An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement s.
Let us try to derive these equations by graphical method.
Equations of motion from velocity-time graph:
Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph.
The initial velocity of the object = u = OD = EA
The final velocity of the object = v = OC = EB
Time = t = OE = DA
Also from the graph we know that, AB = DC
For First equation of motion
By definition, acceleration = change in velocity / time
= (final velocity – initial velocity)/time
= (OC – OD) / OE = DC / OE
a = DC/t
DC = AB = at
From the graph EB = EA + AB
v = u + at …….(1)
This is first equation of motion.
For Second equation of motion
From the graph the distance covered by the object during time t is given by the area of quadrangle DOEB
s = area of the quad-rectangle DOEB
s= area of the rectangle DOEA + area of the triangle DAB
(AE × OE) + (\(\frac{1}{2}\) × AB × DA)
s = ut + \(\frac{1}{2}\)at2 ….(2)
This is second equation of motion.
For Third equation of motion
From the graph the distance covered by the object during time t is given by the area of thequadrangle DOEB.
Here DOEB is a trapezium. Then
s = area of trapezium DOEB
= \(\frac{1}{2}\)× sum of length of parallel side × distance between parallel sides
= \(\frac{1}{2}\) × (OD + BE) × OE
s = \(\frac{1}{2}\) × (u + v) × t
since a = (v – u) / t or t = (v – u)/a
Therefore s = \(\frac{1}{2}\) × (v + u) × (v – u)/a
2as = v2 – u2
v2 = u2 + 2 as ………..(3)
This is third equation of motion.