Let’s derive the second equation of motion with the help of graphical method. This equation is very useful in finding the displacement, acceleration, initial velocity and time.
Consider the body moving in XY plane starts its motion from point P with initial velocity ‘u’ (u ≠ 0) and reaches to point Q in time ‘t’ with uniform acceleration ‘a’ and final velocity ‘v’ shown in figure below.
In graph X- axis represents time and Y-axis represents velocity,
From graph we have,
OP =u (initial velocity)
OQ’ =v (final velocity)
OS=t (time)
Draw QS perpendicular to X axis
Now draw PR ll OS
Then from graph, OS=PR, OP=SR and
OQ’ – OP = RQ = v – u = change in velocity
Second equation of motion:
According to definition, velocity is rate of change of displacement hence,
∴ velocity = displacement/time
∴ displacement = velocity × time
From the above equation, we can say that the displacement (from point P to Q) of object is nothing the area covered object in the quadrilateral OPQS
∴ displacement = area of quadrilateral OPQS
(□OPQS) is combination of rectangle and triangle so that
Let’s discuss the uses of this equation with the help of few numerical as follows,
1) Ram pushed the stone of mass 100 g, from the terrace of his building. He found that the stone takes 6 seconds to reach the ground. What should be height of his building? (use g= 10 m/s2)
Ans: Here as the stone was initially at rest, u=0, t= 6 sec,
Stone is falling in influence of gravity, hence a= g = 10 m/s2
Using second equation of motion, we have
2) Ravindra Jadeja throws a ball from boundary at speed of 20 m/s and uniform acceleration of 0.5 m/s2 takes 2 seconds to settle in gloves of M.S. Dhoni. How far was Jadeja standing from Dhoni?
Ans: u=20 m/s, t= 2 sec, a= 0.5 m/s2
Using second equation of motion, we have