There are three basic equations of motion for bodies having uniformly accelerated rectilinear motion. These three equations relate time t, distance covered S, initial velocity `\ v_i`, final velocity `\ v_f`, and acceleration an of a moving body.



To simplify the derivation of these equations, we assume that:
  • The motion of the body is rectilinear motion (motion in a straight line).
  • The body is moving with uniform acceleration and
  • Assuming only the magnitude of displacements, velocities, and acceleration.

Consider a body moving with initial velocity `\ v_i` in a straight line with uniform acceleration a. Its velocity becomes `\ v_f` after time t. The motion of the body is described by a speed-time graph as shown in the figure by line AB. The slope of line AB is acceleration a. The total distance covered by the body is shown by the shaded area under line AB. Equations of motion can be obtained easily from this graph


FIRST EQUATION OF MOTION ( `\ v_f` = `\ v_i` + at ):

Consider a body moving with initial velocity `\ v_i` in a straight line with uniform acceleration a. Its velocity becomes `\ v_f` after time t. The motion of the body is described by a speed-time graph as shown in the figure by line AB below. 



The slope of line AB gives the magnitude of its acceleration a. Mathematically


acceleration = a = slope of line AB = `\frac {BC}{AC}`

From the graph we have AC = OD and BC = BD - CD, So


a = `\frac {BD - CD}{OD}`


As BD = `\ v_f` , CD = `\ v_i` and OD = t

a = `\frac {v_f - v_i}{t}` = change in velocity / time interval
or

`\ v_f` - `\ v_i` = at

or

`\ v_f` = `\ v_i` + at -----------------(1)





SECOND EQUATION OF MOTION ( S = `\ v_i` t + `\frac {1}{2}` at²):


As we know that the area under the speed-time graph represents the total distance S traveled by the object. So, from the above figure

Total distance traveled = S = area of (rectangle OACD + triangle ABC)

Area of rectangle OACD = OA ｘ OD = `\ v_i` ｘ t

Area of triangle ABC = `\frac {1}{2}` ( AC ｘ BC) = `\frac {1}{2}` ( t ｘ at)


thus

Total distance traveled = S = `\ v_i` ｘ t + `\frac {1}{2}` ( t ｘ at)

or


S = `\ v_i` t + `\frac {1}{2}` at²  ------------------(2)



THIRD EQUATION OF MOTION ( 2aS = `\v_f ^2` - `\v_i^2` ):

As we know that the area under the speed-time graph represents the total distance S traveled by the object. So, from the above figure


Total Distance Travelled = S = area of the trapezium (OABD) of the graph


Total Distance Travelled = S = `\frac {1}{2}` ｘ {sum of the parallel sides} ｘ height of Trapezium

Total Distance Travelled = S = `\frac {OA + BD}{2}` ｘ OD

multiplying both side by `\frac {BC}{OD}` we get

S ｘ `\frac {BC}{OD}` = `\frac {OA + BD}{2}` ｘ OD ｘ `\frac {BC}{OD}`

OD will cancel from RHS 

S ｘ `\frac {BC}{OD}` = `\frac {OA + BD}{2}` ｘ BC

As BC = at, OD = t, OA = `\ v_i`, BD = `\ v_f` and BC = BD - CD = `\ v_f` - `\ v_i` So,

S ｘ `\frac {a \cancel t}{\cancelt}` = `\frac {v_i + v_f}{2}` ｘ `\ v_f` - `\ v_i`

Or

Sa = `\frac {v_f ^2 - v_i^2}{2}`

Or

2aS = `\v_f ^2` - `\v_i^2` ---------------(3)


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