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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.

Speed

Speed is a measure of how fast an object is moving. It is defined as the distance traveled per unit time. Speed is a scalar quantity because it only has magnitude and no direction. The SI unit of speed is meters per second (m/s), but kilometers per hour (km/h) is also commonly used. The formula for speed is Speed = Distance / Time. Speed can be uniform or variable. Uniform speed means the object covers equal distances in equal intervals of time, while variable speed means the distances covered in equal intervals of time are not equal. For example, a car moving at a constant speed of 60 km/h on a highway is an example of uniform speed. Speed is an important concept in daily life and various fields such as transportation, sports, and engineering. Understanding speed helps in planning travel time and analyzing motion. The section also discusses average speed, which is the total distance traveled divided by the total time taken, useful when speed varies during the journey.

📊 Diagram: Graphical representation showing distance covered plotted against time for uniform and non-uniform speed, illustrating straight line for uniform speed and curved line for variable speed.

🧪 Activity: Activity: Students measure the time taken to cover a known distance by walking or running to calculate their speed and observe uniformity.

🔗 Connection: This section leads to the next section, 'Velocity,' where direction is introduced along with speed to describe motion more completely.

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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