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 $(t_1, s_1)$ and $(t_2, s_2)$ on the line.
   • Compute $\Delta s = s_2 - s_1$ and $\Delta t = t_2 - t_1$.
   • 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: $(t_1, s_1) = (0,\ 0)$ and $(t_2, s_2) = (8,\ 40)$.
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

Graph Type	Shape of $s$–$t$ Curve	Velocity	Interpretation
Straight line (positive slope)	Linear increase	Constant $v = \dfrac{\Delta s}{\Delta t} > 0$	Uniform motion in positive direction
Straight line (negative slope)	Linear decrease	Constant $v = \dfrac{\Delta s}{\Delta t} < 0$	Uniform motion in negative direction
Horizontal line	Zero slope	$v = 0$	Object at rest
Curved line	Changing slope	Variable $v$ (gradient of tangent)	Accelerated motion

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.