Kinematics

Last Updated : 24 Mar, 2026

Kinematics is the branch of physics that deals with the study of the motion of objects without considering the forces causing the motion. It focuses only on describing how an object moves, not why it moves. In kinematics, objects are often treated as point particles, and their motion is explained using mathematical relationships.

This part of classical mechanics looks at motion from a geometric and algebraic point of view, using quantities like position, displacement, velocity, and acceleration as functions of time. Since forces and mass are not involved, kinematics is often referred to as the mathematics of motion.

Kinematics is widely used in fields like mechanical engineering, robotics, biomechanics, and astrophysics, where understanding motion is a prerequisite to studying forces.

Kinematic Formulas

The kinematics formulas deal with displacement, velocity, time, and acceleration. In addition, the following are the four kinematic formulas:

1. {v}=u +at

2. s=ut+\frac{1}{2}at^{2}

3. v^{2}=u^{2}+2as

4. Average Velocity Formula  s = \frac{1}{2}(u + v) \times t

5. Distance Travelled in nth Second S_{n^{th}}=u+\frac{a}{2}(2n-1)

Derivation of Kinematic Formulas

Here is the derivation of the four-kinematics formula mentioned above:

1. Derivation of First Kinematic Formula

We have,

Acceleration = Change in Velocity / Time ⇒ a = Δv / Δt

We can now use the definition of velocity change v-v0 to replace Δv ⇒ a = (v-u)/Δt.

v = u + aΔt

This becomes the first kinematic formula if we agree to just use t for Δt.

\boxed {{v}=u +at}

law_1

2. Derivation of Second Kinematic Formula

Average Velocity = Δx/t = (v + u)/2

put v = u + at, we get ⇒ s/t = (u+at+u)/2

s/t = u + at/2

Finally, to obtain the third kinematic formula,

\boxed {s=ut+\frac{1}{2}at^{2}}

law_2

3. Derivation of Third Kinematic Formula

From the second kinematic formula, ⇒ Δx = ((v+u)/2)t

As we know v = u + at   ⇒ t = (v - u)/a

Put the value of t in the second kinematic formula, we get

Δx = ((v+v0)/2) × ((v-v0)/a)

Δx = (v2+v02)/2a

We get the third kinematic formula by solving for v.

\boxed {v^{2}=u^{2}+2as}

law_3

Key Concept in Kinematics

1. Position and Displacement

  • Position is a vector quantity that represents the location of an object in a given frame of reference. It is often described using coordinates in one-, two-, or three-dimensional space.
  • Displacement is a vector quantity that represents the change in position of an object. It is defined as the straight-line distance and direction from the initial position to the final position. [Δx = x_f - x_i ]

Differences Between Position and Displacement

AspectPositionDisplacement
DefinitionLocation of an object in spaceChange in position of an object
TypeVectorVector
CoordinatesDescribed by coordinates in 1D, 2D, or 3D spaceDifference between initial and final position
MagnitudeAbsolute value of coordinatesStraight-line distance between two positions
DirectionNot inherently directional (depends on reference)Always directional (from initial to final)

2. Speed and Velocity

  • Speed is a scalar quantity that measures the rate at which an object covers distance. It only has magnitude and no direction. Speed is always positive or zero.
  • Velocity is a vector quantity that measures the rate of change in displacement with respect to time. It has both magnitude and direction. Velocity can be positive, negative, or zero. [v_{avg} = \frac{\Delta x}{ \Delta t} ]

Differences Between Speed and Velocity

AspectSpeedVelocity
Quantity TypeScalarVector
ComponentsMagnitude onlyMagnitude and direction
Formulav = d/tv = Δx/t

Positive/Negative

Always positive or zero

Can be positive, negative, or zero

Direction

No direction

Specific direction

Example

60 km/hr (without direction)

60 km/hr north

3. Acceleration

Acceleration is the rate at which an object's velocity changes with time. It indicates whether an object is speeding up, slowing down, or changing direction. The SI unit of acceleration is meters per second squared (m/s²).

a = \frac{\Delta v}{\Delta t}

4. Relative Motion

Relative motion describes the motion of an object as observed from a particular frame of reference.

  • Relative Position: \vec{r}_{AB}=\vec{r}_{A}-\vec{r}_{B}
  • Relative Velocity: \vec{v}_{AB}=\vec{v}_{A}-\vec{v}_{B}
  • Relative Acceleration: \vec{a}_{AB}=\vec{a}_{A}-\vec{a}_{B}

Kinematic Equations for Rotational Motion

Equations of motion for rotational motion are

1. ω_f = ω_i + αt

2. θ = ω_it + \frac{1}{2}αt^2

3. ω_f^2 = ω_i^2 + 2αθ

4. Average Angular Velocity Formula  θ = \frac{1}{2}(ω_i + ω_f) \times t

where,

  • ωf​: Final Angular Velocity
  • ωi: Initial Angular Velocity
  • α: Angular Acceleration
  • t: Time
  • θ: Angular displacement

Graphs

1. Displacement-Time Graphs

distance_time_graphs

2. Velocity-Time Graphs

3. Acceleration-Time Graphs

acceleration_time_graphs


Article Related to Kinematics:

Solved Examples

Example 1: For the time span t = 7 s, an automobile with a beginning velocity of zero accelerates uniformly at 16 m/s2. Do you know how far it's traveled?

Solution: Given: t = 7s, v0 = 0 m/s, a = 16 m/s2

Since,

s=v_{0}t+\frac{1}{2}at^{2}

s = 0 × 7 + (1/2) × 16 × 72

= (1/2) × 16 × 49

= 8 × 49

= 392 m

Example 2: A bicycle with initial velocity 2 experiences a uniform acceleration of 20 m/s2 for the time interval 6 s. Determine its final velocity.

Solution: Given: v0 = 2 m/s, a = 20 m/s2, t = 6s

Since,

v=v_{0}+at

v = 2 + 20 × 6

= 122 m/s

Example 3: Assume that the initial velocity is 0 and the final velocity is 5 for the time interval of 4 s, then find its displacement.

Solution: Given: v0 = 0 m/s, v = 5 m/s, t = 4s

Since,

\Delta{x}=(\frac{v+v_{0}}{2})t

Δx = (5+0) × 4

= 20 m

Example 4: A truck with an initial velocity of zero, a constant acceleration of 6 m/s2, and a time interval of 3 s. Find the final velocity.

Solution: Given: v0 = 0 m/s, a = 6 m/s2, t = 3s

Since,

v=v_{0}+at

= 0 + 6 × 3

= 18 m/s

Unsolved Questions

Q1. A car travels at a constant speed of 60 km/h for 2 hours. How far does the car travel in this time?

Q2. A ball is thrown straight up with an initial velocity of 20 m/s from the ground. Assuming the acceleration due to gravity is -9.8 m/s², calculate the time it takes for the ball to reach its maximum height.

Q3. A stone is dropped from a cliff 80 meters high. How long does it take to hit the ground? Assume the acceleration due to gravity is 9.8 m/s ².

Q4. A projectile is launched with an initial velocity of 50 m/s at an angle of 30° to the horizontal. Calculate the maximum height reached by the projectile. Ignore air resistance and use g = 9.8 m/s² for the acceleration due to gravity.

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