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ScienceClass 9Describing Motion

Describing Motion | Class 9 Science Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

Describing Motion – this guide gives you a concise, exam-ready overview of Describing Motion from Class 9 Science, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Distance and Displacement

Distance and displacement are two fundamental quantities used to describe motion. Distance is the total length of the path traveled by an object, irrespective of direction. It is a scalar quantity, meaning it has only magnitude and no direction. For example, if a person walks 3 meters east and then 4 meters west, the total distance traveled is 7 meters. Displacement, on the other hand, is the shortest straight-line distance from the initial position to the final position of the object, along with the direction. It is a vector quantity, meaning it has both magnitude and direction. In the previous example, the displacement would be 1 meter east (3 meters east minus 4 meters west). Displacement can be zero if the initial and final positions are the same, even if the distance traveled is not zero. Understanding the difference between distance and displacement is crucial for describing motion accurately. The section also introduces the concept of path length and how it differs from displacement, emphasizing that displacement depends only on initial and final positions, while distance depends on the actual path taken.

📊 Diagram: Diagram illustrating a path taken by an object from point A to point B along a curved route, showing the total distance traveled as the length of the path and displacement as the straight line connecting A and B with an arrow indicating direction.

🧪 Activity: Activity: Students measure the distance and displacement by walking along a path and marking initial and final positions to understand the difference practically.

🔗 Connection: This section prepares students for the next topic, 'Speed,' where the rate of change of distance with time is introduced.

Frequently asked questions

8. A truck driver driving at the speed of $54\mathrm{km}\mathrm{h}^{-1}$ notices a road sign with a speed limit of $40\mathrm{km}\mathrm{h}^{-1}$ (Fig. 4.29) for trucks. He slows down to $36\mathrm{km}\mathrm{h}^{-1}$ in $36\mathrm{s}$ . What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

Given: Initial speed, u = 54 km/h = 54 × (1000/3600) = 15 m/s Final speed, v = 36 km/h = 36 × (1000/3600) = 10 m/s Time, t = 36 s

Acceleration, a = (v - u)/t = (10 - 15)/36 = -5/36 m/s² (negative sign indicates deceleration)

Distance travelled, s = ut + (1/2) a t² = 15 × 36 + 0.5 × (-5/36) × (36)² = 540 - 0.5 × (5/36) × 1296 = 540 - 0.5 × 180 = 540 - 90 = 450 m

Therefore, the truck travelled 450 meters while slowing down.

9. A car starts from rest and accelerates uniformly to $20\mathrm{ms}^{-1}$ in 5 seconds. It then travels at $20\mathrm{ms}^{-1}$ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.

Given: Initial velocity, u = 0 m/s Final velocity after acceleration, v = 20 m/s Time to accelerate, t1 = 5 s Time at constant velocity, t2 = 10 s Time to stop, t3 = 6 s

Step 1: Calculate acceleration during first phase: a = (v - u)/t1 = (20 - 0)/5 = 4 m/s²

Distance during acceleration, s1 = ut1 + 0.5 a t1² = 0 + 0.5 × 4 × 25 = 50 m

Step 2: Distance during constant velocity, s2 = v × t2 = 20 × 10 = 200 m

Step 3: Calculate acceleration during braking: Initial velocity u = 20 m/s, final veloci

10. A bus is travelling at $36\mathrm{km}\mathrm{h}^{-1}$ when the driver sees an obstacle $30\mathrm{m}$ ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of $2.5\mathrm{ms}^{-2}$ . Will the bus be able to stop before reaching the obstacle?

Given: Initial speed, u = 36 km/h = 36 × (1000/3600) = 10 m/s Distance to obstacle, d = 30 m Reaction time, t_r = 0.5 s Deceleration, a = -2.5 m/s²

Step 1: Distance travelled during reaction time (no deceleration): s_r = u × t_r = 10 × 0.5 = 5 m

Remaining distance to stop = 30 - 5 = 25 m

Step 2: Calculate stopping distance under deceleration: Using v² = u² + 2as, final velocity v = 0 (bus stops) 0 = u² + 2 a s_s s_s = -u² / (2a) = -(10)² / (2 × -2.5) = 100 / 5 = 20 m

Step 3: Total distance c

11. A student said, "The Earth moves around the Sun". In this context, discuss whether an object kept on the Earth can be considered to be at rest.

An object kept on the Earth cannot be considered to be at rest absolutely because the Earth itself is moving around the Sun. However, relative to the Earth’s surface, the object appears to be at rest as it does not change its position with respect to the Earth. Hence, rest and motion are relative concepts depending on the frame of reference chosen.

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