Displacement–Time and Velocity–Time Graphs

AS · M1 · 12 min

A graph of motion turns a story ("the train speeds up, cruises, then brakes") into a picture you can calculate from. Two graphs matter in Paper 4: the displacement–time graph and the velocity–time graph. Three facts about them, two gradients and one area, solve a large family of exam questions, many of which need no formulae beyond the area of a trapezium. Velocity–time graph questions appear on almost every paper, often as the first part of a longer question.

Displacement–time graphs

A displacement–time ( $s$ – $t$ ) graph shows the position of a particle relative to the origin at each moment. Time goes on the horizontal axis, displacement on the vertical.

Key result

The gradient of a displacement–time graph is the velocity.

The reason: velocity is the rate of change of displacement, and the gradient measures exactly how fast $s$ changes as $t$ increases.

How to read an $s$ – $t$ graph:

Straight line: constant velocity. Steeper means faster.
Horizontal line: at rest (velocity zero).
Negative gradient: moving in the negative direction, back towards (or past) the origin.
Curve: velocity changing. The velocity at an instant is the gradient of the tangent at that point.
Crossing the $t$ -axis: passing through the origin.

(0, 0) -- (120, 600) (120, 600) -- (180, 600) (180, 600) -- (330, 0)

The graph above shows a cyclist who rides $600\ \text{m}$ from home in $120\ \text{s}$ , stops for $60\ \text{s}$ , then rides home in $150\ \text{s}$ . Gradients: $\tfrac{600}{120} = 5\ \text{m s}^{-1}$ , then $0$ , then $\tfrac{0 - 600}{150} = -4\ \text{m s}^{-1}$ . The negative sign records that the cyclist is travelling back towards home.

For a particle with constant acceleration from rest, $s$ is proportional to $t^2$ , so the $s$ – $t$ graph is part of a parabola. Its gradient (the velocity) increases steadily. The tangent at $t = 2$ shown below has gradient $4$ , so for $s = t^2$ the velocity at $t = 2$ is $4\ \text{m s}^{-1}$ .

y = x^2 y = 4x - 4 (2, 4)

Velocity–time graphs

A velocity–time ( $v$ – $t$ ) graph shows the velocity at each moment. It carries more information than the $s$ – $t$ graph and is the one most used in exam questions.

The two facts about v–t graphs

The gradient of a velocity–time graph is the acceleration.
The area between a velocity–time graph and the time axis is the displacement.

Why the area is the displacement. For constant velocity $v$ over time $t$ , displacement is $vt$ , which is the area of a rectangle of height $v$ and width $t$ . A changing velocity can be cut into thin strips, each almost a rectangle; adding the strips gives the area. (In the language of calculus, displacement is $\int v\,dt$ , which is the area under the curve. See Variable acceleration (note not yet published).)

How to read a $v$ – $t$ graph:

Straight line: constant acceleration; the gradient is the acceleration.
Horizontal line: constant velocity (zero acceleration). On the $t$ -axis itself: at rest.
Line sloping down while above the axis: slowing down (decelerating).
Below the $t$ -axis: moving in the negative direction.
Crossing the $t$ -axis: momentarily at rest and changing direction.

Areas below the axis

Area below the $t$ -axis counts as negative displacement: the particle is moving backwards. This gives two different answers depending on what is asked:

Key result

Displacement $=$ (area above the axis) $-$ (area below the axis).
Distance travelled $=$ (area above the axis) $+$ (area below the axis).

(0, 8) -- (6, -4) fill 0 4 y = 8 - 2x fill 4 6 y = 8 - 2x

In this graph a particle starts at $8\ \text{m s}^{-1}$ and has constant acceleration $-2\ \text{m s}^{-2}$ . It stops at $t = 4$ and then moves backwards, reaching $-4\ \text{m s}^{-1}$ at $t = 6$ . The triangle above the axis has area $\tfrac12 \times 4 \times 8 = 16$ and the triangle below has area $\tfrac12 \times 2 \times 4 = 4$ . So after $6\ \text{s}$ the displacement is $16 - 4 = 12\ \text{m}$ , but the distance travelled is $16 + 4 = 20\ \text{m}$ .

Speed–time graphs

Some questions draw a speed–time graph instead. Speed is never negative, so the graph never goes below the axis, and the area under it is the distance travelled. If the particle reverses, the speed–time graph "bounces" off the axis where the velocity–time graph would cross it.

The standard trapezium question

The most common Paper 4 graph question describes a journey in three stages: speed up uniformly, travel at constant speed, slow down uniformly. The $v$ – $t$ graph is a trapezium (or a triangle if there is no constant-speed stage).

