Q1–Q2MechanicsFoundation

Motion Graphs

A graph of motion carries more information than a list of values. Two operations extract it: the $\textbf{gradient}$ gives a rate of change, and the $\textbf{area}$ underneath gives an accumulated total. Everything in kinematics follows from those two.

30 min Kinematics

The big picture

It is worth noticing early that gradient and area are differentiation and integration in disguise. For straight-line graphs you can find them with triangles and rectangles; for curves you need calculus, which is exactly what the variable-acceleration lesson introduces. So this lesson is not a preliminary to be got through — it is the geometric picture that makes the calculus later feel inevitable rather than arbitrary. A student who can read a velocity–time graph fluently already understands what $\int v \,\mathrm{d}t$ means before meeting the notation.

What you'll be able to do

Interpret displacement–time and velocity–time graphs
Use gradients to find velocity and acceleration, including from tangents
Use areas under velocity–time graphs to find displacement
Distinguish distance from displacement on a graph with negative regions

Displacement–time graphs

On a displacement–time graph the gradient is the $\textbf{velocity}$ . A straight line means constant velocity; a horizontal line means the object is at rest; a steeper line means faster motion.

A negative gradient means motion in the negative direction — back towards, and possibly past, the starting point. Where the graph crosses the time axis, the displacement is zero and the object has returned to its start.

For a curve, the velocity at an instant is the gradient of the $\textbf{tangent}$ at that point. A curve bending upwards means the velocity is increasing.

v = \dfrac{\mathrm{d}s}{\mathrm{d}t}

Velocity is the gradient of

s

against

t

Example 1A displacement–time graph is a straight line from

(0, 2)

(5, 17)

, with

s

in metres and

t

in seconds. Find the velocity.

1Gradient:

\dfrac{17 - 2}{5 - 0} = \dfrac{15}{5}

Answer

3\ \mathrm{m\,s^{-1}}

, constant.

Tip — The area under a displacement–time graph has no physical meaning. Only velocity–time and acceleration–time graphs have useful areas.

Velocity–time graphs

On a velocity–time graph the gradient is the $\textbf{acceleration}$ and the area between the graph and the time axis is the $\textbf{displacement}$ .

A horizontal line means constant velocity and therefore zero acceleration. A straight sloping line means constant acceleration — which is precisely the situation the suvat equations describe.

Area below the axis counts as negative displacement, because the object is moving backwards during that interval. This is where distance and displacement separate: displacement is the signed total, distance is the sum of the magnitudes.

a = \dfrac{\mathrm{d}v}{\mathrm{d}t}

Acceleration is the gradient.

s = \int v \,\mathrm{d}t

Displacement is the area — signed.

Example 1A particle accelerates uniformly from rest to

8\ \mathrm{m\,s^{-1}}

in 4 s, holds that velocity for 6 s, then decelerates uniformly to rest in 2 s. Find the total distance.

1First phase: triangle of area

\tfrac12 \times 4 \times 8 = 16

2Second phase: rectangle of area

6 \times 8 = 48

3Third phase: triangle of area

\tfrac12 \times 2 \times 8 = 8

4Total:

16 + 48 + 8

Answer

72\ \mathrm{m}

Why this works

Splitting the area into triangles and rectangles works for any straight-line graph and keeps every step visible for method marks. It is also more reliable than trying to apply a trapezium formula to a shape with several sections.

Distance versus displacement on a graph

When a velocity–time graph crosses the time axis, the object reverses direction. The area above the axis is positive displacement and the area below is negative.

For $\textbf{displacement}$ , add the signed areas — the negative region partially or wholly cancels the positive one. For $\textbf{distance}$ , add the magnitudes, treating every area as positive.

Questions distinguish these deliberately, and the difference is often the whole point of including a region below the axis.

Example 1A velocity–time graph gives an area of

+20

m from 0 to 4 s and

-12

m from 4 to 7 s. Find the displacement and the distance travelled.

1Displacement: add signed areas,

20 + \left(-12\right)

2Distance: add magnitudes,

20 + 12

AnswerDisplacement

8

m; distance

32

Tip — If a question asks for "the distance travelled" and the graph dips below the axis, the answer is not the same as the displacement. Check for a crossing before answering.

Curved graphs and tangents

A curved velocity–time graph means the acceleration is changing. The acceleration at an instant is the gradient of the tangent at that point, found by drawing the tangent and computing its gradient.

