Uniform acceleration on a v–t graph

Velocity–time graphs

How do you read acceleration and distance off a velocity–time graph?

Set the starting velocity and the acceleration, then let time run. The slope of the v–t line is the acceleration and the shaded area under it is the displacement.

Drag the ringed points on the board.

High school
Slope = acceleration
Area = displacement
Kinematic equations
Turning back

v_0

2.0

a

uniform

1.0

t

4.00

x = v_0t + \tfrac12 at^2

$t$: 4.00
$v = v_0 + at$: 6.00
$x = v_0t + \tfrac12 at^2$: 16.00
$\text{distance}$: 16.00

Speedincreasing

Guided tour

1Constant velocity
With $a=0$ the graph is flat at $v_0=3\,\text{m/s}$ . After $4\,\text{s}$ the shaded rectangle is $3\times4=12$ : the object has moved $12\,\text{m}$ .
2Starting from rest
Now $v_0=0$ and $a=2\,\text{m/s}^2$ . The line climbs $2$ for every second, so at $t=4\,\text{s}$ the velocity is $8\,\text{m/s}$ . The area is a triangle, $\tfrac12\times4\times8=16\,\text{m}$ .
3A head start
With $v_0=2$ and $a=1$ the velocity at $4\,\text{s}$ is $6\,\text{m/s}$ . The area is a trapezoid, $\tfrac{2+6}{2}\times4=16\,\text{m}$ : average velocity times time.
4Braking, then reversing
Start at $6\,\text{m/s}$ with $a=-2$ . The line crosses zero at $t=3\,\text{s}$ : the object stops after $9\,\text{m}$ , then comes back. At $t=5\,\text{s}$ , $v=-4$ and the area below the axis is $-4$ , so the displacement is $9-4=5\,\text{m}$ while the distance travelled is $13\,\text{m}$ .

Explanation

1
The slope is the acceleration
Acceleration is how fast velocity changes: $a=\dfrac{\Delta v}{\Delta t}$ . On a $v$ – $t$ graph that is rise over run, the slope. Uniform acceleration means the slope never changes, so the graph is a straight line:
$v=v_0+at$
A steeper line is a harder push; a line sloping down is braking.
2
The area is the displacement
Over a short moment $\Delta t$ the object moves about $v\,\Delta t$ : a thin strip of height $v$ and width $\Delta t$ . Adding the strips gives the area under the line. For a straight line the area is a trapezoid:
$x=\frac{v_0+v}{2}\,t=v_0t+\tfrac12at^2$
Area below the axis counts as negative: the object is moving backwards.
3
Three equations, one graph
Everything in uniformly accelerated motion follows from the slope and the area:
$v=v_0+at,\qquad x=v_0t+\tfrac12at^2,\qquad v^2-v_0^2=2ax$
The third comes from eliminating $t$ between the first two. Use it when the time is not given, for example a braking distance: from $6\,\text{m/s}$ at $a=-2$ , the car stops after $x=\dfrac{0-36}{-4}=9\,\text{m}$ .
4
Displacement is not distance
Once the line crosses the axis the object has turned around. Displacement adds the areas with their signs; distance adds their sizes. In the last guide step they differ: $9-4=5\,\text{m}$ of displacement, $9+4=13\,\text{m}$ of distance. Whether the speed is rising depends on both signs: speed grows when $v$ and $a$ point the same way.

Check yourself

Q01

A car starts from rest with $a=3\,\text{m/s}^2$ . How far has it gone after $4\,\text{s}$ ?

Q02

On a $v$ – $t$ graph, what does a line sloping downward but still above the axis mean?

Q03

A train brakes from $20\,\text{m/s}$ at $a=-4\,\text{m/s}^2$ . What is its stopping distance?

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