Velocity–time graph problems

Sketch the graph from the description, even if not asked. Mark every known time, velocity and gradient on it.
Use gradient = acceleration to find unknown times: time $= \dfrac{\text{change in velocity}}{\text{acceleration}}$ .
Use area = displacement to form an equation. Split the area into triangles and rectangles, or use the trapezium formula: area $= \tfrac12(a + b)h$ , with the parallel sides being the total time and the time at constant speed.
Solve for the unknown.
Answer the question asked: total time, distance, an acceleration, or an average speed.

Reading a three-stage journey

A car starts from rest and accelerates uniformly to $15\ \text{m s}^{-1}$ in $6\ \text{s}$ . It travels at this speed for $20\ \text{s}$ , then decelerates uniformly to rest in $4\ \text{s}$ .

(a) Sketch the velocity–time graph.

(b) Find the acceleration and the deceleration.

Solution

(a)

(0, 0) -- (6, 15) (6, 15) -- (26, 15) (26, 15) -- (30, 0) fill 0 30 y = min(2.5x, 15, 3.75(30 - x))

(b) Acceleration $= \dfrac{15}{6} = 2.5\ \text{m s}^{-2}$ . Deceleration $= \dfrac{15}{4} = 3.75\ \text{m s}^{-2}$ (an acceleration of $-3.75\ \text{m s}^{-2}$ ).

Equivalently, the trapezium has parallel sides $30$ (total time) and $20$ (constant-speed time), and height $15$ : $\tfrac12(30 + 20)(15) = 375\ \text{m}$ .

Average speed $= \dfrac{375}{30} = 12.5\ \text{m s}^{-1}$ .

Finding an unknown time from the total distance

A train starts from rest at station $A$ and accelerates uniformly at $0.5\ \text{m s}^{-2}$ until it reaches $20\ \text{m s}^{-1}$ . It travels at this speed for $T$ seconds and then decelerates uniformly at $0.8\ \text{m s}^{-2}$ , coming to rest at station $B$ , $4\ \text{km}$ from $A$ . Find $T$ and the total time for the journey.

Solution

Times from gradients:

t_1 = \frac{20}{0.5} = 40\ \text{s}, \qquad t_3 = \frac{20}{0.8} = 25\ \text{s}

Area under the graph $= 4000\ \text{m}$ :

\tfrac12(40)(20) + 20T + \tfrac12(25)(20) = 4000 \quad\Rightarrow\quad 400 + 20T + 250 = 4000 \quad\Rightarrow\quad T = 167.5

Total time $= 40 + 167.5 + 25 = 232.5\ \text{s}$ .

An unknown top speed

A particle starts from rest and accelerates uniformly to $V\ \text{m s}^{-1}$ in $10\ \text{s}$ . It moves at constant speed $V$ for $30\ \text{s}$ and then decelerates uniformly to rest in $5\ \text{s}$ . The total distance is $1050\ \text{m}$ . Find $V$ .

Solution

The trapezium has parallel sides $10 + 30 + 5 = 45$ and $30$ , height $V$ :

\tfrac12(45 + 30)V = 1050 \quad\Rightarrow\quad 37.5V = 1050 \quad\Rightarrow\quad V = 28

Interpreting a displacement–time graph

The cyclist whose displacement–time graph appears earlier in this note rides from home for $120\ \text{s}$ , rests, and returns home at $t = 330$ . Find the velocity in each stage, the average speed and the average velocity for the whole trip.

Solution

Velocities from gradients: $\dfrac{600}{120} = 5\ \text{m s}^{-1}$ ; $0$ ; $\dfrac{-600}{150} = -4\ \text{m s}^{-1}$ .

Total distance $= 600 + 600 = 1200\ \text{m}$ in $330\ \text{s}$ :

\text{average speed} = \frac{1200}{330} = 3.64\ \text{m s}^{-1}

The cyclist ends at home, so the displacement and the average velocity are both $0$ .

Two particles on the same graph

When two vehicles are compared, draw both $v$ – $t$ graphs on the same axes. One vehicle catches the other when their displacements are equal, which means the areas under their graphs (from the same starting point and time) are equal. It does not happen where the graphs cross: crossing $v$ – $t$ graphs only means the two have the same velocity at that moment.

