Distance–time and velocity–time graphs

Key ideas

Graphs turn motion into pictures, and two rules unlock all of them:

On a distance–time graph, the gradient is the velocity.
On a velocity–time graph, the gradient is the acceleration, and the area under the graph is the distance travelled.

Velocity–time graph rising steadily from 0 to 8 metres per second over 4 seconds then staying flat until 8 seconds, with the gradient labelled as acceleration and the shaded area under the line labelled as distance.

Worked ExampleGradient and area on a v–t graph

A runner accelerates uniformly from rest to $8.0$ m s⁻¹ in $4.0$ s, then holds that velocity for another $4.0$ s (the graph above). Find the acceleration in the first phase and the total distance.

Step 1 — Acceleration = gradient of the first phase

$a = \frac{\Delta v}{\Delta t} = \frac{8.0 - 0}{4.0} = 2.0 \text{ m s}^{-2}$

Step 2 — Distance = area under the whole line

Split the shape into a triangle (rising phase) plus a rectangle (steady phase):

$d = \underbrace{\tfrac{1}{2}(4.0)(8.0)}_{\text{triangle}} + \underbrace{(4.0)(8.0)}_{\text{rectangle}} = 16 + 32 = 48 \text{ m}$

Reading the shape

Graph	Flat (horizontal) means	Straight slope means	Curve means
distance–time	stationary	constant velocity	accelerating
velocity–time	constant velocity	constant acceleration	changing acceleration

Practice question

A car's velocity–time graph shows a straight line from $20$ m s⁻¹ down to $0$ in $5.0$ s. How far does it travel while stopping?

Worked solution: area under the line is a triangle: $d = \tfrac{1}{2}(5.0)(20) = 50$ m.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A car's velocity–time graph is a horizontal line at $12$ m s⁻¹ for $8.0$ s.

State what the area under the line represents, and calculate the distance the car travels.

Merit

A cyclist's velocity–time graph rises in a straight line from 0 to $10$ m s⁻¹ in $4.0$ s, then stays at $10$ m s⁻¹ for a further $6.0$ s.

Calculate the total distance travelled.

Excellence

A car accelerates uniformly from rest, then travels at a constant velocity, then brakes steadily to a stop.

Describe and explain, using physics principles, what the gradient and the area of a velocity–time graph tell you about each of these three phases.

Distance–time and velocity–time graphs

Covers AS91171

Key ideas

Graphs turn motion into pictures, and two rules unlock all of them:

On a distance–time graph, the gradient is the velocity.
On a velocity–time graph, the gradient is the acceleration, and the area under the graph is the distance travelled.

Worked ExampleGradient and area on a v–t graph

A runner accelerates uniformly from rest to $8.0$ m s⁻¹ in $4.0$ s, then holds that velocity for another $4.0$ s (the graph above). Find the acceleration in the first phase and the total distance.

Step 1 — Acceleration = gradient of the first phase

$a = \frac{\Delta v}{\Delta t} = \frac{8.0 - 0}{4.0} = 2.0 \text{ m s}^{-2}$

Step 2 — Distance = area under the whole line

Split the shape into a triangle (rising phase) plus a rectangle (steady phase):

$d = \underbrace{\tfrac{1}{2}(4.0)(8.0)}_{\text{triangle}} + \underbrace{(4.0)(8.0)}_{\text{rectangle}} = 16 + 32 = 48 \text{ m}$

Reading the shape

Graph	Flat (horizontal) means	Straight slope means	Curve means
distance–time	stationary	constant velocity	accelerating
velocity–time	constant velocity	constant acceleration	changing acceleration

Practice question

A car's velocity–time graph shows a straight line from $20$ m s⁻¹ down to $0$ in $5.0$ s. How far does it travel while stopping?

Worked solution: area under the line is a triangle: $d = \tfrac{1}{2}(5.0)(20) = 50$ m.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A car's velocity–time graph is a horizontal line at $12$ m s⁻¹ for $8.0$ s.

State what the area under the line represents, and calculate the distance the car travels.

Merit

A cyclist's velocity–time graph rises in a straight line from 0 to $10$ m s⁻¹ in $4.0$ s, then stays at $10$ m s⁻¹ for a further $6.0$ s.

Calculate the total distance travelled.

Excellence

A car accelerates uniformly from rest, then travels at a constant velocity, then brakes steadily to a stop.

Describe and explain, using physics principles, what the gradient and the area of a velocity–time graph tell you about each of these three phases.