Kinematics – ICSE Class 10 Physics Notes & Guide

Kinematics is the branch of physics that describes how objects move. It deals with quantities such as distance, displacement, speed, velocity and acceleration, and explains their relationships using equations and graphs. Mastering these ideas helps you solve motion problems quickly and accurately in the ICSE exam.

Key concepts

Scalar and vector quantities

• Scalar: has only magnitude (e.g., distance, speed, mass). It is described by a single number and unit.
• Vector: has both magnitude and direction (e.g., displacement, velocity, acceleration). It is represented by an arrow; the length shows magnitude and the arrowhead shows direction. Vectors are added tip‑to‑tail and can be resolved into perpendicular components using right‑angle triangles.

Distance and displacement

• Distance: total length of the path travelled, always positive, measured in metres (m).
• Displacement: straight‑line change in position from the initial point to the final point, having both magnitude and direction. It can be positive, negative or zero depending on the chosen coordinate axis.

Speed, velocity and acceleration

• Speed = distance ÷ time. It is a scalar (e.g., 20 km h⁻¹).
• Velocity = displacement ÷ time. It is a vector (e.g., 20 km h⁻¹ north).
• Acceleration = rate of change of velocity with time. It can be positive (speeding up) or negative (slowing down) and is a vector.

Equations of motion for uniformly accelerated motion

When acceleration (a) is constant, the following three equations relate the five kinematic variables (u – initial velocity, v – final velocity, s – displacement, t – time, a – acceleration):

1. v = u + a t
2. s = u t + ½ a t²
3. v² = u² + 2 a s

These equations are derived from the definitions of velocity and acceleration and are valid only for constant a.

Graphical representation of motion

• s‑t graph: slope = velocity. A straight‑line slope indicates constant velocity; a curved line shows changing velocity.
• v‑t graph: slope = acceleration; area under the curve = displacement.
• a‑t graph: slope = rate of change of acceleration (jerk); area under the curve = change in velocity.

Understanding how to read and draw these graphs is essential for solving many ICSE questions.

Relative velocity

The velocity of an object as observed from another moving object is called relative velocity.

• If A moves with velocity v_A and B with v_B, the velocity of A relative to B is vAB = vA – v_B.
• Use a consistent sign convention (e.g., rightward positive). Relative velocity helps solve problems involving trains, boats, or particles moving in opposite directions.

Important definitions and formulas

Solved examples

Example 1: A car accelerates uniformly from rest to 20 m s⁻¹ in 5 s. Find its acceleration and the distance covered.

Solution

1. Initial velocity, u = 0 (rest). Final velocity, v = 20 m s⁻¹. Time, t = 5 s.
2. Acceleration, a = (v – u) ÷ t = (20 – 0) ÷ 5 = 4 m s⁻².
3. Use the second equation of motion: s = u t + ½ a t² = 0 × 5 + ½ × 4 × (5)² = ½ × 4 × 25 = 2 × 25 = 50 m.

Answer: a = 4 m s⁻², distance = 50 m.

Example 2: A stone is thrown vertically upward with a speed of 30 m s⁻¹. Calculate the maximum height reached and the time taken to reach that height. (Take g = 10 m s⁻² downward.)

Solution

1. Upward direction is taken as positive. Initial velocity, u = +30 m s⁻¹. At the highest point, final velocity, v = 0.
2. Using v = u + a t, where a = –g = –10 m s⁻²: 0 = 30 – 10 t → t = 30 ÷ 10 = 3 s.
3. Use v² = u² + 2 a s to find height (s): 0 = (30)² + 2(–10) s → 0 = 900 – 20 s → 20 s = 900 → s = 45 m.

Answer: Maximum height = 45 m, time to reach = 3 s.

Example 3: Train A moves east at 40 km h⁻¹ and Train B moves west at 30 km h⁻¹ on parallel tracks. Find the relative speed of Train A with respect to Train B.

Solution

1. Choose east as positive. Then v_A = +40 km h⁻¹, v_B = –30 km h⁻¹.
2. Relative velocity of A w.r.t B: v_AB = v_A – v_B = 40 – (‑30) = 70 km h⁻¹.
3. Since the direction is east relative to B, the magnitude of relative speed is 70 km h⁻¹.

Answer: 70 km h⁻¹ east.

Common mistakes to avoid

• Confusing scalar and vector signs – always attach a direction to vectors; forgetting the sign leads to wrong displacement or relative velocity.
• Using the wrong equation – the three motion equations are for constant acceleration only. Do not apply them to variable‑a problems.
• Mixing units – keep all quantities in SI units (m, s) before substituting into formulas.
• Reading graphs incorrectly – the slope of an s‑t graph gives instantaneous velocity, not average speed. Area under a v‑t graph gives displacement, not distance.
• Ignoring the direction in relative velocity – treat one velocity as positive and the other as negative according to the chosen axis.

Exam tips

• Memorise the three equations of motion and the conditions under which they are valid.
• Practice converting between speed, velocity and acceleration quickly; write the sign immediately.
• When a graph is given, identify slope and area first before writing equations.
• For relative velocity problems, draw a simple diagram with arrows; it reduces sign errors.
• Always check the answer’s dimension (m, m s⁻¹, m s⁻²) and whether the sign makes physical sense.

Frequently asked questions

What is the difference between average speed and average velocity?

Average speed = total distance ÷ total time (scalar). Average velocity = total displacement ÷ total time (vector). They are equal only when motion is in a straight line without change of direction.

How do I find displacement from a velocity‑time graph?

The displacement equals the area under the v‑t graph between the two time limits. Positive area (above the time axis) adds, while area below the axis subtracts because it represents motion in the opposite direction.

Can the equations of motion be used for a falling object?

Yes, if you treat the downward direction as positive (or negative) and use a constant acceleration of g = 9.8 m s⁻² (≈10 m s⁻²). The same three equations apply.