Suppose a body having initial velocity 'u' attains a final velocity 'v' in time 't' travels a distance 's' undergoing a uniform acceleration 'a'. Using graphical method, derive the distance - time relation.
Answers
Answer:
hence we can find distance from velocity time graph as its area and acceleration as its slope
Given: A body having initial velocity 'u' attains a final velocity 'v' in time 't' travels a distance 's' undergoing a uniform acceleration 'a'.
To Derive: Distance-time relation using the graphical method
Proof:
The velocity-time graph of the given body is represented by a straight line., AB, as shown in the attached figure.
Distance traveled is given by the area of the space enclosed between the velocity-time graph and the time axis.
So, distance, s = Area enclosed by the figure OABC
= Area of rectangle OADC + Area of triangle ABD
= OA x AC + 1/2(BD x AD)
= u x t + 1/2(BC-CD) x OC
= u x t + 1/2(BC-OA) x OC
= ut + 1/2(v - u)t..........(i)
Also, slope of the velocity-time graph gives acceleration.
tanθ = (v₂ - v₁)/(t₂ - t₁)
⇒ a = (v - u)/t ............................(ii)
Putting the value of eq (ii) in eq(i), we get
s = ut + 1/2 at²
This is the required position-time relation.
