determine displacement from the area under a velocity–time graph

Published by Patrick Mutisya · 8 days ago

Cambridge A-Level Physics 9702 – Equations of Motion: Displacement from \cdot elocity–Time Graph

Equations of Motion – Displacement from a \cdot elocity–Time Graph

Learning Objective

By the end of this lesson you should be able to determine the displacement of an object by calculating the area under its velocity–time graph.

Key Concepts

  • Velocity is the rate of change of displacement: \$v = \frac{dx}{dt}\$.

  • The area under a velocity–time graph between two times \$t1\$ and \$t2\$ gives the displacement \$Δx\$:

    \$Δx = \int{t1}^{t_2} v(t)\,dt\$

  • For constant velocity the area is a rectangle; for uniformly accelerated motion the area is a trapezium.

  • Sign of the area indicates direction of displacement (positive for motion in the chosen positive direction, negative otherwise).

Common \cdot elocity–Time Graph Shapes

ShapeMotion DescriptionArea Formula (Displacement)
Horizontal lineConstant velocity\$Δx = v \, (t2 - t1)\$
Straight line through the originUniform acceleration from rest\$Δx = \frac{1}{2} a (t2^2 - t1^2)\$
Straight line not through the originUniform acceleration with initial velocity \$v0\$\$Δx = v0 (t2 - t1) + \frac{1}{2} a (t2^2 - t1^2)\$
Triangular region above the axisAcceleration then deceleration returning to rest\$Δx = \frac{1}{2} (v{\text{max}}) (t{\text{up}} + t_{\text{down}})\$
Trapezium (mixed positive and negative velocities)Motion changes direction\$Δx = \frac{1}{2}(v1+v2)(t2-t1)\$ (taking sign into account)

Step‑by‑Step Procedure

  1. Identify the time interval \$[t1, t2]\$ over which you need the displacement.

  2. Sketch or examine the velocity–time graph for that interval.

  3. Break the area into simple geometric shapes (rectangles, triangles, trapezia).

  4. Calculate the area of each shape using the appropriate geometric formula.

  5. Assign a positive sign to areas above the time axis and a negative sign to areas below.

  6. Sum all signed areas to obtain the total displacement \$Δx\$.

Worked Example

A car starts from rest, accelerates uniformly to \$20\ \text{m s}^{-1}\$ in \$5\ \text{s}\$, then moves at this constant speed for a further \$10\ \text{s}\$. Determine the displacement during the whole \$15\ \text{s}\$ interval.

  1. Divide the graph into two parts: (a) a triangle for \$0\le t\le5\ \text{s}\$, (b) a rectangle for \$5\le t\le15\ \text{s}\$.

  2. Area of the triangle: \$A_{\text{tri}} = \frac{1}{2}\times \text{base}\times\text{height}= \frac{1}{2}\times5\ \text{s}\times20\ \text{m s}^{-1}=50\ \text{m}\$.

  3. Area of the rectangle: \$A_{\text{rect}} = \text{base}\times\text{height}=10\ \text{s}\times20\ \text{m s}^{-1}=200\ \text{m}\$.

  4. Total displacement: \$Δx = A{\text{tri}} + A{\text{rect}} = 50\ \text{m} + 200\ \text{m}=250\ \text{m}\$.

Practice Questions

  1. Given a velocity–time graph that is a straight line from \$v=5\ \text{m s}^{-1}\$ at \$t=0\$ to \$v=15\ \text{m s}^{-1}\$ at \$t=4\ \text{s}\$, find the displacement between \$t=0\$ and \$t=4\ \text{s}\$.

  2. A particle moves with velocity \$v(t)=4t-6\ \text{m s}^{-1}\$ for \$0\le t\le5\ \text{s}\$. Sketch the graph and calculate the displacement, indicating any change of direction.

  3. For the graph shown below (horizontal line at \$+8\ \text{m s}^{-1}\$ from \$t=0\$ to \$t=3\ \text{s}\$, then a straight line descending to \$-4\ \text{m s}^{-1}\$ at \$t=6\ \text{s}\$), determine the net displacement from \$t=0\$ to \$t=6\ \text{s}\$.

Answers to Practice Questions

  1. Area is a trapezium: \$Δx = \frac{1}{2}(5+15)\times4 = 40\ \text{m}\$.

  2. Integrate: \$Δx = \int{0}^{5}(4t-6)\,dt = [2t^{2}-6t]{0}^{5}= (2\cdot25-30)-0 =20\ \text{m}\$. The velocity becomes zero at \$t=1.5\ \text{s}\$, so the particle reverses direction after that point.

  3. Rectangle area: \$8\ \text{m s}^{-1}\times3\ \text{s}=24\ \text{m}\$. Triangle area (positive to zero): \$\frac{1}{2}\times3\ \text{s}\times8\ \text{m s}^{-1}=12\ \text{m}\$. Triangle area (negative): \$\frac{1}{2}\times3\ \text{s}\times4\ \text{m s}^{-1}=6\ \text{m}\$ (taken as \$-6\ \text{m}\$). Net displacement \$=24+12-6=30\ \text{m}\$.

Common Mistakes to Avoid

  • Forgetting to assign a negative sign to areas below the time axis.

  • Mixing up the units; velocity must be in \$\text{m s}^{-1}\$ and time in \$\text{s}\$ so that area yields metres.

  • Treating a curved velocity–time graph as a straight line; if the curve is not linear, use integration or approximate with smaller geometric shapes.

Suggested diagram: Velocity–time graph for the worked example showing a triangle followed by a rectangle.

Summary

The displacement of an object over a time interval can be obtained directly from the area under its velocity–time graph. By breaking the graph into simple geometric shapes, applying the appropriate area formulas, and respecting the sign convention, you can solve a wide range of A‑Level questions without resorting to algebraic kinematic equations.