ScienceCBSE Class 9 · Describing Motion Around Us

Distance, Displacement, Speed, Velocity and Motion Graphs (Class 9)

Updated 17 October 2026 · Fact-checked

Distance travelled is the total path length (a scalar), while displacement is the net change in position with a direction (a vector). Average speed is total distance ÷ time, average velocity is displacement ÷ time, and average acceleration is change in velocity ÷ time. On a position-time graph the slope is velocity; on a velocity-time graph the slope is acceleration and the area under the graph is displacement.

Distance adds up the whole path (100 m), but displacement is only the straight change from start to end (20 m forward).

Illustration drawn for QuestionBuddy.

Understand Distance, Displacement, Speed, Velocity and Motion Graphs

Position and the reference point. To say where an object is, you need a fixed reference point (origin) and a direction. On a straight line, one side of the origin is taken as positive (+) and the other as negative (−), and this choice must stay the same through a problem. An object is in motion if its position changes with time.

Distance and displacement. Distance travelled is the total length of the path. It has size only (scalar) and never decreases. Displacement is the net change in position, from the first position to the final one; it has a magnitude and a direction (vector). The SI unit of both is the metre. The magnitude of displacement is never greater than the distance, and the two are equal only if the object moves in one direction without turning back. In a round trip, displacement is zero but distance is not.

Speed, velocity and acceleration. Average speed = total distance ÷ time interval. Average velocity = displacement ÷ time interval = s ÷ t. Both are measured in m s⁻¹ (or km h⁻¹). To convert: 1 km h⁻¹ = 5/18 m s⁻¹, so 18 km h⁻¹ = 5 m s⁻¹ and 72 km h⁻¹ = 20 m s⁻¹. Motion is uniform if equal distances are covered in equal time intervals. Average acceleration = (v − u) ÷ time interval, in m s⁻². Acceleration is in the direction of velocity if the object speeds up and opposite to it if the object slows down. An object moving fast at steady velocity has zero acceleration; acceleration measures how quickly the velocity changes, in size or in direction.

Position-time graphs. A graph is not a route map. A straight line parallel to the time axis means the object is at rest; a straight sloping line means constant velocity; a steeper line means a larger velocity; a curve getting steeper means velocity is increasing. Slope = (s₂ − s₁) ÷ (t₂ − t₁) = velocity. A line sloping downwards has a negative slope, so a negative velocity (moving in the negative direction).

Velocity-time graphs. A horizontal line means constant velocity (zero acceleration); a straight rising line means constant positive acceleration; a straight falling line means constant negative acceleration. Slope = acceleration. The area between the graph and the time axis = displacement. For velocity rising from u to v over time t, the area is a rectangle plus a triangle, which equals the trapezium (u + v)/2 × t. Area below the axis (negative velocity) counts as negative displacement, but for distance take its size.

Method for graph questions. Read values with units on both axes; split the graph into straight sections; find the slope or area of each; add areas for total displacement; use sizes of areas to find the total distance.

Distance, Displacement, Speed, Velocity and Motion Graphs: formulas and key facts

Average speed: Total distance ÷ time interval
Average velocity: $D i s pl a ce m e n t \div t im e in t er v a l = s \div t$
Average acceleration: (v − u) ÷ time interval, unit m s⁻²
Unit conversion: $1 k m h^{- 1} = 5/18 m s^{- 1} ∣1 m s^{- 1} = 18/5 k m h^{- 1} = 3.6 k m h^{- 1}$
Graph rules: $S l o p eo f p os i t i o n - t im e = v e l oc i t y ∣ S l o p eo f v e l oc i t y - t im e = a cce l er a t i o n ∣ A r e a u n d er v e l oc i t y - t im e = d i s pl a ce m e n t$

How to solve Distance, Displacement, Speed, Velocity and Motion Graphs questions

Use this method for motion numericals and graph questions.

1Choose an origin and a positive direction and write signs on velocities and positions.
2Convert all units to metre and second before you calculate.
3Separate distance (add all path lengths) from displacement (final position minus initial position).
4For graphs, divide the graph into straight-line sections and handle each separately.
5Find the slope (rise ÷ run) or area (rectangle, triangle or trapezium) of each section, keeping units.
6Check the answer: average speed is never smaller than the magnitude of average velocity, and the areas must add up to the total displacement.

