use graphical methods to represent distance, displacement, speed, velocity and acceleration

3. Graph Types – What Slopes and Areas Represent

4. Uniform (Constant) Acceleration – Deriving the SUVAT Equations from Graphs

When a is constant the v–t graph is a straight line. The area under the line between two times t₁ and t₂ is a trapezium whose area equals the displacement:

Using the definition of slope (a = Δv/Δt) together with the area relation gives the three commonly‑used SUVAT equations:

where u is the initial velocity, v the final velocity, s the displacement (or distance if the motion never reverses) and t the elapsed time.

5. Variable (Non‑Uniform) Acceleration – Using Calculus on Graphs

• Instantaneous acceleration at any instant is the gradient of the tangent to the v–t curve.
• The displacement is still the area under the v–t curve, but it must be evaluated by:
  1. Dividing the curve into simple shapes (triangles, rectangles, trapezia) and adding the areas, or
  2. Performing the integration Δx = ∫ v(t)\,dt. A‑Level students are required to carry out the integration algebraically.

Worked integration example

Given v = 2t² (m s⁻¹) for t ≥ 0:

The corresponding a–t graph is a straight line with gradient da/dt = 4t (jerk), showing that the acceleration is increasing linearly with time.

6. Adding Coplanar Vectors Graphically (Link to the Vector Part of the Syllabus)

• Place the tail of the second vector at the head of the first (head‑to‑tail method).
• The resultant vector is drawn from the tail of the first to the head of the second.
• For perpendicular components (e.g. horizontal velocity vₓ and vertical velocity v_y) the resultant speed is the length of the diagonal of the right‑angled triangle: v = √(vₓ² + v_y²).
• This technique is used when analysing projectile motion, circular motion (tangential vs radial components) and when resolving forces in dynamics.

7. Free‑Fall and Projectile Motion – A Graphical View

• Free fall (no air resistance):
  • a = –g = –9.8 m s⁻² (downward taken as negative).
  • v–t graph: straight line through the origin with slope = –g.
  • y–t graph: downward‑opening parabola y = u t – ½ g t².
• Projectile motion:
  • Horizontal motion: aₓ = 0 → vₓ–t graph is a horizontal line; x–t graph is a straight line.
  • Vertical motion: identical to free fall (see above).
  • The resultant velocity at any instant is obtained by vector addition of the horizontal and vertical components (see Section 6).

8. Connecting Kinematics to Dynamics (Newton’s 2nd Law)

Newton’s second law, ∑F = m a, links the shape of a graph to the underlying forces:

• A horizontal a–t line → constant net force (e.g. a car with a constant engine thrust).
• A sloping a–t line → net force varies with time (e.g. air resistance ∝ v).
• When the v–t graph is curved, the changing slope tells you the acceleration is not constant, so the net force is changing.
• In circular motion the centripetal acceleration a_c = v²/r appears as a constant positive value on an a–t graph if the speed is uniform, even though the direction of the acceleration vector is continually changing.

9. Practical Skills: Reading Graphs, Estimating Uncertainties and Propagation (AO3)

1. Measuring gradients – Use a ruler or digital cursor. The uncertainty Δ(gradient) can be estimated by drawing the steepest and shallowest plausible tangents through the data points.
2. Estimating areas – Count full squares on graph paper; add half‑square contributions. Uncertainty is roughly ±½ square (or ±½ grid‑unit × time‑scale × value‑scale).
3. Scale errors – Record the scale of each axis to 2 s.f. A 1 % error in the time scale produces a 1 % error in any quantity obtained from a slope involving time.
4. Propagation of uncertainties – When a quantity is calculated from measured values, combine relative uncertainties:
   ΔQ/Q = √[(ΔA/A)² + (ΔB/B)² + …] for multiplicative relationships, and add absolute uncertainties for additive relationships.
5. Fitting non‑linear data – Use least‑squares fitting (or a smooth curve drawn by eye) and quote the standard error of the fit as the uncertainty in the derived gradient or area.
6. Assessing the graph – Check for:
   • Systematic curvature that indicates a non‑constant acceleration.
   • Outlying points that may be experimental errors.
   • Consistency between different graphs (e.g. the area under a v–t graph should match the displacement read from the x–t graph).

