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derivation of equations of motion by graphical method
Asked by gauri | 17 Sep, 2013, 12:53: AM

## 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

Answered by | 18 Sep, 2013, 04:34: PM

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