Catching up (exam standard)

Car $A$ starts from rest at a point $O$ and accelerates uniformly at $2\ \text{m s}^{-2}$ until it reaches $24\ \text{m s}^{-1}$ , then continues at this speed. At the instant $A$ starts, car $B$ passes $O$ travelling in the same direction at a constant $18\ \text{m s}^{-1}$ . Find the time at which $A$ overtakes $B$ and the distance from $O$ at which this happens.

Solution

$A$ reaches $24\ \text{m s}^{-1}$ after $\dfrac{24}{2} = 12\ \text{s}$ , having travelled $\tfrac12(12)(24) = 144\ \text{m}$ .

(0, 0) -- (12, 24) (12, 24) -- (30, 24) (0, 18) -- (30, 18) (24, 0) -- (24, 24)

Could $A$ catch $B$ during the first 12 s? In that phase $A$ has travelled $\tfrac12(2)t^2 = t^2$ and $B$ has travelled $18t$ . These are equal at $t = 18$ , which is outside $0 \le t \le 12$ , so no.

After 12 s, $A$ has travelled $144 + 24(t - 12) = 24t - 144$ and $B$ has travelled $18t$ :

24t - 144 = 18t \quad\Rightarrow\quad t = 24\ \text{s}

Distance from $O$ $= 18 \times 24 = 432\ \text{m}$ .

Note that the graphs cross at $t = 9$ , when both cars are travelling at $18\ \text{m s}^{-1}$ . At that moment $B$ is furthest ahead ( $162 - 81 = 81\ \text{m}$ ); it is not where $A$ overtakes.

Sketching graphs

When a question says "sketch", the examiner wants the right shape with the key values marked, not a scale drawing:

straight lines where the acceleration is constant, with the gradients in the right direction;
known velocities and times labelled on the axes;
the graph crossing the $t$ -axis where the particle changes direction;
axes labelled $t$ (s) and $v$ ( $\text{m s}^{-1}$ ) or $s$ (m).

Common mistakes

Graph errors

Reading the gradient of a $v$ – $t$ graph as velocity, or the area under an $s$ – $t$ graph as anything. Gradient of $s$ – $t$ is velocity; gradient of $v$ – $t$ is acceleration; area under $v$ – $t$ is displacement.
Adding area below the axis as positive when displacement is asked, or subtracting it when distance is asked.
Thinking one vehicle catches another where the $v$ – $t$ graphs cross. That is where the velocities are equal; catching up needs equal areas.
Using the total time for the top of the trapezium. The shorter parallel side is the time at constant speed only.
Forgetting to convert $\text{km}$ to $\text{m}$ or minutes to seconds before using areas.

Exam technique

Exam tip

Sketch the $v$ – $t$ graph even when the question does not ask for one. It is the fastest way to see the structure of a multi-stage journey, and the area equation often comes straight from it.
When asked to "sketch" a graph, label the key values on the axes. A sketch without values usually loses the mark.
Show the area calculation explicitly, split into named pieces or as a trapezium, so method marks can be given.
Check that a found time is positive and fits the story (for example, the constant-speed stage cannot have negative length).

Summary

Gradient of a displacement–time graph $=$ velocity. A horizontal line means at rest.
Gradient of a velocity–time graph $=$ acceleration. A horizontal line means constant velocity.
Area under a velocity–time graph $=$ displacement; area below the axis is negative.
Distance $=$ total area, all counted positive; displacement $=$ area above minus area below.
Trapezium journeys: find times from gradients, then form an equation from the area.
Two vehicles meet when the areas under their $v$ – $t$ graphs are equal, not where the graphs cross.