10. Link‑Forward to Later A‑Level Topics

The graphical techniques mastered here are reused throughout the A‑Level syllabus:

• Circular motion – angular‑velocity vs time and angular‑acceleration vs time graphs; use α = dω/dt and ω² = ω₀² + 2αθ analogues of SUVAT.
• Gravitation – g vs altitude graphs; interpretation of the inverse‑square law.
• Energy – kinetic‑energy vs speed (parabolic) and potential‑energy vs height (linear) graphs.
• Waves – displacement‑time graphs for transverse and longitudinal waves; phase and group velocity from slopes.
• Electricity – voltage vs time, charge vs time, and current vs time graphs; area under a current–time graph gives charge (Q = ∫I dt).

Thus, competence with the graphs in this note is a prerequisite for the entire A‑Level physics programme.

11. Summary Table

12. Worked Examples

Example 1 – Constant Acceleration (Car)

A car starts from rest, accelerates uniformly to 20 m s⁻¹ in 5 s, then travels at that speed for a further 5 s.

1. Gradient of the accelerating part: a = Δv/Δt = 20/5 = 4 m s⁻².
2. Area of the triangular region (0–5 s): Δx₁ = ½ × 5 s × 20 m s⁻¹ = 50 m.
3. Area of the rectangular region (5–10 s): Δx₂ = 20 m s⁻¹ × 5 s = 100 m.
4. Total displacement: Δx = 150 m.

Example 2 – Variable Acceleration (Integration)

For a particle moving along a straight line, v = 3t² – 12t + 9 (m s⁻¹).

• Acceleration: a = dv/dt = 6t – 12 m s⁻² (gradient of the v–t curve).
• Displacement between t = 1 s and t = 4 s: \[ \Delta x = \int_{1}^{4} (3t^{2}-12t+9)\,dt = \Big[t^{3}-6t^{2}+9t\Big]_{1}^{4} = (64-96+36)-(1-6+9)=4\ \text{m}. \]

Example 3 – Free‑Fall (Area Method)

A ball is dropped from rest. Find the distance fallen after 3 s using the area under the v–t graph.

1. Acceleration = –g = –9.8 m s⁻², so the v–t graph is a straight line through the origin with slope –g.
2. Velocity after 3 s: v = –g t = –29.4 m s⁻¹.
3. Area (triangle) = ½ × 3 s × (–29.4 m s⁻¹) = –44.1 m. The negative sign indicates downward displacement.

Example 4 – Vector Addition in Projectile Motion

A projectile is launched with vₓ = 15 m s⁻¹ horizontally and v_y = 20 m s⁻¹ upwards at t = 0. Find the speed after 2 s.

• Horizontal component remains constant: vₓ = 15 m s⁻¹.
• Vertical component after 2 s: v_y = 20 – g·2 = 20 – 19.6 = 0.4 m s⁻¹.
• Resultant speed: v = √(15² + 0.4²) ≈ 15.0 m s⁻¹ (graphically, the diagonal of a right‑angled triangle).

13. Practice Questions

1. A distance–time graph is described by s = 2t² (m, s).
   • Write the expression for speed as a function of time.
   • State whether the acceleration is constant and, if not, give its time‑dependence.
2. The velocity of a particle is given by v = 4t – t³ (m s⁻¹).
   • Find the acceleration as a function of time.
   • Determine the total displacement between t = 0 s and t = 2 s using the area under the v–t curve.
   • At what time(s) is the particle momentarily at rest?
3. Sketch an a–t graph for a motion that:
   • starts from rest,
   • accelerates uniformly for 3 s,
   • then experiences a constant retarding force that brings it to rest after a further 2 s.
4. A cart of mass 0.5 kg is pulled by a constant horizontal force of 2 N on a frictionless track.
   • Draw the expected a–t graph and state its numerical value.
   • From the graph, find the velocity after 4 s and the distance travelled in that time.
5. Experimental data for a falling object give the following points on a v–t graph (t s, v m s⁻¹): (0, 0), (1, –9.5), (2, –19.0), (3, –28.8).
   • Estimate the acceleration by measuring the gradient.
   • Calculate the distance fallen after 3 s using the area method.
   • Give the percentage uncertainty if the time scale is uncertain by ±0.1 s and the velocity scale by ±0.2 m s⁻¹.