Derivations of equations of motion (graphically)

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

Hence, v = u + at 1^st equation of motion

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 ΔABC

      … (a)        

But v = u + at            

⇒ v – u = at                … (b)

Substituting this value in the equation (a), we get

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

       …. (1)

But we know that

Or

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

= (v + u) (v – u)

v² – u² = 2as