Class 9 Science · Original EZT practice
Describing Motion Around Us (Class 9 Chapter 4): summary and practice
From a galloping horse to a drifting cloud, motion is everywhere. Chapter 4 describes it with numbers, equations and graphs - speed, velocity, acceleration and the constant-acceleration formulas. All questions are original EZT tutor practice following the chapter's own section order. Edition note: this page covers the new NCERT book Exploration (Grade 9), First Edition April 2026. Older Class 9 books placed motion in a different chapter - check your chapter title, not just the number.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 20 October 2026:
1. Motion in a straight line
Describe position from a reference point. Distance is the full path travelled; displacement is the straight change in position with direction. Average speed is distance over time; average velocity is displacement over time; average acceleration is the change in velocity over time.
2. Graphical representation of motion
A position-time graph shows where the object is at each instant; its slope gives velocity. A velocity-time graph shows how velocity changes; its slope gives acceleration and the area under it gives displacement.
3. The kinematic equations
For constant acceleration in a straight line: v = u + at, s = ut + (1/2)at^2 and v^2 = u^2 + 2as. They hold only while the acceleration is constant, and signs matter when the direction reverses.
4. Motion in a plane
Motion in two dimensions includes vehicles overtaking, kicked balls and satellites. An object moving in a circular path at constant speed is in uniform circular motion - its velocity keeps changing direction.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Separate distance from displacement
Choose the method: Distance adds every leg of the path; displacement looks only at start and end.
EZT practice question
A swimmer crosses a 50 m pool in 50 s, then swims back to the start in another 50 s. For the round trip, find the distance, displacement, average speed and average velocity.
Show the worked solution
- Distance is the full path: 50 m out plus 50 m back is 100 m.
- Displacement is zero - the swimmer ends at the start.
- Average speed is 100 m over 100 s; average velocity is 0 m over 100 s.
Answer: Distance 100 m; displacement 0 m; average speed 1 m/s; average velocity 0 m/s.
Check: Speed uses distance and velocity uses displacement - that is why they differ here.
Watch for this: Do not report the distance as 50 m; the return leg also counts.
Now try it yourself
A runner completes one full 400 m lap in 80 s. Find average speed and average velocity.
Check your answer
Average speed 5 m/s; average velocity 0 m/s.
A full lap ends where it starts, so the displacement is zero.
Convert km/h to m/s
Choose the method: One km/h is 1000 m over 3600 s, which is 5/18 m/s.
EZT practice question
Convert 36 km/h to m/s.
Show the worked solution
- One km is 1000 m and one hour is 3600 s.
- 36 km/h = 36 x 1000 / 3600 = 10 m/s.
Answer: 10 m/s.
Check: Multiplying m/s by 18/5 returns km/h: 10 x 18/5 = 36.
Watch for this: Do not mix units inside one calculation; convert first.
Now try it yourself
Convert 54 km/h to m/s.
Check your answer
15 m/s.
54 x 5/18 = 15.
Find acceleration from a velocity change
Choose the method: Acceleration is the change in velocity divided by the time taken.
EZT practice question
A vehicle speeds up from 10 m/s to 15 m/s in 10 s. Find its acceleration.
Show the worked solution
- The change in velocity is 15 - 10 = 5 m/s.
- Divide by the time: 5/10.
Answer: 0.5 m/s^2.
Check: In 10 s at 0.5 m/s^2 the velocity indeed rises by 5 m/s.
Watch for this: Use final minus initial; swapping them flips the sign.
Now try it yourself
A car brakes from 15 m/s to rest in 5 s. Find its acceleration.
Check your answer
-3 m/s^2.
(0 - 15)/5 = -3; the negative sign marks slowing down.
Use v = u + at from rest
Choose the method: With u = 0 the equations shrink to v = at and s = (1/2)at^2.
EZT practice question
An object starts from rest and accelerates at 2 m/s^2 for 5 s. Find its final velocity and the distance covered.
Show the worked solution
- u = 0, so v = at = 2 x 5 = 10 m/s.
- s = (1/2)at^2 = (1/2) x 2 x 25 = 25 m.
- Check with v^2 = u^2 + 2as: 100 = 2 x 2 x 25.
Answer: v = 10 m/s and s = 25 m.
