PhysicsICSE Class 9 · Motion in One Dimension

Distance–Time and Velocity–Time Graphs: Reading Slope and Area (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

On a distance–time graph the slope is the speed (or velocity on a displacement–time graph); a horizontal line means rest. On a velocity–time graph the slope is the acceleration and the area between the graph and the time axis is the displacement (the distance if the velocity does not change sign). Split the graph into triangles and rectangles.

Slope of a distance–time line gives speed, slope of a velocity–time line gives acceleration, and the area under a velocity–time graph gives displacement.

Illustration drawn for QuestionBuddy.

Understand Distance–Time and Velocity–Time Graphs: Slope and Area

Distance–time (or displacement–time) graph.

A straight line sloping up: uniform speed; slope $= \frac{Δ s}{Δ t}$ = speed (velocity).
A line parallel to the time axis: the body is at rest.
A curve that gets steeper: speed is increasing (accelerated motion); a curve that flattens: speed is decreasing.
A steeper straight line means a greater speed.
A distance–time graph never slopes downward (distance cannot decrease), but a displacement–time graph can, meaning the body is coming back.
Two lines crossing means the two bodies are at the same position at that time.

Velocity–time graph.

A line parallel to the time axis: uniform velocity (acceleration zero).
A straight line sloping up: uniform acceleration = slope $= \frac{Δ v}{Δ t}$ . Sloping down: uniform retardation (negative slope).
A line below the time axis: velocity is in the negative direction (the body has reversed).
Area between the graph and the time axis = displacement. Areas above the axis count positive, those below negative. Distance is the sum of the magnitudes of all areas.

Areas to remember. Rectangle: $v \times t$ . Triangle: $\frac{1}{2} \times base \times height$ . Trapezium: $\frac{1}{2} (sum of parallel sides) \times width$ .

Average speed and average velocity. Average speed $= \frac{total distance}{total time}$ ; average velocity $= \frac{total displacement}{total time}$ . A body that returns to its start has average velocity zero but non-zero average speed.

Reading a graph step by step.
1. Split the time axis wherever the graph changes shape.
2. For each piece, state the motion (rest, uniform velocity, uniform acceleration, retardation).
3. Find the slope (rate) and area (change in position) of each piece.
4. Add pieces for totals; do not average rates over unequal intervals.

Cautions. A velocity–time graph is not a picture of the path. A rising line on it does not mean the body is going uphill, only that it is speeding up.

Distance–Time and Velocity–Time Graphs: Slope and Area: formulas and key facts

Slope of s–t graph: $speed = \frac{Δ s}{Δ t}$; Velocity for displacement.
Slope of v–t graph: $a = \frac{Δ v}{Δ t}$; Negative slope = retardation.
Area under v–t graph: $displacement = area (above) - area (below)$; Distance: add the magnitudes.
Average values: $avg speed = \frac{distance}{time}; avg velocity = \frac{displacement}{time}$; Over the whole interval.

How to solve Distance–Time and Velocity–Time Graphs: Slope and Area questions

Convert the picture to numbers piece by piece.

1Mark the points where the graph changes shape.
2For each piece compute the slope using two clean points.
3Compute each area as a triangle, rectangle or trapezium.
4Treat areas below the axis as negative for displacement and positive for distance.
5Divide the totals by the full time for averages.

Distance–Time and Velocity–Time Graphs: Slope and Area in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A lift ride

Suppose a lift in a building starts from the ground floor, speeds up for 4 s at a steady rate, travels at constant speed for 6 s and then slows to a stop in 3 s.

If the top speed is 6 m s⁻¹, how high does the lift rise?

Try it first, then show the solutionSolution

Area $= \frac{1}{2} (4) (6) + (6) (6) + \frac{1}{2} (3) (6) = 12 + 36 + 9 = 57$ m.

Answer: 57 m.

Memory hook: Trapezium: add the pieces.

Tricks to remember Distance–Time and Velocity–Time Graphs: Slope and Area

Short memory aids. Say each one aloud once, then use it on a question the same day.