Distance, Displacement, Speed, Velocity and Motion Graphs 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.

Distance versus displacement on a morning walk

Suppose a boy walks 600 m east from his gate to a park in 400 s, rests, and then walks 200 m back west in 200 s to buy a newspaper, taking the gate as origin and east as positive.

Find (a) the distance travelled, (b) his final position, (c) the average speed and (d) the average velocity for the 600 s of the walk, not counting rest.

Try it first, then show the solutionSolution

(a) Distance = 600 + 200 = 800 m.
(b) Final position = +600 − 200 = +400 m, i.e. 400 m east of the gate.
(c) Average speed = 800 m ÷ 600 s = 1.33 m s⁻¹ (to two decimal places).
(d) Average velocity = displacement ÷ time = 400 m ÷ 600 s = 0.67 m s⁻¹ towards the east.

Answer: (a) 800 m; (b) 400 m east; (c) 1.33 m s⁻¹; (d) 0.67 m s⁻¹ east.

Memory hook: Distance counts every step; displacement counts only start to finish.

Converting speedometer readings

Suppose a school bus has a speed limit of 54 km h⁻¹ on a campus road that is 1.2 km long.

Find the speed limit in m s⁻¹ and the least time the bus takes to cross the road at the limit.

Try it first, then show the solutionSolution

54 km h⁻¹ = 54 × 5/18 m s⁻¹ = 15 m s⁻¹.
1.2 km = 1200 m, so time = 1200 m ÷ 15 m s⁻¹ = 80 s.

Answer: 15 m s⁻¹; 80 s.

Memory hook: Multiply by 5/18 going to m s⁻¹.

Tricks to remember Distance, Displacement, Speed, Velocity and Motion Graphs

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

Distance: odometer; displacement: a straight arrow
Distance is what the odometer counts. Displacement is a straight arrow from where you started to where you ended.
Use it when: Separating distance from displacement in round-trip questions.
Slope tells the story
Position-time slope = velocity. Velocity-time slope = acceleration. Area under velocity-time = displacement.
Use it when: Every graph question.
5 over 18
km h⁻¹ to m s⁻¹: multiply by 5/18. 18 km h⁻¹ is 5 m s⁻¹, 36 is 10, 72 is 20.
Use it when: Quick unit conversion.
Speeding up: a and v agree
If the acceleration points the same way as the velocity, speed rises; opposite, speed falls.
Use it when: Sign questions.

Distance, Displacement, Speed, Velocity and Motion Graphs: solved examples

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

Example 1: Distance, displacement, average speed and average velocity · 3 marks · Short answerBoard level

A runner goes 300 m east in 60 s and then 100 m west in 40 s along a straight track. Take east as positive and the starting point as origin.
(a) Find the distance and the displacement. (1)
(b) Find the average speed. (1)
(c) Find the average velocity, with direction. (1)

Try it first, then show the solutionSolution

(a) Distance = 300 + 100 = 400 m. Displacement = +300 − 100 = +200 m, i.e. 200 m east. (1)
(b) Total time = 60 + 40 = 100 s. Average speed = 400 ÷ 100 = 4 m s⁻¹. (1)
(c) Average velocity = 200 ÷ 100 = 2 m s⁻¹ towards the east. (1)

Answer: (a) 400 m and 200 m east; (b) 4 m s⁻¹; (c) 2 m s⁻¹ east.

Example 2: Unit conversion and time of travel · 3 marks · Short answerBoard level

A car moves at 54 km h⁻¹ and a cyclist at 5 m s⁻¹ along the same straight road.
(a) Convert 54 km h⁻¹ to m s⁻¹ and 5 m s⁻¹ to km h⁻¹. (1)
(b) How long does the car take to travel 1.2 km? (1)
(c) How far does the cyclist travel in that time? (1)

Try it first, then show the solutionSolution

(a) 54 × 5/18 = 15 m s⁻¹. 5 × 18/5 = 18 km h⁻¹. (1)
(b) Time = 1200 m ÷ 15 m s⁻¹ = 80 s. (1)
(c) Distance = 5 m s⁻¹ × 80 s = 400 m. (1)

Answer: (a) 15 m s⁻¹ and 18 km h⁻¹; (b) 80 s; (c) 400 m.

