Chapter 2 · ICSE Class 9 Physics (Selina Concise)

Motion in One Dimension

Distance, speed, acceleration, the equations of motion and what graphs tell you.

🏃 Chapter 2

Velocity-time graph Straight line from the origin showing constant acceleration; the shaded triangle under it equals the distance travelled. Area = ½ × 10 × 20 = 100 m 20 m/s at 10 s time t (s) v Slope = a = 20 ÷ 10 = 2 m/s² Area under a v–t graph = distance. From rest: v = at, s = ½at² = 100 m. Free fall from rest for 3 s (g = 10): v = 30 m/s, h = 45 m.

↔ Swipe the diagram sideways to see all of it

2.1

Distance, speed, velocity and acceleration

Distance is the length of the path travelled (scalar); displacement is the shortest straight line from start to end, with direction (vector). Speed = distance ÷ time; velocity = displacement ÷ time. Average speed = total distance ÷ total time. Acceleration = change in velocity ÷ time, measured in m/s²; negative acceleration is called retardation.

1

A car covers 120 km in 4 h. Find its average speed.

120 ÷ 4 = 30 km/h.

2

A body speeds up from 10 m/s to 30 m/s in 5 s. Find its acceleration.

a = (30 − 10) ÷ 5 = 4 m/s².

3

A runner completes one lap of a 400 m track. State distance and displacement.

Distance 400 m; displacement 0, as the start and end are the same.

2.2

Equations of uniformly accelerated motion

For constant acceleration a, with initial velocity u, final velocity v, time t and distance s: v = u + at, s = ut + ½at², v² = u² + 2as. A freely falling body has a = g = 10 m/s² (9.8 m/s² more exactly) downward, regardless of mass if air resistance is ignored.

1

A body starts from rest with a = 2 m/s². Find v and s after 10 s.

v = 0 + 2 × 10 = 20 m/s. s = ½ × 2 × 10² = 100 m.

2

Using v² = u² + 2as with u = 10, v = 30, a = 4, find the distance.

s = (30² − 10²) ÷ (2 × 4) = 100 m.

3

A stone dropped from a tower takes 3 s. Find the final speed and height (g = 10).

v = 10 × 3 = 30 m/s. h = ½ × 10 × 3² = 45 m.

2.3

Motion graphs

On a distance–time graph the slope is speed; a horizontal line means at rest and a straight sloping line means uniform speed. On a velocity–time graph the slope is acceleration and the area under the graph is the distance travelled. A horizontal v–t line is uniform velocity; a sloping straight line is uniform acceleration.

1

A v–t graph is a straight line from 0 to 20 m/s in 10 s. Find the acceleration and distance.

Slope = 20 ÷ 10 = 2 m/s². Area = ½ × 10 × 20 = 100 m.

2

What does a horizontal line on a distance–time graph show?

The body is at rest.

3

What does a steeper d–t line show?

A greater speed.

Practice

Exercises

1. Differentiate distance and displacement.

2. Define acceleration.

3. A bus goes 120 km in 4 h. State its average speed.

4. A car starts from rest and reaches 30 m/s in 5 s. Find acceleration.

5. Write the three equations of motion.

6. A ball is dropped from rest. Find its speed after 2 s (g = 10).

7. What is the slope of a v–t graph?

8. What does the area under a v–t graph give?

9. A body moves with u = 5 m/s and a = 2 m/s² for 4 s. Find s.

10. Does a heavy body fall faster than a light one in a vacuum?

Next chapter — Laws of Motion →