Slope speeds, area travels
On v–t: slope = a, area = distance. On s–t: slope = speed.
Use it when: Which quantity?
Flat means different things
Flat s–t: rest. Flat v–t: constant velocity, not rest.
Use it when: Common trick questions.
Cross means meet
Two s–t lines crossing: same place, same time.
Use it when: Two-body graphs.

Distance–Time and Velocity–Time Graphs: Slope and Area: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Flat distance–time line · 1 mark · MCQBoard level

A horizontal line on a distance–time graph means the body is:
(a) moving with uniform speed
(b) accelerating
(c) at rest
(d) retarding

Try it first, then show the solutionSolution

Distance does not change with time: the body is at rest.

Answer: (c) at rest

Example 2: Speed from a straight line · 2 marks · Short answerBoard level

A distance–time graph is a straight line from the origin to the point (8 s, 20 m). Find the speed and the distance at $t = 5$ s.

Try it first, then show the solutionSolution

Speed $= \frac{20}{8} = 2.5$ m s⁻¹. (1)
Distance at 5 s $= 2.5 \times 5 = 12.5$ m. (1)

Answer: 2.5 m s⁻¹; 12.5 m

Example 3: A trapezium graph · 3 marks · Short answerHarder board level

A body accelerates uniformly from rest to 6 m s⁻¹ in 3 s, moves at 6 m s⁻¹ for 5 s, then retards uniformly to rest in 2 s. Find (a) the two accelerations, (b) the total distance, (c) the average speed.

Try it first, then show the solutionSolution

(a) $a_{1} = \frac{6}{3} = 2$ m s⁻²; $a_{2} = \frac{0 - 6}{2} = - 3$ m s⁻². (1)
(b) $\frac{1}{2} (3) (6) + 6 (5) + \frac{1}{2} (2) (6) = 9 + 30 + 6 = 45$ m. (1)
(c) $\frac{45}{10} = 4.5$ m s⁻¹. (1)

Answer: (a) 2 and −3 m s⁻² (b) 45 m (c) 4.5 m s⁻¹

Example 4: Velocity that reverses · 3 marks · Short answerHarder board level

The velocity of a body falls uniformly from $+ 8$ m s⁻¹ at $t = 0$ to $- 4$ m s⁻¹ at $t = 6$ s. Find (a) the acceleration, (b) the instant it is momentarily at rest, (c) its displacement and (d) the distance over 6 s.

Try it first, then show the solutionSolution

(a) $a = \frac{- 4 - 8}{6} = - 2$ m s⁻². (½)
(b) $0 = 8 - 2 t$ , $t = 4$ s. (½)
(c) Area above $= \frac{1}{2} (4) (8) = 16$ m; area below $= \frac{1}{2} (2) (4) = 4$ m (negative). Displacement $= 16 - 4 = 12$ m. (1)
(d) Distance $= 16 + 4 = 20$ m. (1)

Answer: (a) −2 m s⁻² (b) 4 s (c) 12 m (d) 20 m

Example 5: Two cars on one graph · 4 marks · Case-basedHOTS / competency

Car A and car B move along the same road. On a distance–time graph, A is a straight line from the origin with slope 3 m s⁻¹; B is a straight line starting at 24 m on the distance axis with slope 1 m s⁻¹, both moving away.
(a) Which car is faster?
(b) At what time and place do the lines cross?
(c) What does the crossing mean?
(d) What is the gap between the cars at $t = 20$ s?

Try it first, then show the solutionSolution

(a) A, since its slope (3 m s⁻¹) is greater. (1)
(b) $3 t = 24 + t$ , so $t = 12$ s; position $3 \times 12 = 36$ m. (1)
(c) A overtakes B: they are at the same position at 12 s. (1)
(d) A at 60 m, B at $24 + 20 = 44$ m; gap $= 16$ m (A ahead). (1)

Answer: (a) A (b) 12 s, 36 m (c) overtaking (d) 16 m

Where students lose marks in Distance–Time and Velocity–Time Graphs: Slope and Area

The slips examiners see most often, and what to write instead.

