The slope of a position-time graph is not just a number — it is the velocity of the moving object. The triangle method from NCERT Fig. 4.14 makes this extraction precise, repeatable, and geometrically meaningful. From NCERT Chapter 4 (Exploration edition) Class 9 Science. Aligned with CBSE syllabus 2026-27.
Aim: To calculate the velocity of an object by finding the slope of its position-time graph using the triangle method described in NCERT Fig. 4.14.
This activity uses the graph plotted in Activity 4.3 (C18) — specifically the straight-line graph from Table 4.3 data. You need that graph already drawn on graph paper before beginning this activity.
The triangle method is the standard technique NCERT uses to extract velocity from a p-t graph. Follow these four steps:
Using the Table 4.3 graph (straight line), choose two points from the data:
Point A: $(t_1, s_1) = (2\,\text{s},\; 40\,\text{m})$
Point B: $(t_2, s_2) = (4\,\text{s},\; 80\,\text{m})$
Form the triangle: draw a horizontal line from A to directly below B, meeting at C = $(4\,\text{s},\; 40\,\text{m})$.
BC (vertical side) = $s_2 - s_1 = 80 - 40 = 40\,\text{m}$
CA (horizontal side) = $t_2 - t_1 = 4 - 2 = 2\,\text{s}$
$$v = \frac{BC}{CA} = \frac{40\,\text{m}}{2\,\text{s}} = \mathbf{20\,\text{m s}^{-1}}$$Does it matter which two points you choose? No — for a straight-line graph, the slope is the same everywhere. You can verify this by picking a different pair, say A = (0, 0) and B = (6, 120):
$$v = \frac{120 - 0}{6 - 0} = \frac{120}{6} = 20\,\text{m s}^{-1} \checkmark$$The result is always 20 m s⁻¹, regardless of which two points on the straight line you use.
Q. What exactly is "slope" in geometry, and why does it equal velocity here?
In geometry, the slope (or gradient) of a straight line is the ratio of the vertical rise to the horizontal run between any two points on that line:
$$\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x}$$On a position-time graph, the y-axis is position ($s$) and the x-axis is time ($t$). So:
$$\text{slope} = \frac{\Delta s}{\Delta t} = \frac{s_2 - s_1}{t_2 - t_1}$$But from the very definition of average velocity: $v = \Delta s / \Delta t$. The two expressions are identical. Therefore, the slope of a position-time graph is always equal to the velocity — this is not a coincidence or a formula to memorise, it follows directly from the definition of velocity.
For a curved (non-straight) p-t graph, velocity at any instant equals the slope of the tangent drawn to the curve at that point. Drawing and measuring a tangent is the technique shown in the p-t graph notes (C04) for Activity 4.4's extension to accelerated motion.
| Slope of p-t graph | Velocity | Type of motion |
|---|---|---|
| Zero (horizontal line) | v = 0 | Object at rest |
| Positive, small | Low positive velocity | Slow uniform motion (forward) |
| Positive, steep | High positive velocity | Fast uniform motion (forward) |
| Negative | Negative velocity | Moving backward (returning toward origin) |
| Increasing (curve bending upward) | Increasing velocity | Accelerating (non-uniform motion) |
Comparing two objects (NCERT Example 4.7): If objects A and B both have straight-line p-t graphs through the origin, but A's line is steeper, then A has a higher slope and therefore a higher velocity. This is a direct visual comparison — no calculation needed, just compare the steepness.
Q1. A p-t graph has points (0, 0) and (10 s, 50 m) on a straight line. Find the velocity using the triangle method.
Q2. Two students plot p-t graphs for the same vehicle but choose different pairs of points for their triangles. Will they get the same velocity? Why?
Q3. A p-t graph shows a straight line starting at s = 10 m (at t = 0) and reaching s = 70 m at t = 6 s. (a) What is the velocity? (b) Is the object at the origin at t = 0?