Kinematics graphs: displacement, velocity and acceleration

Kinematics graphs describe how motion changes with time. The central ideas are:

$$\boxed{\text{gradient gives rate of change}} \qquad\text{and}\qquad \boxed{\text{area gives accumulated change}}.$$

For straight line motion, the gradient of a displacement time graph is velocity, the gradient of a velocity time graph is acceleration, and the signed area under a velocity time graph is displacement.

Prerequisites

You should be able to:

• distinguish displacement from distance, and velocity from speed, using kinematics language;
• calculate the gradient of a straight line;
• find areas of rectangles, triangles and trapezia;
• interpret negative coordinates and choose a positive direction;
• use SI units from quantities and units in mechanics.

This lesson concerns motion along a straight line. Time is plotted horizontally and cannot run backwards.

The three graph types

Let $s$, $v$ and $a$ denote displacement, velocity and acceleration. Their connections are:

Graph	Gradient represents	Signed area between $t=p$ and $t=q$ represents
displacement time	velocity, $v$	no standard kinematic quantity
velocity time	acceleration, $a$	displacement, $s(q)-s(p)$
acceleration time	rate of change of acceleration	change in velocity, $v(q)-v(p)$

In compact form,

$$v=\frac{\mathrm ds}{\mathrm dt}, \qquad a=\frac{\mathrm dv}{\mathrm dt},$$

and

$$s(q)-s(p)=\int_p^q v\,\mathrm dt, \qquad v(q)-v(p)=\int_p^q a\,\mathrm dt.$$

You do not need calculus to use straight line segments: gradient and elementary area are enough. The notation explains why those methods work and prepares you for calculus in kinematics.

Read the vertical axis first

The area under a displacement time graph is not distance, and the gradient of a velocity time graph is not velocity. Identify the plotted quantity before doing any calculation.

Displacement time graphs

A point $(t,s)$ gives the particle’s displacement from the chosen origin at time $t$. The sign of $s$ tells you which side of the origin the particle occupies. It does not tell you its direction of travel.

The gradient gives velocity:

$$v=\frac{\text{change in displacement}}{\text{change in time}}.$$

Therefore:

• a positive gradient means motion in the positive direction;
• a negative gradient means motion in the negative direction;
• a horizontal section means the particle is stationary;
• a steeper graph means a greater speed;
• a changing gradient means the velocity is changing.

A straight line has constant velocity. A curve that becomes steeper has increasing speed, but the sign of its gradient still determines the direction.

Worked example 1: reading piecewise displacement

A particle’s displacement time graph joins the points

$$(0,2),\quad (3,8),\quad (5,8),\quad (9,-4)$$

with straight line segments. Distances are in metres and times in seconds. Describe the motion and find the velocity on each interval.

From $0\leq t\leq3$,

$$v=\frac{8-2}{3-0}=2\ \mathrm{m\,s^{-1}}.$$

The particle moves in the positive direction from $s=2$ to $s=8$.

From $3\leq t\leq5$, the graph is horizontal, so

$$v=0.$$

The particle remains at $s=8$ for $2$ seconds.

From $5\leq t\leq9$,

$$v=\frac{-4-8}{9-5}=-3\ \mathrm{m\,s^{-1}}.$$

It moves in the negative direction, passes the origin, and finishes at $s=-4$. The negative displacement at the end describes position, not distance travelled.

Its total distance is

$$|8-2|+|8-8|+|-4-8|=6+0+12=\boxed{18\ \mathrm m}.$$

Curvature and acceleration

On a smooth displacement time curve, acceleration describes how the gradient changes:

• increasing gradient means $a>0$;
• decreasing gradient means $a<0$;
• constant gradient means $a=0$.

For example, a graph can slope downwards while becoming less steep. Then $v<0$ but $a>0$, so the particle moves in the negative direction while slowing down.

Turning points

At a smooth local maximum or minimum of displacement, the tangent is horizontal, so $v=0$. A change from positive to negative gradient, or vice versa, shows a reversal of direction.

Self-check 1

A displacement time graph is a straight line from $(2,7)$ to $(8,-5)$. Find the velocity and the time at which the particle crosses the origin.

Answer

The velocity is the gradient:

$$v=\frac{-5-7}{8-2}=-2\ \mathrm{m\,s^{-1}}.$$

Starting at $s=7$ when $t=2$, the displacement is $s=7-2(t-2)$. Set $s=0$:

$$0=7-2(t-2) \quad\Longrightarrow\quad t=5.5\ \mathrm s.$$

Velocity time graphs

A point $(t,v)$ gives the instantaneous velocity. The graph’s height gives velocity, while its distance from the time axis gives speed:

$$\text{speed}=|v|.$$

Above the time axis, $v>0$ and motion is in the positive direction. Below it, $v<0$ and motion is in the negative direction. Crossing the axis usually indicates a change of direction.