Mistake: Reading a flat v–t line as rest
It shows constant velocity, not zero velocity.
Wrong working
The body is at rest.
Right working
Uniform velocity; acceleration zero.
Do this: Check the axis labels first.
Usually costs: The reasoning mark.
Mistake: Adding areas below the axis for displacement
Areas below the axis are negative.
Wrong working
Displacement = 16 + 4 = 20 m
Right working
Displacement = 16 − 4 = 12 m; distance = 20 m
Do this: Separate displacement from distance.
Usually costs: 1 mark.
Mistake: Average speed taken as mean of the two end speeds for unequal intervals
Average = total distance ÷ total time.
Wrong working
(0 + 6 + 0)/3 = 2 m s⁻¹.
Right working
Total area ÷ total time.
Do this: Compute total distance first.
Usually costs: The answer.

Exam tips

State "slope of the v–t graph = acceleration" in words before using it.
Draw the pieces separately when a graph has several parts.
Show each area as a formula with numbers before adding.
For a reversing velocity, find the instant when v = 0 first.

Practice questions on Distance–Time and Velocity–Time Graphs: Slope and Area

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A distance–time graph is a curve that bends upward (gets steeper). The body is moving with:
(a) uniform speed
(b) increasing speed
(c) decreasing speed
(d) zero speed
Hint
Slope increasing.
Answer
(b) increasing speed
Q2 · 1 mark · Very short answerBoard level
What does a horizontal straight line on a velocity–time graph, above the time axis, represent?
Hint
Think of the slope.
Answer
Uniform velocity (zero acceleration).
Q3 · 3 marks · Short answerHarder board level
A body moves 10 m away from the start in 5 s, rests for 5 s, then returns to the start in 5 s. Find its average speed and average velocity over the 15 s.
Hint
Distance and displacement differ.
Answer
Distance $= 20$ m, so average speed $= \frac{20}{15} \approx 1.33$ m s⁻¹. Displacement $= 0$ , so average velocity $= 0$ .
Q4 · 2 marks · Short answerHarder board level
A velocity–time graph is a straight line from (0 s, 12 m s⁻¹) to (6 s, 0). Find the acceleration and the distance covered in the first 4 s.
Hint
Slope, then the velocity at 4 s and a trapezium.
Answer
$a = \frac{0 - 12}{6} = - 2$ m s⁻²; $v_{4} = 12 - 8 = 4$ m s⁻¹; distance $= \frac{12 + 4}{2} \times 4 = 32$ m.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A velocity–time graph that lies wholly below the time axis shows a body moving in the negative direction.
Reason (R): Velocity is a vector, so its sign tells its direction.
(a) Both true and R explains A.
(b) Both true but R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Sign of a vector.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A goods lift accelerates uniformly from rest for 5 s to a speed of 4 m s⁻¹, climbs at that speed for 8 s and then stops uniformly in 2 s. Find the acceleration in the first part, the retardation in the last part and the height gained. (Do the same sum a different way by using the trapezium formula.)
Hint
Trapezium: parallel sides are the total time and the constant-speed time.
Answer
$a = \frac{4}{5} = 0.8$ m s⁻²; retardation $= \frac{4}{2} = 2$ m s⁻². Height $= \frac{1}{2} (15 + 8) \times 4 = 46$ m, which equals $10 + 32 + 4 = 46$ m.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Slope of a velocity–time graph?Show answer
Acceleration.
2.Which part of a v–t graph gives the distance travelled?Show answer
The area between the graph and the time axis.
3.Meaning of crossing d–t lines?Show answer
The bodies are at the same position at that time.
4.Flat line on an s–t graph?Show answer
Rest.
5.Negative slope on a v–t graph?Show answer
Retardation.

Distance–Time and Velocity–Time Graphs: Slope and Area: frequently asked questions

What does the slope of a distance–time graph show?

The speed of the body; a steeper line means a greater speed, and a horizontal line means the body is at rest.

How do you find distance from a velocity–time graph?

Find the area between the graph and the time axis, splitting it into rectangles, triangles and trapezia.

Can a distance–time graph slope downward?

No; distance never decreases. A displacement–time graph can slope downward when the body returns.