Example 3: Reading a position-time table · 5 marks · Long answerHarder board level

The position of a toy car on a straight track (origin at the start, positive to the right) is: t = 0, 2, 4, 6, 8, 10 s; x = 0, 6, 12, 12, 4, −4 m.
(a) Find the velocity in each interval 0 to 4 s, 4 to 6 s and 6 to 10 s. (2)
(b) Find the total displacement and the total distance in 10 s. (1)
(c) Find the average speed and average velocity for the 10 s. (1)
(d) At what time does the car pass through the origin on its return? (1)

Try it first, then show the solutionSolution

(a) 0 to 4 s: (12 − 0) ÷ 4 = +3 m s⁻¹. 4 to 6 s: no change, 0 m s⁻¹ (at rest). 6 to 10 s: (−4 − 12) ÷ 4 = −4 m s⁻¹ (check 6 to 8 s: (4 − 12) ÷ 2 = −4, so the velocity is constant). (2)
(b) Displacement = −4 − 0 = −4 m. Distance = 12 m out + (12 − (−4)) = 16 m back = 28 m. (1)
(c) Average speed = 28 ÷ 10 = 2.8 m s⁻¹. Average velocity = −4 ÷ 10 = −0.4 m s⁻¹ (0.4 m s⁻¹ in the negative direction). (1)
(d) From x = 12 m at t = 6 s moving at −4 m s⁻¹, it reaches 0 after 12 ÷ 4 = 3 s, i.e. at t = 9 s. Check: at t = 8 s x = 4, so at 9 s x = 0. (1)

Answer: (a) +3, 0 and −4 m s⁻¹; (b) −4 m and 28 m; (c) 2.8 m s⁻¹ and −0.4 m s⁻¹; (d) 9 s.

Example 4: Velocity-time graph with three stages · 5 marks · Long answerHarder board level

A trolley's velocity is: it rises uniformly from 0 to 12 m s⁻¹ in the first 4 s, stays at 12 m s⁻¹ from 4 s to 10 s, and falls uniformly to 0 between 10 s and 14 s.
(a) Find the acceleration in each stage. (1)
(b) Find the total displacement using areas. (2)
(c) Find the average velocity over the 14 s. (1)
(d) At what times is the velocity 6 m s⁻¹? (1)

Try it first, then show the solutionSolution

(a) 0 to 4 s: (12 − 0) ÷ 4 = +3 m s⁻²; 4 to 10 s: 0 m s⁻²; 10 to 14 s: (0 − 12) ÷ 4 = −3 m s⁻². (1)
(b) Triangle 0 to 4 s: ½ × 4 × 12 = 24 m. Rectangle 4 to 10 s: 12 × 6 = 72 m. Triangle 10 to 14 s: ½ × 4 × 12 = 24 m. Total = 24 + 72 + 24 = 120 m. Check with the trapezium: ½ × (14 + 6) × 12 = 120 m. (2)
(c) Average velocity = 120 ÷ 14 = 8.57 m s⁻¹ (to two decimal places). (1)
(d) Rising stage: 3t = 6, so t = 2 s. Falling stage: 12 − 3(t − 10) = 6 gives t − 10 = 2, so t = 12 s. (1)

Answer: (a) +3, 0, −3 m s⁻²; (b) 120 m; (c) 8.57 m s⁻¹; (d) 2 s and 12 s.

