Velocity-time graphs

A distance-time graph tells you how fast something is going. A velocity-time graph goes one step further: its slope tells you how the speed itself is changing (acceleration), and the area trapped under it tells you the distance covered — no calculus required. This single idea is the graphical engine behind all three equations of motion taught in CBSE Class 9: $v = u+at$ , $s = ut+\tfrac12at^2$ , and $v^2 = u^2+2as$ all fall out of reading slopes and areas off a velocity-time plot. Once you can tell a horizontal line from a sloped one, and a rectangle from a triangle or a trapezium, the graph does the algebra for you.

The core insight

On a velocity-time graph, slope equals acceleration and the area under the curve between two times equals the distance (or displacement) covered in that interval.

The core idea

A velocity-time graph plots velocity on the vertical axis against time on the horizontal axis. Two rules link the picture to the physics.

\text{slope} = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} = a

where: $u$ is velocity at the start of the interval, $v$ is velocity at the end, $t$ is the time elapsed, and $a$ is the acceleration — the same acceleration you'd compute from the equations of motion.

\text{area under the graph (between two times)} = \text{distance (or displacement)}

Distance is, roughly, velocity multiplied by time — and multiplying a y-axis quantity by an x-axis quantity is exactly what "finding an area" means on a graph. Four shapes of the graph, four physical situations:

Horizontal line (zero slope) — uniform velocity, zero acceleration. The object's speed never changes.
Straight line, positive slope — uniform acceleration. Velocity increases steadily; equal time intervals add equal velocity increments.
Straight line, negative slope — uniform retardation (negative acceleration). Velocity falls steadily; this is the graph of a body braking at a constant rate.
Curved line — non-uniform acceleration. The rate of change of velocity is itself changing, so no single slope value describes the whole motion.

The area is always broken into shapes you already know how to measure: a rectangle when velocity is constant, and a triangle or trapezium when velocity changes at a steady rate. For a composite motion — say, constant velocity followed by uniform deceleration — you split the region into a rectangle plus a triangle and add the two areas; there is no single formula that covers a two-stage journey in one step.

Figure

A velocity-time graph showing a horizontal segment for constant velocity followed by a downward sloping segment for uniform deceleration, with the area under each segment shaded to represent distance covered

Derivation

Derivation ADistance covered under uniform acceleration, from area under the graph

The goal is to derive $s = ut + \tfrac12at^2$ purely from the geometry of a velocity-time graph — the trapezium rule NCERT uses to build all three equations of motion without calculus — starting from a body that has initial velocity $u$ and moves with constant acceleration $a$ for time $t$ , reaching final velocity $v$ .

Step 1. Draw the graph. Since acceleration is constant, the velocity-time graph is a straight line rising from $u$ at $t=0$ to $v$ at time $t$ . The region under this line, down to the time axis, is a trapezium with two parallel sides of length $u$ and $v$ , and "height" (along the time axis) equal to $t$ .

Step 2. Split the trapezium into a rectangle and a triangle. The rectangle has height $u$ and width $t$ , contributing area $ut$ . The triangle sits on top, with base $t$ and height equal to the velocity gained, $(v-u)$ , contributing area $\tfrac12 t(v-u)$ .

Step 3. Use the slope relation to replace $(v-u)$ . Since slope equals acceleration, $a = \dfrac{v-u}{t}$ , so $(v-u) = at$ . Substituting into the triangle's area gives $\tfrac12 t(at) = \tfrac12 at^2$ .

Step 4. Add the two areas. Total distance is the sum of the rectangle and triangle areas, since the area under the whole graph equals total distance covered:

s = ut + \frac{1}{2}at^2

where: $u$ is initial velocity, $a$ is the (constant) acceleration, $t$ is the elapsed time, and $s$ is the distance (or displacement) covered — exactly the second equation of motion, obtained here without a single line of calculus, purely by decomposing a trapezium into shapes whose areas are known.

