Question

Derive the equations of motion by graphical method.

Answer 1

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 quadrangle DOEB
s = area of the rectangle DOEA + area of the triangle DAB
= (AE × OE) + (1/2 × AB × DA)
s = ut + 1/2at² ….(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 the quadrangle DOEB. Here DOEB is a trapezium. Then
s = area of trapezium DOEB
= 1/2 × sum of length of parallel side × distance between parallel sides
= 1/2 × (OD + BE) × OE
s = 1/2 × (u + v) × t
Since a = Since a = (v – u) / t or t = (v – u)/a
Therefore = 1/2 × (v + u) × (v – u)/a .
2as = v² = u² + 2as
v² = u² + 2as ………(3)
This is third equation of motion.