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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.

Velocity

Velocity is a vector quantity that describes the rate of change of displacement with respect to time. Unlike speed, velocity includes both magnitude and direction. The magnitude of velocity is called speed, but velocity also specifies the direction in which the object is moving. The SI unit of velocity is meters per second (m/s). Velocity can be uniform or variable. Uniform velocity means the object covers equal displacements in equal intervals of time in the same direction. Variable velocity means the displacement or direction changes over time. For example, a car moving north at 60 km/h has a velocity of 60 km/h north. If the car changes direction but maintains the same speed, its velocity changes. The formula for velocity is Velocity = Displacement / Time. Understanding velocity is essential for analyzing motion in two or three dimensions and for studying acceleration and forces. The section also introduces the concept of average velocity, which is the total displacement divided by total time, useful when velocity varies during the journey.

📊 Diagram: Diagram showing displacement vector from initial to final position with an arrow indicating direction, along with time interval, illustrating velocity as displacement/time.

🧪 Activity: Activity: Students track a moving object’s position at regular time intervals to calculate velocity and observe changes in direction.

🔗 Connection: This section connects to the next topic, 'Acceleration,' which deals with changes in velocity over time.

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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