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Motion is usually the first physics chapter of Class 9, and it sets habits you will use until Class 12. Marks here are often lost not for a missing formula but for reading the height of a graph when the question wants its slope, forgetting to convert km/h, or using v = u + at when the acceleration is not constant. This post covers what the new syllabus asks, how to read motion graphs, a fully worked problem, and a weekly practice plan.
What changed in the Class 9 syllabus for 2026–27?
From 2026–27, CBSE Class 9 uses the new NCERT textbook *Exploration*. Motion is Chapter 4, Describing Motion Around Us. The CBSE Class IX Science syllabus for 2026–27 puts it in Unit III (Motion, Force, Work and Sound, 23 marks) and asks you to plot and interpret position-time and velocity-time graphs, derive the kinematic equations by the graphical method, and derive the speed in uniform circular motion.
The new book says "position-time graph" where older books said "distance-time graph"; NCERT notes they match for motion in one direction starting at the origin. Check your school's syllabus for the current year.
What is the difference between distance and displacement?
Distance is the total path length and has no direction. Displacement is the net change in position, with a direction. NCERT's example: an athlete runs 100 m forward and 60 m back. She travels 160 m but her displacement is 40 m forward. Average speed uses distance; average velocity uses displacement. A swimmer who does one 25 m length and back in 50 s has an average speed of 1 m/s and an average velocity of zero.
How do you read a motion graph?
Three rules cover almost every question:
Slope of a position-time graph = velocity. A straight line means constant velocity; a curve means the velocity is changing; a flat line means the object is at rest.
Slope of a velocity-time graph = acceleration. A line sloping down means the velocity is decreasing, so the acceleration is opposite to the velocity.
Area under a velocity-time graph = displacement. Split the shape into rectangles and triangles.
These are easy to state and easy to mix up. Beichner (1994) tested 895 high school and college students on kinematics graphs and found the same difficulties again and again: reading a graph as a picture of the path, confusing slope with height, and not knowing what the area under a curve means. McDermott, Rosenquist and van Zee (1987) found similar errors among university students.
Worked example: a scooter trip
A scooter starts from rest and speeds up uniformly to 12 m/s in 4 s. It then moves at 12 m/s for 6 s, and finally brakes uniformly to rest in 3 s. Find the acceleration in each part, the total displacement and the average speed.
Accelerations (slopes):
Part 1: a = (12 − 0)/4 = 3 m/s².
Part 2: velocity is constant, so a = 0.
Part 3: a = (0 − 12)/3 = −4 m/s². The minus sign means the acceleration is opposite to the motion.
Displacement (area under the velocity-time graph):
Triangle, part 1: ½ × 4 × 12 = 24 m.
Rectangle, part 2: 6 × 12 = 72 m.
Triangle, part 3: ½ × 3 × 12 = 18 m.
Total: 24 + 72 + 18 = 114 m.
Check with the equations. Part 1: s = ut + ½at² = 0 + ½ × 3 × 4² = 24 m. Part 3: v² = u² + 2as gives 0 = 144 + 2(−4)s, so s = 18 m. Both agree with the areas.
Average speed: 114 m ÷ 13 s ≈ 8.8 m/s. Note that this is not the average of 0, 12 and 0.
Why does stopping distance grow so fast with speed?
NCERT links the equations to road safety. Take a car braking at 5 m/s² with a driver who takes 0.5 s to react.
At 36 km/h (10 m/s): reaction distance 10 × 0.5 = 5 m; braking distance from v² = u² + 2as is 10²/(2 × 5) = 10 m; total 15 m.
At 72 km/h (20 m/s): reaction distance 10 m; braking distance 20²/(2 × 5) = 40 m; total 50 m.
Doubling the speed doubles the reaction distance but quadruples the braking distance, because braking distance depends on u². For more practice choosing the right equation, see our guide to equations of motion problems.
What about uniform circular motion?
An object moving in a circle at constant speed covers 2πR in one revolution of time T, so v = 2πR/T. A merry-go-round seat 3 m from the centre that goes round once in 12 s moves at 2π × 3/12 ≈ 1.57 m/s. After one full turn its displacement is zero. Its speed is constant, yet it is accelerating, because the direction of its velocity keeps changing.
Which mistakes should you watch for?
Units. Convert km/h to m/s by multiplying by 5/18 before using the equations: 54 km/h = 15 m/s.
Signs. Choose a positive direction at the start and keep it. A slowing object has acceleration opposite to its velocity.
Wrong situation. The kinematic equations work only for constant acceleration. For a trip with several stages, as in the scooter example, treat each stage separately.
Graph as a map. A position-time graph shows how position changes with time, not the shape of the road.
Seeing a graph build up while an object moves helps the rules stick. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read, so you can predict a graph first and then watch it form.
A weekly practice plan
Day 1: Redo NCERT Examples 4.3 and 4.8 without looking, then check.
Day 2: Draw the velocity-time graph for three of the chapter-end questions before using any equation.
Day 3: Do three questions that give a graph and ask for slope, height or area, and write which one each needs.
Day 4: Mixed set of five problems with unit conversions; check every answer by a second method.
Start today by sketching the scooter's velocity-time graph from memory and finding the three areas.
