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## Derivations of Equations of Motion (Graphically)

## First Equation of Motion

Graphical Derivation of First Equation

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**

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)

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**

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Derivations of Equations of Motion (Graphically)
First Equation of Motion
Graphical Derivation of First Equation
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
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)
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 a2 - b2 = (a+b) (a-b)]
v2 - u2 = 2aS III Equation of Motion

You have given Saddy to this answer

You have given Gladdy to this answer

A mail has been sent to the Administrator, who will approve the reporting

Derivations of Equations of Motion (Graphically)
First Equation of Motion
Graphical Derivation of First Equation
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
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)
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 a2 - b2 = (a+b) (a-b)]
v2 - u2 = 2aS III Equation of Motion

You have given Saddy to this answer

You have given Gladdy to this answer

A mail has been sent to the Administrator, who will approve the reporting

Derivations of Equations of Motion (Graphically)
First Equation of Motion
Graphical Derivation of First Equation
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
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)
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 a2 - b2 = (a+b) (a-b)]
v2 - u2 = 2aS III Equation of Motion

You have given Saddy to this answer

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