Check: All three equations agree: 100 = 100.
Watch for this: Do not use these equations if the acceleration changes midway.
Now try it yourself
From rest, what velocity does 3 m/s^2 give in 4 s?
Check your answer
12 m/s.
v = at = 3 x 4.
Find a braking distance
Choose the method: Use v^2 = u^2 + 2as with a negative acceleration.
EZT practice question
A car moving at 20 m/s brakes with acceleration -4 m/s^2 until it stops. Find the stopping distance and time.
Show the worked solution
- From v = u + at: 0 = 20 - 4t, so t = 5 s.
- From v^2 = u^2 + 2as: 0 = 400 - 8s, so s = 50 m.
Answer: Stopping distance 50 m in 5 s.
Check: Average velocity during braking is 10 m/s, and 10 x 5 = 50 m.
Watch for this: Keep the negative sign on the acceleration; dropping it doubles the error.
Now try it yourself
From 10 m/s, braking at -2 m/s^2, how far to stop?
Check your answer
25 m.
0 = 100 - 4s gives s = 25 m.
Read free-fall values with g = 9.8 m/s^2
Choose the method: Free fall from rest means u = 0 and a = g downward.
EZT practice question
A stone falls freely from rest with g = 9.8 m/s^2. What is its velocity after 3 s?
Show the worked solution
- v = u + gt with u = 0.
- v = 9.8 x 3 = 29.4 m/s.
Answer: 29.4 m/s downward.
Check: After 4 s it is 39.2 m/s - the velocity grows by 9.8 every second.
Watch for this: Do not add an initial velocity; from rest it is zero.
Now try it yourself
What is the velocity after 2 s of free fall from rest?
Check your answer
19.6 m/s.
v = 9.8 x 2.
Get displacement from a velocity-time graph
Choose the method: The area between the graph and the time axis is the displacement.
EZT practice question
An object moves at a constant 20 m/s for 6 s. Using the velocity-time graph, find the displacement.
Show the worked solution
- The graph is a horizontal line at 20 m/s.
- The area under it is a rectangle: 20 x 6.
Answer: 120 m.
Check: Constant velocity means displacement is velocity times time.
Watch for this: The area under a velocity-time graph is displacement, not the reading on the axis.
Now try it yourself
Constant 15 m/s for 10 s - displacement?
Check your answer
150 m.
The rectangle area is 15 x 10.
Explain uniform circular motion
Choose the method: Constant speed around a circle still means changing velocity.
EZT practice question
A satellite moves in a circular path at constant speed. Is its velocity constant?
Show the worked solution
- The speed is constant, but the direction changes at every point.
- Velocity includes direction, so it keeps changing.
Answer: No - the velocity changes because the direction changes, even at constant speed.
Check: A changing velocity means the motion is accelerated, toward the centre.
Watch for this: Do not say the acceleration is zero just because the speed is constant.
Now try it yourself
Give one example of motion in a plane.
Check your answer
A kicked ball, or a vehicle overtaking another.
Both need two dimensions to describe.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Distance versus displacement?
Show answer and reason
Distance is the full path length; displacement is the straight change in position with direction.
A closed loop has zero displacement but positive distance.
2. When do the kinematic equations apply?
Show answer and reason
Only when the acceleration is constant.
They are derived assuming a steady rate of velocity change.
3. What does the area under a velocity-time graph give?
Show answer and reason
The displacement.
The area is velocity times time.
4. Convert km/h to m/s.
Show answer and reason
Multiply by 5/18.
One m/s is 18/5 km/h.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Convert 72 km/h to m/s.
- A car starts from rest and reaches 20 m/s in 8 s. Find its acceleration and the distance covered.
- A bus brakes from 25 m/s to rest over 50 m. Find its acceleration.
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Questions about this practice page
- What is the difference between distance and displacement?
- Distance is the total path travelled; displacement is the straight change in position with direction.
- Speed versus velocity?
- Speed is distance over time; velocity is displacement over time and carries direction.
- When can I use the kinematic equations?
- Only for constant acceleration in a straight line.
- How do I find displacement from a velocity-time graph?
- Take the area between the graph and the time axis.
- Are these the NCERT exercise questions for Describing Motion Around Us (Class 9)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.