The area under a curve cannot be split into exact triangles, so it is estimated — by counting squares, or by using the trapezium rule. Both give an approximation, and questions usually say which to use.

A curve that is getting steeper means the acceleration is increasing; one that is flattening means the acceleration is decreasing, even while the velocity itself continues to rise.

Example 1A velocity–time curve is flattening out as time increases, approaching a horizontal asymptote. Describe the motion.

1The gradient is positive but decreasing, so acceleration is positive but falling.

2Velocity continues to increase, but by less each second.

3Approaching a horizontal asymptote means acceleration tends to zero.

AnswerThe object speeds up at a decreasing rate, approaching a constant terminal velocity.

Why this works

That description is exactly terminal velocity, and it shows why reading gradients matters: "velocity increasing" and "acceleration decreasing" are simultaneously true and not contradictory.

Think like an examiner

Label axes with both the quantity and its units — it is frequently a mark.
For a curve, draw the tangent on the graph before computing its gradient so the method is visible.
Check whether the graph crosses the time axis before answering a distance question.
Split areas into triangles and rectangles and show each one, rather than quoting a single total.

Common misconceptions

Reading the gradient of a displacement–time graph as acceleration.

That gradient is velocity. Acceleration is the gradient of a velocity–time graph.

Treating the area under a displacement–time graph as meaningful.

It has no physical interpretation. Areas matter under velocity– and acceleration–time graphs.

Adding all areas as positive when finding displacement.

Regions below the axis are negative displacement. Only distance uses magnitudes.

Saying a flattening velocity–time curve means the object is slowing down.

It is still speeding up, just less rapidly. Deceleration needs a negative gradient.

Reading motion graphs

v = \tfrac{\mathrm{d}s}{\mathrm{d}t}

Gradient of

s

–

t

a = \tfrac{\mathrm{d}v}{\mathrm{d}t}

Gradient of

v

–

t

s = \int v\,\mathrm{d}t

Area under

v

–

t

\text{below axis} \Rightarrow \text{negative displacement}

Distance uses magnitudes.

Stretch yourself

A particle starts at rest and accelerates uniformly at $2\ \mathrm{m\,s^{-2}}$ for 5 s. It then decelerates uniformly, coming to rest after a further 5 s, and continues to accelerate in the negative direction at the same rate for 5 s more. Sketch the velocity–time graph and find the displacement and total distance.

Hint — Work out the velocity at each key time first, then compute the areas separately, watching signs.

Reveal solution

SolutionAfter 5 s the velocity is

2 \times 5 = 10\ \mathrm{m\,s^{-1}}

. It then decelerates to rest over the next 5 s, so the deceleration is

-2\ \mathrm{m\,s^{-2}}

, and continuing at that rate for 5 s more gives a velocity of

-10\ \mathrm{m\,s^{-1}}

t = 15

s. The graph is a triangle rising to

10

over

0

–

5

, a triangle falling to

0

over

5

–

10

, and a triangle descending to

-10

over

10

–

15

. Areas:

\tfrac12(5)(10) = 25

m, then

\tfrac12(5)(10) = 25

m, then

\tfrac12(5)(10) = 25

m but

\textbf{below}

the axis, so

-25

m. Displacement

= 25 + 25 - 25 = 25

m. Distance

= 25 + 25 + 25 = 75

m. The particle ends 25 m from its start having travelled 75 m — and note the velocity graph is a straight line throughout, since the acceleration only changes sign once.

\blacksquare

Questions students ask

How do I find acceleration from a curved velocity–time graph?

Draw a tangent at the point of interest and find its gradient. Show the tangent on the diagram — the method is credited even if the reading is slightly off.

Why does area below the axis count as negative?

Because the velocity is negative there, so the object is moving in the negative direction and its displacement is decreasing.

Can a displacement–time graph be vertical?

No. That would mean infinite velocity — the object being in two places at once. A vertical section indicates an error in the graph.

What does the area under an acceleration–time graph give?

The change in velocity, by the same logic that makes the area under a velocity–time graph the change in displacement.

Key takeaways

Gradient of displacement–time is velocity; gradient of velocity–time is acceleration.
Area under a velocity–time graph is displacement, signed.
Distance sums the magnitudes of areas; displacement sums them with signs.
For curves, use tangents for gradients and estimation for areas.
Gradient and area are differentiation and integration seen geometrically.

How this fits the course

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