Worked example

Question: A body's velocity increases uniformly from $u = 5\ \text{m/s}$ at $t=0$ to $v = 25\ \text{m/s}$ at $t = 10\ \text{s}$ . Using the velocity-time graph, find (a) the acceleration and (b) the distance covered.

Step 1. Find the acceleration from the slope. Since the graph is a straight line,

a = \frac{v-u}{t} = \frac{25-5}{10} = 2\ \text{m/s}^2

Step 2. Identify the shape under the graph. The region under this rising straight line, down to the time axis, is a trapezium with parallel sides $u = 5\ \text{m/s}$ and $v = 25\ \text{m/s}$ , and height $t = 10\ \text{s}$ along the time axis.

Step 3. Compute the area (distance) using the trapezium formula.

s = \frac{1}{2}(u+v)\times t = \frac{1}{2}(5+25)\times 10 = \frac{1}{2}\times 30\times 10 = 150\ \text{m}

Step 4. Cross-check with the derived formula $s = ut+\tfrac12at^2$ : $s = 5\times10 + \tfrac12\times2\times10^2 = 50+100 = 150\ \text{m}$ . Both routes agree. The body accelerates uniformly at $2\ \text{m/s}^2$ and covers $150\ \text{m}$ in $10\ \text{s}$ .

Competitive stretch

For the same graph, distance covered only in the last $2$ seconds (from $t=8\ \text{s}$ to $t=10\ \text{s}$ ) needs velocities at those instants: $v(8) = u+at = 5+2\times8 = 21\ \text{m/s}$ , $v(10) = 25\ \text{m/s}$ . That strip is a smaller trapezium: $s = \tfrac12(21+25)\times2 = 46\ \text{m}$ — a reminder that "area between two times" always means the area over just that time window, not the whole graph.

Common traps

Confusing slope (acceleration) with area (distance/displacement) — they are two different readings of the same graph and answer different questions.
Applying the rectangle area formula when the graph is actually sloped (a triangle or trapezium) — always check the shape before picking a formula.
Treating a graph that dips below the time axis as "impossible" — negative velocity is meaningful; it represents motion in the opposite direction along the same line, and the area there still counts as displacement (with sign).
Trying to compute the area of a composite region (e.g., rectangle plus triangle) with one combined formula instead of splitting it into simple shapes first and adding.
Mixing up distance-time and velocity-time graph rules: on a distance-time graph slope is speed; on a velocity-time graph slope is acceleration and area is distance.

Ask the tutor: Stuck on whether a shaded region under a velocity-time graph is a triangle, a trapezium, or a composite shape you need to split up — ask the AI tutor to walk through identifying and computing the area with you.

Why does area under a velocity-time graph give distance, not something else?

Because distance is (roughly) velocity multiplied by time, and multiplying the quantity on the y-axis (velocity) by the quantity on the x-axis (time) is exactly what computing an "area" on that graph means. For a rectangle this is obvious — constant velocity times time is just length times width. For a sloped line, the trapezium or triangle split works because we are still, in effect, multiplying an average velocity by the time over which it acted; NCERT uses this trick specifically so Class 9 students can derive $s = ut+\tfrac12at^2$ without calculus.

What if the velocity-time graph is a curve, not a straight line?

Then the acceleration is non-uniform — it changes value from moment to moment, so there is no single "slope" for the whole graph. At Class 9 level you are not asked to find exact areas under curves; you are only expected to recognise the shape (curved means non-uniform acceleration) and, at most, estimate area by counting grid squares. Exact curved-area calculations belong to calculus, taught in later classes.

Can the area under a velocity-time graph come out negative?

Yes, when the graph lies below the time axis — meaning the object's velocity is in the negative direction (opposite to whatever you defined as positive). The area calculated there is still a valid distance in magnitude, but as a displacement it carries a negative sign, since the object is moving backward relative to its starting direction. Always check whether the question wants total distance travelled (add magnitudes of all pieces) or net displacement (add pieces with their signs).