Practice

Question

A cyclist starts from rest, accelerates uniformly to $12\ \text{m s}^{-1}$ in $4\ \text{s}$ , rides at $12\ \text{m s}^{-1}$ for $10\ \text{s}$ , then decelerates uniformly to rest in $6\ \text{s}$ . Find the acceleration, the deceleration and the total distance.
A displacement–time graph consists of straight-line segments joining $(0, 0)$ , $(10, 40)$ , $(25, 40)$ and $(40, -20)$ , with $t$ in seconds and $s$ in metres. Find the velocity in each stage, the total distance travelled, the average speed and the average velocity for the $40\ \text{s}$ .
A cyclist starts from rest and accelerates at $0.4\ \text{m s}^{-2}$ to $8\ \text{m s}^{-1}$ , travels at this speed for $T\ \text{s}$ , then decelerates at $0.5\ \text{m s}^{-2}$ to rest. The total distance is $1\ \text{km}$ . Find $T$ .
A particle moves in a straight line. Its velocity decreases uniformly from $6\ \text{m s}^{-1}$ at $t = 0$ to $-6\ \text{m s}^{-1}$ at $t = 8$ . Find its displacement and the distance it travels in the $8\ \text{s}$ .
A car starts from rest, accelerates uniformly for $8\ \text{s}$ , travels at constant speed $V\ \text{m s}^{-1}$ for $40\ \text{s}$ , and decelerates uniformly to rest in $12\ \text{s}$ . The total distance is $1.2\ \text{km}$ . Find $V$ , the acceleration and the deceleration.
Car $P$ starts from rest at a point $O$ and accelerates at $1.5\ \text{m s}^{-2}$ for $20\ \text{s}$ , then continues at constant speed. As $P$ starts, car $Q$ passes $O$ in the same direction at a constant $25\ \text{m s}^{-1}$ . Find when and where $P$ overtakes $Q$ .
A lift starts from rest and rises with acceleration $0.8\ \text{m s}^{-2}$ until its speed is $4\ \text{m s}^{-1}$ . It continues at this speed and then decelerates at $0.5\ \text{m s}^{-2}$ , coming to rest $50\ \text{m}$ above its starting point. Find the total time taken.
A particle leaves a point $O$ with velocity $10\ \text{m s}^{-1}$ and has constant acceleration $-2\ \text{m s}^{-2}$ . (a) Sketch the velocity–time graph for $0 \le t \le 12$ . (b) Find the time at which the particle returns to $O$ . (c) Find the total distance travelled in the first $12\ \text{s}$ .

Answers

Acceleration $= \tfrac{12}{4} = 3\ \text{m s}^{-2}$ ; deceleration $= \tfrac{12}{6} = 2\ \text{m s}^{-2}$ . Distance $= \tfrac12(4)(12) + 10(12) + \tfrac12(6)(12) = 24 + 120 + 36 = 180\ \text{m}$ .
Velocities: $\tfrac{40}{10} = 4\ \text{m s}^{-1}$ ; $0$ ; $\tfrac{-20 - 40}{15} = -4\ \text{m s}^{-1}$ . Distance $= 40 + 60 = 100\ \text{m}$ . Average speed $= \tfrac{100}{40} = 2.5\ \text{m s}^{-1}$ . Final displacement $-20\ \text{m}$ , so average velocity $= -0.5\ \text{m s}^{-1}$ .
Times: $\tfrac{8}{0.4} = 20\ \text{s}$ and $\tfrac{8}{0.5} = 16\ \text{s}$ . Area: $80 + 8T + 64 = 1000$ , so $T = 107$ .
The velocity is zero at $t = 4$ . Area above $= \tfrac12(4)(6) = 12$ , area below $= 12$ . Displacement $= 0$ ; distance $= 24\ \text{m}$ .
Trapezium: $\tfrac12(60 + 40)V = 1200$ , so $V = 24$ . Acceleration $= \tfrac{24}{8} = 3\ \text{m s}^{-2}$ ; deceleration $= \tfrac{24}{12} = 2\ \text{m s}^{-2}$ .
$P$ reaches $30\ \text{m s}^{-1}$ at $t = 20$ , having travelled $300\ \text{m}$ . During the first phase $0.75t^2 = 25t$ gives $t = 33.3 > 20$ , so no overtaking then. After: $300 + 30(t - 20) = 25t$ , so $5t = 300$ and $t = 60\ \text{s}$ , at $25 \times 60 = 1500\ \text{m}$ from $O$ .
Accelerating: $5\ \text{s}$ , $\tfrac12(5)(4) = 10\ \text{m}$ . Decelerating: $8\ \text{s}$ , $\tfrac12(8)(4) = 16\ \text{m}$ . Constant speed: $50 - 26 = 24\ \text{m}$ , taking $6\ \text{s}$ . Total time $= 5 + 6 + 8 = 19\ \text{s}$ .
(a) A straight line from $(0, 10)$ through $(5, 0)$ to $(12, -14)$ . (b) Area above the axis from $0$ to $5$ is $25$ . After $t = 5$ the area below the axis up to time $t$ is $\tfrac12(t - 5)\cdot 2(t - 5) = (t - 5)^2$ . Back at $O$ when $(t - 5)^2 = 25$ , so $t = 10\ \text{s}$ . (c) Distance $= 25 + \tfrac12(7)(14) = 25 + 49 = 74\ \text{m}$ .

Displacement–time graphs

Velocity–time graphs

Areas below the axis

Speed–time graphs

The standard trapezium question

Two particles on the same graph

Sketching graphs

Common mistakes

Exam technique

Summary

Practice

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