1Reference point, distance and displacement
An object is in motion if its position changes with time relative to a reference point (origin). Rest and motion are relative: a passenger in a moving bus is at rest relative to the bus but moving relative to the road.
Distance is the total length of the path travelled. It has only size (magnitude), so it is a scalar. Displacement is the shortest straight-line change in position from start to finish, with direction, so it is a vector.
Displacement can be zero even when distance is not — run one full lap of a 400 m track and your distance is 400 m but your displacement is 0. The magnitude of displacement is never greater than the distance.
- Distance: path length, scalar, always ≥ 0.
- Displacement: final − initial position, vector, can be 0 or negative.
A boy walks 4 m east and then 3 m north. Find the distance and displacement.
- Distance m.
- Displacement is the hypotenuse: m, towards the north-east (from his start).
Answer: Distance 7 m; displacement 5 m north of east (about 37° north of east).
- Giving displacement without a direction.
- Adding displacements in different directions as plain numbers.
2Speed and velocity
Speed is distance travelled per unit time. Average speed . Its SI unit is m/s; km/h is common in daily life.
Velocity is speed in a given direction: average velocity . It is a vector and can be negative (moving in the negative direction).
Motion is uniform if equal distances are covered in equal intervals of time (in a straight line, this means constant velocity) and non-uniform otherwise. When velocity changes uniformly, average velocity .
- Average speed
- Average velocity ; for uniform acceleration
- ;
A car goes 60 km at 30 km/h and returns at 60 km/h. Find the average speed and average velocity for the round trip.
- Time out h, time back h, total 3 h; distance 120 km.
- Average speed km/h (not 45!).
- Displacement , so average velocity .
Answer: 40 km/h and 0.
To convert km/h → m/s multiply by 5/18 (72 km/h = 20 m/s); m/s → km/h multiply by 18/5.
- Averaging speeds directly when times differ.
- Forgetting to convert km/h to m/s before using SI formulas.
3Acceleration
Acceleration is the rate of change of velocity: , where is the initial velocity and the final velocity after time . Its SI unit is m/s² (metres per second, per second).
If velocity increases in the direction of motion, acceleration is positive. If it decreases, acceleration is negative — called retardation or deceleration (a braking car).
Velocity also changes when direction changes. So an object moving in a circle at constant speed is accelerating — its velocity keeps turning. Uniform acceleration means equal changes in velocity in equal time intervals, such as a freely falling stone (about 9.8 m/s² downward).
- , unit m/s²
- Negative (opposite to motion) = retardation.
- Uniform circular motion: constant speed, changing velocity ⇒ accelerated.
A bus moving at 54 km/h brakes and stops in 5 s. Find its acceleration.
- m/s, , s.
- m/s².
Answer: m/s² (a retardation of 3 m/s²).
- Thinking an object at constant speed cannot be accelerating.
- Dropping the minus sign for retardation.
4Motion graphs
On a distance–time (or position–time) graph, the slope gives the speed (velocity). A straight sloping line = uniform motion; a steeper line = faster; a horizontal line = at rest; a curve = changing speed.
On a velocity–time graph, the slope gives the acceleration, and the area under the graph gives the displacement. A horizontal line = constant velocity; a straight sloping line = uniform acceleration; a line going down = retardation.
Read graphs carefully: check the axes, the units and the scale first. A graph is not a picture of the path — a straight line on a d–t graph does not mean the road is straight.
- d–t graph: slope = speed
- v–t graph: slope = acceleration; area = displacement
- Area of a trapezium under a v–t line:
A car's velocity rises uniformly from 0 to 20 m/s in 10 s, stays at 20 m/s for 20 s, then falls uniformly to 0 in 5 s. Find the total distance from the v–t graph.
- Triangle (0–10 s): m.
- Rectangle (10–30 s): m.
- Triangle (30–35 s): m.
- Total m.
Answer: 550 m.
- Reading the slope of a v–t graph as speed.
- Ignoring the scale (e.g. time in minutes, not seconds).
5Equations of motion
For motion in a straight line with uniform acceleration, three equations connect (initial velocity), (final velocity), (acceleration), (time) and (displacement).
They come straight from the v–t graph: the slope gives ; the area of the trapezium gives ; eliminating gives .
Choose the equation that contains the three quantities you know and the one you want. Pick a positive direction and keep signs consistent. These equations do not apply when acceleration is changing.
A train starts from rest and accelerates uniformly at 0.5 m/s² for 2 minutes. Find its final speed and the distance covered.
- , m/s², s.
- m/s.
- m km.
Answer: 60 m/s; 3.6 km.
A car moving at 20 m/s brakes with a retardation of 4 m/s². What is its stopping distance? If the driver's reaction time is 0.5 s, what is the total stopping distance?
- Braking: m.
- During the reaction time the car moves at 20 m/s: m.
- Total m.
Answer: 50 m braking; 60 m in total.
List u, v, a, s, t, tick the three you know, cross the one not asked — the equation without that missing letter is the one to use.
- Using when the velocity is changing.
- Not converting minutes to seconds or km/h to m/s.
Important questions
Try each question on paper first, then tap it to check the answer.
Basic Recall and direct use
Convert 90 km/h into m/s.Answer
State the SI unit of acceleration and what a negative acceleration means.Answer
An athlete runs one full round of a circular track of radius 50 m. Find the distance and displacement. ()Answer
Medium Two or three steps
A scooter accelerates from 5 m/s to 25 m/s in 8 s. Find the acceleration and distance covered.Answer
What does the area under a velocity–time graph represent? Prove it for uniform velocity.Answer
A ball is thrown vertically up at 20 m/s. Taking m/s², find the maximum height and the time to reach it.Answer
Advanced Reasoning and HOTS
The tip of a clock's minute hand is 7 cm from the centre. Find its speed, and explain why its motion is accelerated. ()Answer
Two cars start together from rest. A accelerates at 2 m/s²; B moves at a steady 10 m/s from the same point. When and where does A catch B?Answer
Self-check quiz
Five quick questions. Tap an option to see if you are right.
FAQ
Can distance and displacement be equal?
Yes, when an object moves in a straight line without changing direction. Otherwise the displacement is smaller than the distance.
How do I choose the right equation of motion?
Write down which of u, v, a, s, t you know and which you need. Use the equation that contains exactly those four quantities.
Why is average speed for a round trip not the average of the two speeds?
Because you spend more time at the slower speed. Average speed is total distance divided by total time, which weights each speed by the time spent at it.
What factors increase a car's stopping distance?
Higher speed (braking distance grows with the square of speed), longer reaction time (tired or distracted driver), wet or icy roads, worn tyres, and a heavier load.
Book references
- NCERT Exploration, Chapter 4: Describing Motion Around UsRead the chapter and do every exercise set first.
- Science for Class 9 (Physics, Chemistry, Biology) — Lakhmir Singh & Manjit Kaur (S. Chand)Extra numericals and short answers; chapter order follows the old syllabus, so match by topic.
- S. Chand's Science for Class 9 — S. ChandUseful extra reading and practice; match by topic.
- Exemplar Problems in Science, Class IX — NCERTConcept-testing MCQs and short answers (written for the old book).
These notes, examples and questions are written by PrepRung in our own words to help you study the NCERT Exploration book. NCERT and CBSE are not affiliated with PrepRung. Always study from your textbook too.