SEAB Original-style practice: A speed-time graph shows a car speeding up steadily from 0 to 12\ m s^-1 in the first 4.0\ s, then staying at 12\ m s^-1 for the next 6.0\ s. (a) Find the acceleration in the first 4.0\ s. (b) Find the total distance travelled in the 10\ s.

(a) Acceleration is the gradient of the speed-time graph: a = 12 - 04.0 = 3.0\ m s^-2. (b) Distance is the area under the graph. The first part is a triangle: 12 × 4.0 × 12 = 24\ m. The second part is a rectangle: 12 × 6.0 = 72\ m. Total = 24 + 72 = 96\ m. What markers reward: gradient for acceleration, area under the graph for distance, splitting the area into a triangle and a rectangle, and adding them.

SingaporePhysicsSyllabus dot point

What do distance-time and speed-time graphs tell us about motion, and how do objects fall under gravity?

Interpret distance-time and speed-time graphs, and describe free fall and the effect of air resistance

Read distance-time and speed-time graphs, find speed from a gradient and distance from an area, and describe how objects fall under gravity with and without air resistance at N(A)-Level.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to read distance-time and speed-time graphs, to get speed from the gradient of a distance-time graph and acceleration from the gradient of a speed-time graph, to get distance from the area under a speed-time graph, and to describe how objects fall under gravity with and without air resistance. The big idea is that the shape of a motion graph tells the whole story of the motion.

The answer

Distance-time graphs

On a distance-time graph, the distance is on the vertical axis and time on the horizontal axis.

A horizontal line means the object is not moving (distance is not changing).
A straight sloping line means a steady speed.
The gradient (steepness) of the line is the speed: a steeper line means a faster speed.

So speed $= \dfrac{\text{change in distance}}{\text{change in time}}$ , which is the gradient of the line.

Speed-time graphs

On a speed-time graph, speed is on the vertical axis and time on the horizontal axis. These graphs carry two pieces of information.

The gradient of the line is the acceleration. A horizontal line means steady speed (zero acceleration); a sloping line means the object is speeding up or slowing down.
The area under the line is the distance travelled.

For a triangle the area is $\dfrac{1}{2} \times \text{base} \times \text{height}$ , and for a rectangle it is $\text{length} \times \text{width}$ . Real graphs are often split into triangles and rectangles, and you add the parts.

Free fall

Free fall is motion under gravity alone, with no air resistance. Near the Earth all objects fall with the same acceleration, the acceleration of free fall:

g \approx 10\ \text{m s}^{-2}

This means a falling object gains about $10\ \text{m s}^{-1}$ of speed every second. A heavy object and a light object dropped together in the absence of air resistance hit the ground at the same time, because the acceleration does not depend on mass.

Air resistance and terminal velocity

In real air there is a drag force that pushes up against the motion. Drag grows as the object speeds up.

At the start, weight is much bigger than drag, so the object accelerates.
As it speeds up, drag rises, so the resultant force and the acceleration fall.
Eventually drag equals weight, the resultant force is zero, and the object falls at a steady speed called the terminal velocity.

This is why a feather, with a large drag for its weight, falls slowly, while a stone falls fast.

Worked example

A cyclist's speed-time graph is a straight line from $4.0\ \text{m s}^{-1}$ at time $0$ up to $10\ \text{m s}^{-1}$ at $6.0\ \text{s}$ . Find (i) the acceleration, and (ii) the distance travelled in the $6.0\ \text{s}$ .

Step 1: Find the acceleration from the gradient

a = \frac{\text{change in speed}}{\text{time}} = \frac{10 - 4.0}{6.0} = 1.0\ \text{m s}^{-2}

Step 2: Set up the area under the graph

The area is a trapezium. Split it into a rectangle (height $4.0$ , width $6.0$ ) and a triangle (base $6.0$ , height $10 - 4.0 = 6.0$ ).

Step 3: Work out each part

Rectangle: $4.0 \times 6.0 = 24\ \text{m}$ . Triangle: $\dfrac{1}{2} \times 6.0 \times 6.0 = 18\ \text{m}$ .

Step 4: Add the parts

\text{distance} = 24 + 18 = 42\ \text{m}

The acceleration is $1.0\ \text{m s}^{-2}$ and the cyclist travels $42\ \text{m}$ .

Examples in context

Example 1. A journey with a stop. A distance-time graph for a walk to a shop rises steadily, then is flat while the person waits at a crossing, then rises again. The flat section shows zero speed; the steeper sections show faster walking. Reading the gradient of each sloping part gives the speed during that part.

Example 2. A skydiver. A skydiver speeds up after jumping, then drag grows until it balances weight and they fall at a terminal velocity. Opening the parachute suddenly increases drag, so they decelerate to a new, slower terminal velocity safe for landing.

Try this

Cue. A distance-time graph is a straight line from the origin reaching $30\ \text{m}$ at $5.0\ \text{s}$ . Find the speed. [2 marks] Speed is the gradient: $\dfrac{30}{5.0} = 6.0\ \text{m s}^{-1}$ .
Cue. A speed-time graph shows a steady speed of $8.0\ \text{m s}^{-1}$ for $5.0\ \text{s}$ . Find the distance. [2 marks] Distance is the area: $8.0 \times 5.0 = 40\ \text{m}$ .
Cue. Explain why a stone and a feather dropped in a vacuum land together, but not in air. [3 marks] In a vacuum both have the same acceleration of free fall with no drag; in air the feather has large drag for its weight, so it reaches a low terminal velocity and lands later.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA speed-time graph shows a car speeding up steadily from

0

12\ \text{m s}^{-1}

in the first

4.0\ \text{s}

, then staying at

12\ \text{m s}^{-1}

for the next

6.0\ \text{s}

. (a) Find the acceleration in the first

4.0\ \text{s}

. (b) Find the total distance travelled in the

10\ \text{s}

Show worked answer →

(a) Acceleration is the gradient of the speed-time graph: $a = \dfrac{12 - 0}{4.0} = 3.0\ \text{m s}^{-2}$ .

(b) Distance is the area under the graph. The first part is a triangle: $\dfrac{1}{2} \times 4.0 \times 12 = 24\ \text{m}$ . The second part is a rectangle: $12 \times 6.0 = 72\ \text{m}$ . Total $= 24 + 72 = 96\ \text{m}$ .

What markers reward: gradient for acceleration, area under the graph for distance, splitting the area into a triangle and a rectangle, and adding them.

Original3 marksA ball is dropped from rest and falls freely. (a) State the value of its acceleration. (b) Describe how its speed changes as it falls. (c) Explain what changes if there is significant air resistance.

Show worked answer →

(a) The acceleration of free fall is about $10\ \text{m s}^{-2}$ (taking $g = 10\ \text{m s}^{-2}$ ).

(b) Its speed increases steadily because gravity gives it a constant acceleration downwards.

(c) With air resistance, the upward drag grows as the ball speeds up. The resultant force and so the acceleration fall, until drag equals weight and the ball falls at a constant terminal velocity.

What markers reward: stating $g \approx 10\ \text{m s}^{-2}$ , speed increasing under constant acceleration, and the idea of drag rising to give a terminal velocity.