Example 5: Two postmen walking towards each other · 4 marks · Case-basedHOTS / competency

Suppose two postmen start together from two posts 1200 m apart on a straight road. Postman A starts from the left post and walks to the right at 2 m s⁻¹; postman B starts from the right post and walks to the left at 3 m s⁻¹. Take the left post as origin and right as positive.
(a) Write the velocities of A and B with signs. (1)
(b) After how long do they meet and where? (1)
(c) What is the distance between them at t = 300 s, and who is further from the left post? (1)
(d) If A turns back at the meeting point and walks to his post at the same speed, find his total distance and displacement when he is back. (1)

Try it first, then show the solutionSolution

(a) A: +2 m s⁻¹. B: −3 m s⁻¹ (moving towards the origin). (1)
(b) They close the gap at 2 + 3 = 5 m s⁻¹, so time = 1200 ÷ 5 = 240 s. A has walked 2 × 240 = 480 m, so they meet 480 m from the left post (B has walked 3 × 240 = 720 m; 480 + 720 = 1200 m). (1)
(c) At t = 300 s, A is at 2 × 300 = 600 m and B is at 1200 − 3 × 300 = 300 m. They have crossed, and the distance between them is 600 − 300 = 300 m, with A further from the left post. (1)
(d) A walks 480 m out and 480 m back, so distance = 960 m and displacement = 0, since he is back at his post. (1)

Answer: (a) +2 and −3 m s⁻¹; (b) 240 s, 480 m from the left post; (c) 300 m apart, A further; (d) 960 m and 0 m.

Where students lose marks in Distance, Displacement, Speed, Velocity and Motion Graphs

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

Mistake: Using distance when displacement is asked, or the reverse.
Students treat both as the same 'how far'.
Wrong working
Round trip of 400 m: displacement = 400 m.
Right working
A round trip returns to the start, so displacement = 0; distance = 400 m.
Do this: Ask: does the question want the path length or the net change in position?
Usually costs: 1 to 2 marks.
Mistake: Not converting km h⁻¹ to m s⁻¹ before using with seconds.
Units are mixed in one line.
Wrong working
a = (72 − 36) ÷ 20 = 1.8 m s⁻²
Right working
a = (20 − 10) ÷ 20 = 0.5 m s⁻²
Do this: Convert at the start and write the unit with every number.
Usually costs: 2 marks.
Mistake: Reading a position-time graph as a path.
A graph of x against t looks like a hill, so it is mistaken for the shape of the route.
Wrong working
The car goes up a hill and then comes down.
Right working
The car moves away from the origin and then returns; the graph shows position against time.
Do this: Label the axes and describe the motion in words.
Usually costs: 1 mark.
Mistake: Ignoring the sign of velocity or acceleration.
Magnitudes are used when the direction matters.
Wrong working
Slowing down from 20 m s⁻¹ to 0 in 4 s: a = 5 m s⁻².
Right working
a = (0 − 20) ÷ 4 = −5 m s⁻², opposite to the velocity.
Do this: Always compute (v − u) ÷ t in that order.
Usually costs: 1 mark.
Mistake: Using the area of the whole graph as distance when the graph dips below the axis.
Areas below the time axis are negative displacement.
Wrong working
Distance = net area.
Right working
Displacement = net area; distance = sum of the sizes of all the areas.
Do this: Add the magnitudes for distance, the signed areas for displacement.
Usually costs: 1 to 2 marks.

Exam tips

Write every quantity with its unit and sign; marks in graph questions are often for 'slope = ...' written with units.
Find the distance and displacement separately and say which one the question needs.
In a velocity-time graph, divide into triangles, rectangles and trapezia and check with one other method.
For a position-time graph, write 'velocity = slope = (s₂ − s₁) ÷ (t₂ − t₁)' before substituting.
For 'average speed' over a journey with different speeds, use total distance ÷ total time, never the average of the speeds.

