A stone is dropped from rest. Using g = 10\ m s^-2, find its speed after 2.0\ s. [2 marks]

SEAB Original-style practice: A velocity-time graph shows a car accelerating uniformly from rest to 20\ m s^-1 in 8.0\ s, then travelling at 20\ m s^-1 for 12\ s. (a) Find the acceleration in the first stage. (b) Find the total distance travelled.

(a) Acceleration is the gradient of a velocity-time graph: a = 20 - 08.0 = 2.5\ m s^-2. (b) Distance is the area under the graph. First stage (triangle): 12 × 8.0 × 20 = 80\ m. Second stage (rectangle): 20 × 12 = 240\ m. Total = 80 + 240 = 320\ m. Markers reward gradient as acceleration, area under the graph as distance, splitting the shape into a triangle and a rectangle, and the correct total with units.

SingaporePhysicsSyllabus dot point

How do distance-time and velocity-time graphs describe motion, and what happens to a body in free fall?

Interpret distance-time and velocity-time graphs and describe free fall and the effect of air resistance on a falling body

A focused answer to the O-Level Physics outcome on motion graphs and free fall. Reading gradients and areas on distance-time and velocity-time graphs, acceleration of free fall, and how air resistance leads to terminal velocity.

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

SEAB wants you to read and interpret distance-time and velocity-time graphs, knowing that a gradient and an area each have a physical meaning, and to describe free fall under gravity and how air resistance changes the motion of a real falling object until it reaches terminal velocity.

The answer

Distance-time graphs

On a distance-time graph the gradient is the speed.

A horizontal line means the object is at rest (distance not changing).
A straight slope means constant speed.
A curve that gets steeper means the object is speeding up.

\text{speed} = \text{gradient} = \frac{\text{change in distance}}{\text{change in time}}

Velocity-time graphs

On a velocity-time graph two things have meaning. The gradient is the acceleration, and the area under the line is the distance travelled.

A horizontal line means constant velocity (zero acceleration).
A straight slope means constant acceleration.
A line sloping down means deceleration.

\text{acceleration} = \text{gradient}, \qquad \text{distance} = \text{area under the graph}

For a triangle the area is $\tfrac{1}{2} \times \text{base} \times \text{height}$ ; for a rectangle it is base times height.

Free fall

Free fall is motion under gravity alone, with no air resistance. Near the Earth's surface every object in free fall has the same acceleration, the acceleration of free fall:

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

(the value used at O-Level). So a dropped object gains $10\ \text{m s}^{-1}$ of speed every second, whatever its mass. In a vacuum a feather and a hammer fall together.

Air resistance and terminal velocity

Real falling objects meet air resistance, which acts upward and grows as the object speeds up. At first weight is much bigger than air resistance, so the object accelerates. As speed rises, air resistance rises until it equals the weight. The resultant force is then zero, the acceleration is zero, and the object falls at a steady maximum speed called the terminal velocity.

Worked example

A ball is dropped from rest and falls freely. Take $g = 10\ \text{m s}^{-2}$ . (a) Find its velocity after $3.0\ \text{s}$ . (b) Find the distance it has fallen in that time.

Step 1: Find the velocity

In free fall the acceleration is $g$ . Use $v = u + at$ with $u = 0$ , $a = 10\ \text{m s}^{-2}$ , $t = 3.0\ \text{s}$ :

v = 0 + 10 \times 3.0 = 30\ \text{m s}^{-1}

Step 2: Find the distance using the area idea

On a velocity-time graph the motion is a triangle from $0$ to $30\ \text{m s}^{-1}$ over $3.0\ \text{s}$ . Distance is the area:

\text{distance} = \tfrac{1}{2} \times 3.0 \times 30 = 45\ \text{m}

Step 3: State the answers

After $3.0\ \text{s}$ the ball moves at $30\ \text{m s}^{-1}$ and has fallen $45\ \text{m}$ . The same answers come from the area under a velocity-time graph, which is why the graph method is so useful.

Examples in context

Example 1. A bus journey graph. A distance-time graph for a bus shows sloping lines while it moves and flat sections at bus stops. The steepest slope is where the bus moves fastest, and a flat section means it is stationary, so the shape tells the whole story of the trip at a glance.

Example 2. A parachute opening. A skydiver reaches a high terminal velocity in free fall, then opens a parachute. The large area of the parachute greatly increases air resistance, so for a moment air resistance exceeds weight and she decelerates to a new, much lower terminal velocity safe for landing.

Try this

Q1. State what the gradient of a distance-time graph represents and what the area under a velocity-time graph represents. [2 marks]

Cue. Gradient of distance-time graph is speed; area under a velocity-time graph is distance travelled.

Q2. A stone is dropped from rest. Using $g = 10\ \text{m s}^{-2}$ , find its speed after $2.0\ \text{s}$ . [2 marks]

Cue. $v = u + at = 0 + 10 \times 2.0 = 20\ \text{m s}^{-1}$ .

Q3. Explain why a falling object reaches a terminal velocity rather than speeding up forever. [3 marks]

Cue. Air resistance grows with speed until it equals the weight; the resultant force is then zero, so the object stops accelerating and falls at a constant maximum speed.

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.

Original5 marksA velocity-time graph shows a car accelerating uniformly from rest to

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

8.0\ \text{s}

, then travelling at

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

for

12\ \text{s}

. (a) Find the acceleration in the first stage. (b) Find the total distance travelled.

Show worked answer →

(a) Acceleration is the gradient of a velocity-time graph: $a = \dfrac{20 - 0}{8.0} = 2.5\ \text{m s}^{-2}$ .

(b) Distance is the area under the graph. First stage (triangle): $\tfrac{1}{2} \times 8.0 \times 20 = 80\ \text{m}$ . Second stage (rectangle): $20 \times 12 = 240\ \text{m}$ . Total $= 80 + 240 = 320\ \text{m}$ .

Markers reward gradient as acceleration, area under the graph as distance, splitting the shape into a triangle and a rectangle, and the correct total with units.

Original4 marksA skydiver jumps from a plane. (a) Describe and explain how her velocity changes from the moment she jumps until she reaches terminal velocity. (b) State the resultant force on her at terminal velocity.

Show worked answer →

(a) At first her weight is much larger than air resistance, so she accelerates and speeds up. As she speeds up, air resistance increases. Eventually air resistance grows until it equals her weight, so the resultant force is zero and she falls at a constant maximum speed, the terminal velocity.

(b) At terminal velocity the resultant force is zero, because air resistance now balances weight.

Markers reward the idea that air resistance increases with speed, that acceleration falls as the forces approach balance, and that terminal velocity is reached when air resistance equals weight giving zero resultant force.

What this dot point is asking

The answer

Distance-time graphs

Velocity-time graphs

Free fall

Air resistance and terminal velocity

Examples in context

Try this

Exam-style practice questions

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