The gradient gives acceleration:

$$a=\frac{\text{change in velocity}}{\text{change in time}}.$$

A horizontal velocity time segment means constant velocity and zero acceleration. It does not mean the particle is stationary unless the segment lies on $v=0$.

Signed area and displacement

Area above the time axis counts positively and area below counts negatively:

$$\text{displacement}=\text{area above axis}-\text{area below axis}.$$

Total distance counts both positively:

$$\text{distance}=\text{area above axis}+\bigl|\text{area below axis}\bigr|.$$

More precisely, distance is the area between the graph and the time axis. Split the interval wherever $v=0$.

Worked example 2: acceleration, displacement and distance

A particle has a velocity time graph made of straight lines joining

$$(0,4),\quad (3,10),\quad (7,10),\quad (12,-5).$$

Find the acceleration on each stage, the time at which the particle reverses, its displacement, and its total distance during the $12$ seconds.

The gradients give the accelerations:

$$0\leq t\leq3: \qquad a=\frac{10-4}{3}=2\ \mathrm{m\,s^{-2}},$$ $$3\leq t\leq7: \qquad a=0,$$

and

$$7\leq t\leq12: \qquad a=\frac{-5-10}{12-7}=-3\ \mathrm{m\,s^{-2}}.$$

On the last stage, velocity falls from $10$ at $3\ \mathrm{m\,s^{-1}}$ each second. It reaches zero after $10/3$ seconds, at

$$t=7+\frac{10}{3}=\frac{31}{3}\ \mathrm s.$$

The particle reverses there. Calculate signed areas:

$$A_1=\frac{4+10}{2}(3)=21,$$ $$A_2=10(4)=40,$$ $$A_3=\frac{10+(-5)}{2}(5)=\frac{25}{2}.$$

The third trapezium formula already includes the negative portion, so the displacement is

$$21+40+\frac{25}{2}=\boxed{73.5\ \mathrm m}.$$

For distance, split the final stage at $v=0$. Its positive triangular area is

$$\frac12\left(\frac{10}{3}\right)(10)=\frac{50}{3},$$

and its negative triangular area has magnitude

$$\frac12\left(\frac{5}{3}\right)(5)=\frac{25}{6}.$$

Hence

$$\text{distance}=21+40+\frac{50}{3}+\frac{25}{6} =\boxed{\frac{491}{6}\ \mathrm m}.$$

The distance, about $81.8$ m, is greater than the displacement because the particle reverses.

Worked example 3: reconstructing a velocity graph

A car starts from rest, accelerates uniformly at $3\ \mathrm{m\,s^{-2}}$ for $4$ seconds, travels at constant velocity for $5$ seconds, then decelerates uniformly to rest in $6$ seconds. Find its maximum velocity and total distance.

After $4$ seconds,

$$v=0+3(4)=12\ \mathrm{m\,s^{-1}}.$$

The velocity time graph joins

$$(0,0),\quad(4,12),\quad(9,12),\quad(15,0).$$

The total distance is the area under the graph:

$$\begin{aligned} \text{distance} &=\frac12(4)(12)+(5)(12)+\frac12(6)(12)\\ &=24+60+36\\ &=\boxed{120\ \mathrm m}. \end{aligned}$$

This is also a derivation of the constant acceleration result that displacement equals average velocity multiplied by time. See constant acceleration and SUVAT equations.

Self-check 2

A velocity time graph joins $(0,-6)$ to $(4,2)$ with a straight line. Find the acceleration, when the particle changes direction, the displacement, and the distance travelled.

Answer

The acceleration is

$$a=\frac{2-(-6)}{4}=2\ \mathrm{m\,s^{-2}}.$$

Since $v=-6+2t$, the velocity is zero at $t=3$ seconds. The signed trapezium area is

$$\text{displacement}=\frac{-6+2}{2}(4)=-8\ \mathrm m.$$

For distance, split at $t=3$:

$$\text{distance}=\frac12(3)(6)+\frac12(1)(2)=\boxed{10\ \mathrm m}.$$

Acceleration time graphs

The height of an acceleration time graph gives acceleration. Its signed area gives change in velocity:

$$\Delta v=v_{\text{final}}-v_{\text{initial}}.$$

Area does not give the final velocity unless the initial velocity is known:

$$v_{\text{final}}=v_{\text{initial}}+\text{signed area under the acceleration time graph}.$$

An acceleration graph below the axis means $a<0$. Whether the particle speeds up or slows down still depends on the sign of $v$.

Worked example 4: change in velocity

A particle has initial velocity $5\ \mathrm{m\,s^{-1}}$. Its acceleration is $4\ \mathrm{m\,s^{-2}}$ for $0\leq t\leq3$, then $-2\ \mathrm{m\,s^{-2}}$ for $3\leq t\leq8$. Find its velocity at $t=3$ and $t=8$.

