Class 9 Science Motion: Distance, Displacement, Velocity, Acceleration and Equations of Motion (Notes + Numericals)

Motion is the first physics chapter of Class 9 and the foundation for everything you will study about forces and energy. Learn the definitions, the three equations and the graphs, and you can solve almost every numerical in this chapter.

Distance and displacement#

Example: A student walks 3 m east and then 4 m north. Distance = 3 + 4 = 7 m. Displacement = √(3² + 4²) = 5 m towards the north-east.

Speed and velocity#

• Speed = distance ÷ time (scalar). SI unit: m/s.
• Velocity = displacement ÷ time (vector).
• Average speed = total distance ÷ total time.

Example: A bus travels 60 km in the first hour and 40 km in the next hour. Average speed = 100 km ÷ 2 h = 50 km/h.

Unit conversion: km/h → m/s multiply by 5/18; m/s → km/h multiply by 18/5. So 72 km/h = 72 × 5/18 = 20 m/s.

Acceleration#

Acceleration is the rate of change of velocity:

where $u$ is the initial velocity, $v$ the final velocity and $t$ the time. SI unit: m/s². Negative acceleration (slowing down) is called retardation or deceleration.

Example: A scooter speeds up from 10 m/s to 20 m/s in 5 s. a = (20 − 10)/5 = 2 m/s².

Uniform and non-uniform motion#

• Uniform motion: equal distances in equal intervals of time (constant speed).
• Non-uniform motion: unequal distances in equal intervals (a bus in city traffic).

Graphs#

The three equations of motion#

For motion in a straight line with uniform acceleration:

How to choose: list what is given and what is asked, then pick the equation that contains exactly those quantities.

Solved numericals#

1. A train starting from rest attains a velocity of 72 km/h in 5 minutes. Find its acceleration and the distance travelled.

u = 0, v = 72 km/h = 20 m/s, t = 5 min = 300 s.

a = (v − u)/t = 20/300 = 1/15 m/s² (about 0.067 m/s²)

s = ut + ½at² = 0 + ½ × (1/15) × 300² = 3000 m = 3 km

2. A car moving at 20 m/s is braked and stops in 4 s. Find the retardation and the stopping distance.

a = (0 − 20)/4 = −5 m/s² → retardation 5 m/s²

v² = u² + 2as → 0 = 400 + 2(−5)s → s = 40 m

3. A stone is thrown vertically upward at 5 m/s. If the acceleration is 10 m/s² downward, how high does it go and how long does it take?

At the top v = 0, a = −10 m/s².

v² = u² + 2as → 0 = 25 − 20s → s = 1.25 m

v = u + at → 0 = 5 − 10t → t = 0.5 s

Uniform circular motion#

When an object moves in a circle at constant speed, its velocity changes because the direction keeps changing. So it is accelerated motion, even though the speed is constant.

Example: An athlete runs one round of a circular track of radius 14 m in 40 s. Speed = 2 × (22/7) × 14 / 40 = 88/40 = 2.2 m/s.

Quick practice#

1. A cyclist covers 2 km in 10 minutes. Find the speed in m/s.
2. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find its acceleration.
3. Can an object have zero displacement but non-zero distance? Give an example.

Answers: 1) 2000/600 ≈ 3.33 m/s. 2) (16.67 − 22.22)/5 ≈ −1.11 m/s². 3) Yes, for example running one full lap of a track.