Physics 9702 · Paper 2

Velocity-Time Graphs

The velocity-time (v-t) graph is the graph this chapter uses most: it holds two pieces of information at once.
The gradient is the acceleration. The area between the line and the time axis is the displacement.

The gradient is the acceleration

Rise over run here is $\Delta v / \Delta t$ , and that is the definition of acceleration.
A straight line means constant acceleration. A flat line means constant velocity (zero acceleration). A line sloping down means the velocity is decreasing.
If the v-t line curves, the acceleration is changing. The gradient of a tangent gives the acceleration at that instant, the same tangent method as lesson 2.2.

Worked example

Warm-up: acceleration from a v-t line

A v-t line rises from 0 to 15 m s⁻¹ in 6.0 s. Find the acceleration.

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{15}{6.0} = 2.5 \text{ m s}^{-2}$ .

Answer

a = 2.5 \text{ m s}^{-2}

The area is the displacement

In a short time, displacement = velocity × time. On the graph that product is a thin strip of area under the line. Adding all the strips gives the whole area, so area = displacement.
Split the area into triangles and rectangles, find each one, and add them.

Area under a v–t graph is the displacement: the triangle (45 m) plus the rectangle (60 m) gives 105 m.

Worked example

The standard two-part journey

An object accelerates from rest at 2.5 m s⁻² for 6.0 s, then travels at constant velocity for 4.0 s. Find the total displacement.

Velocity reached: $v = at = 2.5 \times 6.0 = 15 \text{ m s}^{-1}$ .
Triangle (0 to 6 s): $s_1 = \tfrac{1}{2} \times 6.0 \times 15 = 45 \text{ m}$ .
Rectangle (6 to 10 s): $s_2 = 4.0 \times 15 = 60 \text{ m}$ .
Total: $s = 45 + 60 = 105 \text{ m}$ .

Answer

s = 105 \text{ m}

Drag the sliders below to build your own v-t journey and read the areas:

v–t Graph Explorer

Adjust the sliders — the gold triangle shows acceleration (gradient Δv/Δt), and the shaded area shows displacement.

Initial velocity u5 m s⁻¹

Acceleration a2 m s⁻²

Time range t4 s

Gradient = acceleration

2 m s⁻²

Shaded area = displacement

s = 36.0 m

Area below the axis is negative

When the line drops below the time axis, the velocity is negative: the object is moving backward.
Area below the axis counts as negative displacement. Add the areas with their signs to get the displacement; add their sizes to get the total distance.

Area above the axis is forward displacement (+16 m); area below is backward (−4 m). Displacement = 16 − 4 = 12 m, but total distance = 16 + 4 = 20 m.

Worked example

A harder step: displacement and distance

A v-t line runs from 8.0 m s⁻¹ at t = 0 down through zero at t = 4.0 s to −4.0 m s⁻¹ at t = 6.0 s. Find the displacement and the total distance.

Area above the axis (0 to 4 s): $\tfrac{1}{2} \times 4.0 \times 8.0 = +16 \text{ m}$ .
Area below the axis (4 to 6 s): $\tfrac{1}{2} \times 2.0 \times 4.0 = 4.0 \text{ m}$ , counted as $-4.0 \text{ m}$ .
Displacement: $16 - 4.0 = 12 \text{ m}$ . Distance: $16 + 4.0 = 20 \text{ m}$ .

Answer

s = 12 \text{ m}, \quad d = 20 \text{ m}

One v-t graph answers two different questions. Gradient gives the acceleration; area gives the displacement, with area below the axis negative.

Common mistake

Adding every area as positive. If part of the line is below the axis, the object came back: the displacement is smaller than the distance. Check which side of the axis each area is on before adding.

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Velocity-Time Graphs

The velocity-time (v-t) graph is the graph this chapter uses most: it holds two pieces of information at once.
The gradient is the acceleration. The area between the line and the time axis is the displacement.

The gradient is the acceleration

Rise over run here is $\Delta v / \Delta t$ , and that is the definition of acceleration.
A straight line means constant acceleration. A flat line means constant velocity (zero acceleration). A line sloping down means the velocity is decreasing.
If the v-t line curves, the acceleration is changing. The gradient of a tangent gives the acceleration at that instant, the same tangent method as lesson 2.2.

Worked example

Warm-up: acceleration from a v-t line

A v-t line rises from 0 to 15 m s⁻¹ in 6.0 s. Find the acceleration.

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{15}{6.0} = 2.5 \text{ m s}^{-2}$ .

Answer

a = 2.5 \text{ m s}^{-2}

The area is the displacement

In a short time, displacement = velocity × time. On the graph that product is a thin strip of area under the line. Adding all the strips gives the whole area, so area = displacement.
Split the area into triangles and rectangles, find each one, and add them.

Area under a v–t graph is the displacement: the triangle (45 m) plus the rectangle (60 m) gives 105 m.

Worked example

The standard two-part journey

An object accelerates from rest at 2.5 m s⁻² for 6.0 s, then travels at constant velocity for 4.0 s. Find the total displacement.

Velocity reached: $v = at = 2.5 \times 6.0 = 15 \text{ m s}^{-1}$ .
Triangle (0 to 6 s): $s_1 = \tfrac{1}{2} \times 6.0 \times 15 = 45 \text{ m}$ .
Rectangle (6 to 10 s): $s_2 = 4.0 \times 15 = 60 \text{ m}$ .
Total: $s = 45 + 60 = 105 \text{ m}$ .

Answer

s = 105 \text{ m}

Drag the sliders below to build your own v-t journey and read the areas:

v–t Graph Explorer

Adjust the sliders — the gold triangle shows acceleration (gradient Δv/Δt), and the shaded area shows displacement.

Initial velocity u5 m s⁻¹

Acceleration a2 m s⁻²

Time range t4 s

Gradient = acceleration

2 m s⁻²

Shaded area = displacement

s = 36.0 m

Area below the axis is negative

When the line drops below the time axis, the velocity is negative: the object is moving backward.
Area below the axis counts as negative displacement. Add the areas with their signs to get the displacement; add their sizes to get the total distance.

Area above the axis is forward displacement (+16 m); area below is backward (−4 m). Displacement = 16 − 4 = 12 m, but total distance = 16 + 4 = 20 m.

Worked example

A harder step: displacement and distance

A v-t line runs from 8.0 m s⁻¹ at t = 0 down through zero at t = 4.0 s to −4.0 m s⁻¹ at t = 6.0 s. Find the displacement and the total distance.

Area above the axis (0 to 4 s): $\tfrac{1}{2} \times 4.0 \times 8.0 = +16 \text{ m}$ .
Area below the axis (4 to 6 s): $\tfrac{1}{2} \times 2.0 \times 4.0 = 4.0 \text{ m}$ , counted as $-4.0 \text{ m}$ .
Displacement: $16 - 4.0 = 12 \text{ m}$ . Distance: $16 + 4.0 = 20 \text{ m}$ .

Answer

s = 12 \text{ m}, \quad d = 20 \text{ m}

One v-t graph answers two different questions. Gradient gives the acceleration; area gives the displacement, with area below the axis negative.

Common mistake

Adding every area as positive. If part of the line is below the axis, the object came back: the displacement is smaller than the distance. Check which side of the axis each area is on before adding.