Describing Motion Around Us Class 9 Notes and Questions

These Describing Motion Around Us Class 9 notes cover the chapter step by step: how to state a position, the difference between distance and displacement, speed and velocity, acceleration, graphs, the equations of motion and uniform circular motion. Every formula comes with a worked example, and the important questions at the end have full answers.

Position, rest and motion

To say where something is, we give its distance and direction from a fixed reference point, called the origin (O). "The school is 2 km North of the railway station" uses the station as the reference point.

• An object is in motion if its position relative to the reference point changes with time.
• It is at rest if its position relative to the reference point does not change with time.
• Motion is relative: something can be moving relative to one reference point and at rest relative to another.

For motion along a straight line, use a number line: positions to the right of the origin are positive, positions to the left are negative. Also note the difference between an instant of time (a single clock reading) and a time interval (the time between two readings).

Distance and displacement

Both are measured in metres (m). The magnitude of displacement is always less than or equal to the distance. They are equal only when the object moves in one direction without turning back.

Example: walk 5 m forward and 3 m back. Distance = 5 + 3 = 8 m. Displacement = 5 − 3 = 2 m forward.

Example: one full lap of a 400 m circular track. Distance = 400 m. Displacement = 0 m, because you end where you started.

Average speed and average velocity

• Average speed = total distance ÷ time interval (scalar).
• Average velocity vav = s ÷ t, where s is displacement (vector).
• SI unit of both: m/s (m s⁻¹).

Equal distances in equal time intervals means uniform motion; otherwise it is non-uniform motion. In straight-line motion in one direction, the magnitude of average velocity equals average speed.

Worked example (from the chapter): Sarang swims 25 m to one end of a pool and 25 m back in 50 s.

• Distance = 50 m, so average speed = 50 m ÷ 50 s = 1 m/s.
• Displacement = 0 m, so average velocity = 0 m ÷ 50 s = 0 m/s.

Average acceleration

a = (v − u) ÷ t, where u is initial velocity, v is final velocity and t is time. SI unit: m s⁻².

• If velocity increases, acceleration is in the direction of velocity.
• If velocity decreases (retardation or deceleration), acceleration is opposite to velocity, so it comes out negative.
• A fast object moving with constant velocity has zero acceleration. Acceleration needs a change in speed, direction or both.

Convert first: km/h × 5/18 = m/s.

Worked example (bus): a bus goes from 36 km/h to 54 km/h in 10 s.

• 36 × 5/18 = 10 m/s; 54 × 5/18 = 15 m/s.
• a = (15 − 10) ÷ 10 = 0.5 m s⁻².

It then brakes from 15 m/s to 0 in 5 s: a = (0 − 15) ÷ 5 = −3 m s⁻². The minus sign shows deceleration.

Reading motion graphs

Position-time graph (time on x-axis, position on y-axis):

• Horizontal line: at rest, velocity zero.
• Straight sloping line: uniform motion, constant velocity.
• Curved line: non-uniform motion, velocity changing.
• Slope = velocity: v = (s₂ − s₁) ÷ (t₂ − t₁). Steeper slope means higher velocity.

Velocity-time graph (time on x-axis, velocity on y-axis):

• Horizontal line: constant velocity, zero acceleration.
• Straight sloping line: constant acceleration; sloping down means constant deceleration.
• Slope = acceleration.
• Area between the line and the time axis = displacement. For constant acceleration this area is a trapezium (a rectangle plus a triangle).

Equations of motion (constant acceleration only)

1. v = u + at
2. s = ut + ½at² (this is the area under the velocity-time graph)
3. v² = u² + 2as

Useful extra forms: s = vt − ½at² and s = ½(u + v)t.

Worked example (stopping distance, from the chapter): a car at 54 km/h (15 m/s) brakes with a = −4 m/s². How far does it go before stopping (v = 0)? Time is not given, so use the third equation.

• 0 = 15² + 2(−4)s
• 8s = 225
• s = 225 ÷ 8 = 28.1 m (approximately)

Worked example (bus again): u = 10 m/s, a = 0.5 m s⁻², t = 10 s.

• s = ut + ½at² = (10 × 10) + (½ × 0.5 × 10²) = 100 + 25 = 125 m.
• Check with s = ½(u + v)t = ½ × (10 + 15) × 10 = 125 m. Both agree.

Uniform circular motion

When an object moves in a circle at constant speed, it is in uniform circular motion. Its direction changes at every point, so its velocity changes, and the motion is accelerated.

• Velocity at any point is along the tangent to the circle. If released, the object moves in a straight line along that tangent.
• One revolution of a circle of radius R covers the circumference 2πR. If it takes time T, average speed = 2πR ÷ T.
• Displacement after one full revolution is zero.

An athlete on a rectangular track changes direction 4 times per round, on a hexagonal track 6 times, and on a circular track continuously at every point.

Remember this

• Distance is scalar; displacement is vector.
• Speed uses distance; velocity uses displacement.
• a = (v − u) ÷ t; negative a means slowing down.
• Position-time slope = velocity. Velocity-time slope = acceleration; area = displacement.
• No time given? Use v² = u² + 2as.

Important questions with answers

1. What is the origin when describing motion? A fixed reference point from which we state an object's distance and direction.

2. A runner completes one lap of a 400 m circular track. Find distance and displacement. Distance = 400 m. Displacement = 0 m, since the runner returns to the start.

3. When is average velocity equal to average speed? When the object moves in a straight line in one direction without turning back.

4. What is the acceleration of a car moving at a steady 80 km/h on a straight road? Zero, because neither the magnitude nor the direction of velocity changes.

5. Convert 72 km/h to m/s. 72 × 5/18 = 20 m/s.

6. A cyclist speeds up from 2 m/s to 8 m/s in 3 s. Find the acceleration. a = (8 − 2) ÷ 3 = 6 ÷ 3 = 2 m s⁻².

7. How do you find displacement from a velocity-time graph? Find the area between the line and the time axis.

8. An object moves once around a circle of radius 7 m in 22 s. Find its average speed (take π = 22/7). Distance = 2πR = 2 × 22/7 × 7 = 44 m. Average speed = 44 ÷ 22 = 2 m/s.

9. Why is uniform circular motion called accelerated motion? Its direction, and so its velocity, keeps changing, and any change in velocity is acceleration.

Common mistakes to avoid

• Using km/h directly in a = (v − u) ÷ t. Convert to m/s first.
• Dropping the minus sign for deceleration.
• Saying constant speed means zero acceleration in circular motion.
• Using the equations of motion when acceleration is not constant.
• Reading velocity from the area of a position-time graph. On that graph, use the slope.

Want to practise more motion numericals step by step? Study this chapter with Joy.