determine velocity using the gradient of a displacement–time graph

Equations of Motion

Objective

Determine the instantaneous velocity of an object by analysing the gradient (slope) of its displacement–time graph.

Key Concepts

• Displacement (\$s\$) is the position of an object relative to a chosen origin.
• Time (\$t\$) is measured uniformly and increases in the forward direction.
• The gradient of a straight line on a graph is given by the ratio of the vertical change to the horizontal change.
• For a displacement–time graph, the gradient represents velocity.

Mathematical Relationship

The average velocity over a time interval \$\Delta t\$ is

\$v_{\text{avg}} = \frac{\Delta s}{\Delta t}\$

In the limit as \$\Delta t \to 0\$, the average velocity becomes the instantaneous velocity:

\$v = \frac{ds}{dt}\$

Procedure to Find \cdot elocity from a Displacement–Time Graph

1. Identify the portion of the graph for which the velocity is required.
2. If the graph is a straight line, the velocity is constant. Calculate the gradient directly:

   • Choose two points \$(t1, s1)\$ and \$(t2, s2)\$ on the line.
   • Compute \$\Delta s = s2 - s1\$ and \$\Delta t = t2 - t1\$.
   • Use \$v = \dfrac{\Delta s}{\Delta t}\$.

3. If the graph is curved, the velocity at a particular instant is the gradient of the tangent to the curve at that point.

   • Draw a tangent line at the point of interest.
   • Measure the rise and run of the tangent (use a ruler or grid).
   • Calculate the gradient as above.

4. Record the sign of the gradient:

   • Positive gradient → motion in the positive direction.
   • Negative gradient → motion in the opposite direction.

5. Check units: displacement in metres (m), time in seconds (s), so velocity in metres per second (m s⁻¹).

Worked Example

A car moves along a straight road. The displacement–time graph for the interval \$0 \le t \le 8\ \text{s}\$ is a straight line passing through the points \$(0\ \text{s},\ 0\ \text{m})\$ and \$(8\ \text{s},\ 40\ \text{m})\$.

Find the velocity of the car.

1. Identify two points: \$(t1, s1) = (0,\ 0)\$ and \$(t2, s2) = (8,\ 40)\$.
2. Calculate \$\Delta s = 40\ \text{m} - 0\ \text{m} = 40\ \text{m}\$.
3. Calculate \$\Delta t = 8\ \text{s} - 0\ \text{s} = 8\ \text{s}\$.
4. Apply the gradient formula:

   \$v = \frac{\Delta s}{\Delta t} = \frac{40\ \text{m}}{8\ \text{s}} = 5\ \text{m s}^{-1}\$

5. The positive sign indicates motion in the positive direction.

Common Pitfalls

• Mixing up displacement with distance travelled – displacement can be negative, distance cannot.
• Reading the gradient from a curved segment without drawing a proper tangent.
• Forgetting to keep consistent units when calculating \$\Delta s\$ and \$\Delta t\$.
• Interpreting the area under a velocity–time graph as displacement, not the gradient of a displacement–time graph.

Summary Table

Practice Questions

1. From a displacement–time graph, a particle moves from \$s = 2\ \text{m}\$ at \$t = 1\ \text{s}\$ to \$s = 14\ \text{m}\$ at \$t = 5\ \text{s}\$. Calculate its average velocity.
2. A graph shows a curved \$s\$–\$t\$ relationship. At \$t = 3\ \text{s}\$ the tangent to the curve has a rise of \$6\ \text{m}\$ for a run of \$2\ \text{s}\$. Determine the instantaneous velocity at that instant.
3. Explain why the gradient of a horizontal segment of a displacement–time graph is zero, and what physical situation this represents.