# Derivation of equations of uniformly accelerated motion from velocity time graph

### Asked by Abhishek Poonia | 9th Sep, 2011, 12:00: AM

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## First Equation of Motion

Consider a particle moving along a straight line with uniform acceleration 'a'. At t = 0, let the particle be at A and u be its initial velocity and when t = t, v be its final velocity.

v = u + at ** I equation of motion**

### Graphical Derivation of First Equation of Motion

Consider an object moving with a uniform velocity u in a straight line. Let it be given a uniform acceleration a at time t = 0 when its initial velocity is u. As a result of the acceleration, its velocity increases to v (final velocity) in time t and S is the distance covered by the object in time t.

The figure shows the velocity-time graph of the motion of the object.

Slope of the v - t graph gives the acceleration of the moving object.

Thus, acceleration = slope = AB =

v - u = at

v = u + at **I equation of motion**

## Second Equation of Motion

From equations (1) and (2)

The first equation of motion is v = u + at.

Substituting the value of v in equation (3), we get

### Graphical Derivation of Second Equation of Motion

Let u be the initial velocity of an object and 'a' the acceleration produced in the body. The distance travelled S in time t is given by the area enclosed by the velocity-time graph for the time interval 0 to t.

Graphical Derivation of Second Equation

Distance travelled S = area of the trapezium ABDO

= area of rectangle ACDO + area of DABC

(v = u + at I eqn of motion; v - u = at)

## Third Equation of Motion

The first equation of motion is v = u + at.

v - u = at ... (1)

From equation (2) and equation (3) we get,

Multiplying equation (1) and equation (4) we get,

(v - u) (v + u) = 2aS

[We make use of the identity a^{2} - b^{2} = (a + b) (a - b)]

v^{2} - u^{2} = 2aS **III equation of motion**

### Graphical Derivation of Third Equation of Motion

Let 'u' be the initial velocity of an object and a be the acceleration produced in the body. The distance travelled 'S' in time 't' is given by the area enclosed by the v - t graph.

Graphical Derivation of Third Equation

S = area of the trapezium OABD.

Substituting the value of t in equation (1) we get,

2aS = (v + u) (v - u)

(v + u)(v - u) = 2aS [using the identity a^{2} - b^{2} = (a+b) (a-b)]

v^{2} - u^{2} = 2aS ** III Equation of Motion**

## First Equation of Motion

Consider a particle moving along a straight line with uniform acceleration 'a'. At t = 0, let the particle be at A and u be its initial velocity and when t = t, v be its final velocity.

v = u + at ** I equation of motion**

### Graphical Derivation of First Equation of Motion

Consider an object moving with a uniform velocity u in a straight line. Let it be given a uniform acceleration a at time t = 0 when its initial velocity is u. As a result of the acceleration, its velocity increases to v (final velocity) in time t and S is the distance covered by the object in time t.

The figure shows the velocity-time graph of the motion of the object.

Slope of the v - t graph gives the acceleration of the moving object.

Thus, acceleration = slope = AB =

v - u = at

v = u + at **I equation of motion**

## Second Equation of Motion

From equations (1) and (2)

The first equation of motion is v = u + at.

Substituting the value of v in equation (3), we get

### Graphical Derivation of Second Equation of Motion

Let u be the initial velocity of an object and 'a' the acceleration produced in the body. The distance travelled S in time t is given by the area enclosed by the velocity-time graph for the time interval 0 to t.

Graphical Derivation of Second Equation

Distance travelled S = area of the trapezium ABDO

= area of rectangle ACDO + area of DABC

(v = u + at I eqn of motion; v - u = at)

## Third Equation of Motion

The first equation of motion is v = u + at.

v - u = at ... (1)

From equation (2) and equation (3) we get,

Multiplying equation (1) and equation (4) we get,

(v - u) (v + u) = 2aS

[We make use of the identity a^{2} - b^{2} = (a + b) (a - b)]

v^{2} - u^{2} = 2aS **III equation of motion**

### Graphical Derivation of Third Equation of Motion

Let 'u' be the initial velocity of an object and a be the acceleration produced in the body. The distance travelled 'S' in time 't' is given by the area enclosed by the v - t graph.

Graphical Derivation of Third Equation

S = area of the trapezium OABD.

Substituting the value of t in equation (1) we get,

2aS = (v + u) (v - u)

(v + u)(v - u) = 2aS [using the identity a^{2} - b^{2} = (a+b) (a-b)]

v^{2} - u^{2} = 2aS ** III Equation of Motion**

### Answered by | 10th Sep, 2011, 12:53: PM

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