During the first stage,

$$\Delta v=4(3)=12\ \mathrm{m\,s^{-1}},$$

so

$$v(3)=5+12=17\ \mathrm{m\,s^{-1}}.$$

During the second stage,

$$\Delta v=(-2)(5)=-10\ \mathrm{m\,s^{-1}},$$

so

$$v(8)=17-10=\boxed{7\ \mathrm{m\,s^{-1}}}.$$

Although acceleration is negative for the last $5$ seconds, velocity remains positive. The particle slows down but does not reverse.

Self-check 3

A particle begins with velocity $-3\ \mathrm{m\,s^{-1}}$. Its acceleration increases linearly from $0$ at $t=0$ to $6\ \mathrm{m\,s^{-2}}$ at $t=4$. Find its velocity at $t=4$.

Answer

The change in velocity is the triangular area:

$$\Delta v=\frac12(4)(6)=12\ \mathrm{m\,s^{-1}}.$$

Therefore

$$v(4)=-3+12=\boxed{9\ \mathrm{m\,s^{-1}}}.$$

Moving between graphs

To sketch a velocity graph from a displacement graph, track the displacement graph’s gradient, not its height. To sketch an acceleration graph from a velocity graph, track the velocity graph’s gradient.

For example, if a velocity time graph is:

• a straight line rising from $v=-4$ to $v=2$, acceleration is a positive constant throughout, including when $v=0$;
• horizontal at $v=2$, acceleration is zero;
• a straight line falling, acceleration is a negative constant.

At a sharp corner, the gradient changes instantaneously. In an idealised piecewise model, acceleration jumps and is not defined at that single instant. Do not invent a sloping transition unless the question supplies one.

Conversely, moving from an acceleration graph to a velocity graph requires an initial velocity. Moving from velocity to displacement requires an initial displacement. Area gives change, so without an initial value there are infinitely many vertically shifted graphs.

Common misconceptions

Position is not direction

A particle can have $s<0$ and $v>0$. It is then left of the origin but moving in the positive direction.

Negative velocity is not negative speed

Velocity may be negative, but speed is $|v|$ and is never negative.

Negative acceleration is not always deceleration

If $v<0$ and $a<0$, velocity becomes more negative and speed increases. Speed decreases only when $v$ and $a$ have opposite signs.

Zero velocity is not zero acceleration

At a reversal, $v$ may be zero while the velocity graph has a nonzero gradient, so $a\ne0$.

Area must carry units

Under a velocity time graph,

$$(\mathrm{m\,s^{-1}})(\mathrm s)=\mathrm m.$$

Under an acceleration time graph,

$$(\mathrm{m\,s^{-2}})(\mathrm s)=\mathrm{m\,s^{-1}}.$$

These unit products help identify what an area means.

Exam method

1. Read both axes, scales and units.
2. Mark any axis crossings and stage boundaries.
3. For a gradient, use two points on the relevant straight line or a tangent to a curve.
4. For displacement, use signed area.
5. For distance, split wherever velocity changes sign and add magnitudes.
6. State signs and units, then interpret negative answers in context.

Mixed self-check

A particle starts at displacement $s=4$ m with velocity $2\ \mathrm{m\,s^{-1}}$. Its velocity increases linearly to $8\ \mathrm{m\,s^{-1}}$ at $t=3$, then decreases linearly to $-4\ \mathrm{m\,s^{-1}}$ at $t=9$.

1. Find the acceleration on each stage.
2. Find when the particle changes direction.
3. Find its displacement from the origin at $t=9$.
4. Find the total distance travelled.

Answer

On the first stage,

$$a=\frac{8-2}{3}=2\ \mathrm{m\,s^{-2}}.$$

On the second,

$$a=\frac{-4-8}{9-3}=-2\ \mathrm{m\,s^{-2}}.$$

Starting from $v=8$ at $t=3$, velocity reaches zero $4$ seconds later, so the particle reverses at $t=7$.

The signed area under the velocity graph is

$$\frac{2+8}{2}(3)+\frac{8+(-4)}{2}(6)=15+12=27\ \mathrm m.$$

Therefore its displacement from the origin is

$$s(9)=4+27=\boxed{31\ \mathrm m}.$$

For distance, split the second stage at $t=7$:

$$\text{distance}=15+\frac12(4)(8)+\frac12(2)(4) =\boxed{35\ \mathrm m}.$$

Next steps

• Use straight line velocity graphs to understand and derive the constant acceleration equations.
• Extend gradients and areas to functions and curves in calculus in kinematics.
• Apply the same sign discipline to vertical and horizontal components in two dimensional motion and projectiles.