\)).

3.4 Non‑Uniform Acceleration

If the v‑t graph is curved, the acceleration varies. Two practical ways to obtain the instantaneous acceleration at a specific time:

1. Draw a tangent at the point of interest; the gradient of the tangent is the instantaneous acceleration.
2. Take a very small interval \(\Delta t\) and use \(\displaystyle a \approx \frac{\Delta v}{\Delta t}\) (the smaller the interval, the better the approximation).

4. Determining Acceleration from a v‑t Graph (Gradient Method)

Step‑by‑Step Procedure

1. Confirm constant acceleration – the segment must be a straight line.
2. Read two precise points on that line: \((t_{1},v_{1})\) and \((t_{2},v_{2})\).
3. Calculate the gradient: \[ a = \frac{\Delta v}{\Delta t} = \frac{v_{2}-v_{1}}{t_{2}-t_{1}} \] (ensure \(\Delta v\) in m s⁻¹ and \(\Delta t\) in s).
4. State the direction:
   • Positive gradient → acceleration in the chosen positive direction.
   • Negative gradient → acceleration opposite to the chosen positive direction.
5. Use the obtained \(a\) in subsequent calculations (e.g., equations of motion, \(F=ma\), kinetic‑energy change).

Worked Example 1 – Uniform Acceleration

A car starts from rest and reaches 12 m s⁻¹ after 4 s.

1. Points: \((0\ \text{s},0\ \text{m s}^{-1})\) and \((4\ \text{s},12\ \text{m s}^{-1})\).
2. Gradient: \[ a = \frac{12-0}{4-0}=3\ \text{m s}^{-2} \]
3. Positive slope → acceleration is +3 m s⁻².

Worked Example 2 – Deceleration (Negative Gradient)

A cyclist slows uniformly from 15 m s⁻¹ to 5 m s⁻¹ in 2 s.

1. Points: \((0\ \text{s},15\ \text{m s}^{-1})\) and \((2\ \text{s},5\ \text{m s}^{-1})\).
2. Gradient: \[ a = \frac{5-15}{2-0}= -5\ \text{m s}^{-2} \]
3. Magnitude = 5 m s⁻²; direction opposite to the chosen positive direction.

Worked Example 3 – Instantaneous Acceleration on a Curved v‑t Graph

A falling object follows a parabolic v‑t curve. Estimate the acceleration at \(t = 3\ \text{s}\).

1. Draw a tangent at \((3\ \text{s},v)\). Choose two points on the tangent, e.g. \((2.5\ \text{s},24\ \text{m s}^{-1})\) and \((3.5\ \text{s},32\ \text{m s}^{-1})\).
2. Gradient: \[ a \approx \frac{32-24}{3.5-2.5}=8\ \text{m s}^{-2} \]
3. The value is close to the theoretical \(g = 9.8\ \text{m s}^{-2}\), showing the method’s usefulness for instantaneous acceleration.

5. Link‑in Box – From Graphs to Dynamics

6. Extension – Using Acceleration to Find the Change in Kinetic Energy

From the work‑energy principle:

Example: A 1500 kg car accelerates at \(3\ \text{m s}^{-2}\) for a distance of 20 m.

7. Common Mistakes to Avoid

• Using a curved portion of a v‑t graph as if the acceleration were constant.
• Mixing units – always convert velocities to m s⁻¹ and times to s before applying the gradient formula.
• Reading gradients from a poorly scaled graph; check axis divisions and, if needed, use a ruler or calculate the gradient algebraically.
• Ignoring direction – a negative gradient is a real acceleration opposite to the chosen positive direction.
• Confusing the area under a v‑t graph (displacement) with its gradient (acceleration).

8. Summary

• Acceleration equals the gradient of a straight‑line segment on a velocity–time graph: \(a = \Delta v / \Delta t\).
• Consistent SI units give the result in m s⁻².
• The three equations of motion are derived directly from the same relationship and can be used interchangeably with the graphical method.
• Linking the gradient to Newton’s 2nd law allows a seamless transition from kinematics to dynamics and to kinetic‑energy calculations.
• Always verify that the graph represents constant acceleration before applying the simple gradient formula.

9. Practice Questions (Answers Provided Separately)

1. A cyclist speeds up uniformly from \(5\ \text{m s}^{-1}\) to \(15\ \text{m s}^{-1}\) in \(5\ \text{s}\). Determine the acceleration using the gradient method.
2. On a v‑t graph a line falls from \(20\ \text{m s}^{-1}\) at \(t = 2\ \text{s}\) to \(0\ \text{m s}^{-1}\) at \(t = 6\ \text{s}\). State the magnitude and direction of the acceleration.
3. Explain why a horizontal line on a v‑t graph corresponds to zero acceleration.
4. A 1200 kg car accelerates from rest to 25 m s⁻¹ in 8 s.
   1. Find the acceleration from the v‑t graph.
   2. Calculate the net force acting on the car.
   3. Determine the increase in kinetic energy.
5. A ball is thrown vertically upwards with an initial speed of \(15\ \text{m s}^{-1}\). Its v‑t graph is a straight line with a gradient of \(-9.8\ \text{m s}^{-2}\) until the velocity reaches zero. Using the graph, find:
   1. the time taken to reach the highest point, and
   2. the maximum height reached.
6. (A‑Level extension) A satellite in circular orbit moves at a constant speed of \(7.6\times10^{3}\ \text{m s}^{-1}\). Sketch the v‑t and a‑t graphs for one complete orbit and comment on the gradients.

10. Further Reading & A‑Level Links

• Newton’s laws – how a constant net force produces a constant‑gradient v‑t graph.
• Uniform circular motion – speed is constant (horizontal v‑t line) but acceleration is non‑zero (centripetal, shown on an a‑t graph).
• Free fall – graphs illustrate the use of \(g\) as the gradient of the v‑t curve.
• Electric fields – motion of charged particles in uniform fields can be analysed with the same kinematic techniques.