How the Area Under a Velocity-Time Graph Gives the Distance Travelled

A train leaving a station, a ball dropped from a balcony and an autorickshaw braking at a signal all move in a straight line. Physics describes such motion by separating distance from displacement, reading graphs, and linking everything with three simple equations.

This lesson covers displacement and velocity with their graphs, the graphical derivation of the equations of motion, and relative velocity in one dimension.

Distance is the total path length travelled, a scalar, while displacement is the straight-line change in position with a direction, a vector; speed is distance divided by time, while velocity is displacement divided by time.

Key definitions:

- Distance is always positive; displacement can be positive, negative or zero, and its size never exceeds the distance
- Average speed = total distance ÷ total time; average velocity = displacement ÷ total time
- Instantaneous velocity is $v=\frac{dx}{dt}$, and acceleration is $a=\frac{dv}{dt}$

Worked example 1. A cyclist rides 400 m east in 100 s, then 300 m west in 50 s:

$$\text{average speed}=\frac{700}{150}=4.67{\text{ m s}}^{-1}\text{average velocity}=\frac{400-300}{150}=0.667{\text{ m s}}^{-1}\text{ east}$$

Reading graphs:

- On a position-time graph, the slope gives velocity; a horizontal line means rest
- On a velocity-time graph, the slope gives acceleration and the area under the graph gives displacement

Worked example 2. A car's velocity rises steadily from 0 to 20 m s${}^{-1}$ in 10 s:

$$a=\frac{20-0}{10}=2.0{\text{ m s}}^{-2}\text{displacement}=\frac{1}{2}\times 10\times 20=100\text{ m}$$

An everyday example. A metro train running between two stations in Delhi gives a velocity-time graph with a rising line as it accelerates, a flat line at steady speed and a falling line as it brakes.

The substance. Velocity can be negative while speed cannot — a negative velocity simply means motion in the direction chosen as negative.

**For uniform acceleration the velocity-time graph is a straight line, and its slope and the area beneath it give the three equations $v=u+at$, $s=ut+\frac{1}{2}a{t}^2$ and $v^2=u^2+2as$.

Set-up. A body starts with velocity u and accelerates uniformly at a, reaching velocity v at time t. On the graph, O is the origin, A is (0, u), B is (t, v), C is (t, 0) and D is (t, u).

First equation — from the slope:**

$$a=\frac{BD}{AD}=\frac{v-u}{t}\Rightarrow v=u+at$$

Second equation — from the area. Displacement is the area of rectangle OADC plus triangle ADB:

$$s=ut+\frac{1}{2}t(v-u)=ut+\frac{1}{2}t(at)=ut+\frac{1}{2}a{t}^2$$

Third equation — from the trapezium. Displacement is the area of trapezium OABC, $s=\frac{1}{2}(u+v)t$. Substituting $t=\frac{v-u}{a}$:

$$s=\frac{(v+u)(v-u)}{2a}\Rightarrow v^2=u^2+2as$$

Worked example 1 — a train leaving a station. Starting from rest with $a=0.50$ m s${}^{-2}$ for 40 s:

$$v=0.50\times 40=20{\text{ m s}}^{-1}s=\frac{1}{2}\times 0.50\times 40^2=400\text{ m}$$

Worked example 2 — a vertical throw. Taking upwards as positive, a ball thrown up at 14.7 m s${}^{-1}$ has $a=-9.8$ m s${}^{-2}$ and stops rising when $v=0$:

$$t=\frac{14.7}{9.8}=1.5\text{ s}h=\frac{14.7^2}{2\times 9.8}=11.0\text{ m}$$

An everyday example. A fielder tossing a cricket ball straight up sees it follow these numbers, rising for about 1.5 s before falling back.

The substance. The three equations hold only for uniform acceleration — when acceleration changes, use the graph's area and slope, or calculus, instead.

**The velocity of body A relative to body B is $v_{AB}=v_A-v_B$, the velocity A appears to have to an observer moving along with B.

Rules:

- Choose one direction as positive and give every velocity its sign
- In the same direction, relative speed is the difference of the speeds; in opposite directions**, it is their sum
- $v_{BA}=-v_{AB}$: each body sees the other moving at the same speed in the opposite direction

Worked example 1 — overtaking. A car at 25 m s${}^{-1}$ overtakes a bus at 15 m s${}^{-1}$. Relative to the bus, the car moves at $25-15=10$ m s${}^{-1}$, so gaining 50 m takes

$$t=\frac{50}{10}=5.0\text{ s}$$

Worked example 2 — trains crossing. Two trains, each 200 m long, approach on parallel tracks at 20 and 30 m s${}^{-1}$. Their relative speed is 50 m s${}^{-1}$, and to pass completely they must cover $200+200=400$ m relative to each other:

$$t=\frac{400}{50}=8.0\text{ s}$$

An everyday example. Passengers on a Mumbai local train watching a train on the next track at the same speed feel as though both are standing still, because their relative velocity is zero.

The substance. Relative velocity depends only on velocities, not positions — two cars at the same velocity keep the same gap whether they are 10 m or 10 km apart.