Practice questions on Distance, Displacement, Speed, Velocity and Motion Graphs

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

Q1 · 2 marks · Very short answerBoard level
Classify distance, speed, displacement, velocity and acceleration as scalar or vector, and give an example of a motion in which the displacement is zero but the distance travelled is not.
Hint
Which quantities have a direction?
Answer
Scalars: distance and speed. Vectors: displacement, velocity and acceleration (1 mark). Example: a runner who completes one lap of a circular track and ends where she started has displacement zero but distance equal to the lap length (1 mark).
Q2 · 3 marks · Short answerHarder board level
A train speeds up uniformly from 36 km h⁻¹ to 72 km h⁻¹ in 20 s. Find (a) its initial and final velocities in m s⁻¹, (b) its acceleration, and (c) the distance covered in this time.
Hint
Convert before substituting.
Answer
(a) u = 36 × 5/18 = 10 m s⁻¹; v = 72 × 5/18 = 20 m s⁻¹ (1 mark). (b) a = (20 − 10) ÷ 20 = 0.5 m s⁻² (1 mark). (c) Distance = average velocity × time = (10 + 20) ÷ 2 × 20 = 300 m (1 mark).
Q3 · 2 marks · Assertion-reasonBoard level
Assertion (A): A car moving at a constant speed of 20 m s⁻¹ round a circular track has zero acceleration. Reason (R): Its direction of motion keeps changing, so its velocity keeps changing. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.
Hint
Acceleration is about change in velocity, not only in speed.
Answer
(d). R is true: the direction of velocity changes at every instant. A is false: a changing velocity means non-zero acceleration even when the speed is constant.
Q4 · 3 marks · Short answerHarder board level
A cyclist rides along a straight road, and her position-time graph is a straight line through the points (2 s, 4 m) and (8 s, 28 m). Find (a) her velocity, (b) her position at t = 0, which is not the origin, and (c) her position at t = 5 s.
Hint
Slope gives velocity; extend the line back to t = 0.
Answer
(a) v = (28 − 4) ÷ (8 − 2) = 24 ÷ 6 = 4 m s⁻¹ (1 mark). (b) In 2 s the cyclist moves 4 × 2 = 8 m, so at t = 0 the position was 4 − 8 = −4 m (1 mark). (c) At t = 5 s: −4 + 4 × 5 = 16 m (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
A cart's velocity rises uniformly from 0 to 12 m s⁻¹ in 6 s, stays at 12 m s⁻¹ for the next 3 s, and then falls uniformly to 0 in 4 s. (a) Find the acceleration in each stage. (b) Find the total distance. (c) Find the average speed for the whole motion. (d) A student says the cart is back at its starting point because the final velocity is zero. Is she right? Give a reason.
Hint
Area under the graph is displacement; the velocity never became negative.
Answer
(a) +2 m s⁻², 0, −3 m s⁻² (1 mark). (b) ½ × 6 × 12 = 36 m; 12 × 3 = 36 m; ½ × 4 × 12 = 24 m; total 96 m (1 mark). (c) Time = 6 + 3 + 4 = 13 s, so average speed = 96 ÷ 13 = 7.38 m s⁻¹ (1 mark). (d) No: the velocity was always positive, so the cart moved in one direction and is 96 m from the start; zero final velocity only means it has stopped (1 mark).

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.What is the SI unit of acceleration?Show answer
m s⁻².
2.What does the slope of a velocity-time graph give?Show answer
Acceleration.
3.A velocity-time graph is a trapezium with parallel sides 4 and 10 m s⁻¹ and base 6 s. What is the displacement?Show answer
½ × (4 + 10) × 6 = 42 m.
4.Convert 72 km h⁻¹ to m s⁻¹.Show answer
20 m s⁻¹.
5.When are distance and the magnitude of displacement equal?Show answer
When the object moves in a straight line in one direction without turning back.
6.On a position-time graph, what does a downward-sloping straight line mean?Show answer
The object moves with constant negative velocity, back towards the origin if its position is positive.

Distance, Displacement, Speed, Velocity and Motion Graphs: frequently asked questions

Can average velocity be zero while average speed is not?

Yes. In a round trip the displacement is zero, so the average velocity is zero, but the distance, and therefore the average speed, is not zero.

Can speed be negative?

No. Speed is a magnitude. Velocity can be negative, which shows the direction.

Does a steeper position-time graph mean larger acceleration?

No. A steeper straight line means larger velocity. Acceleration appears as a curve in a position-time graph.

Is a car at constant speed on a bend accelerating?

Yes. Its direction changes, so its velocity changes, and acceleration is the rate of change of velocity.

Why can't we average two speeds to get the average speed?

Because the time spent at each speed may differ. Average speed is always total distance ÷ total time.