Derive the following equations for uniformly accelerated motion by graphical method. a) Velocity -time relation b) Position - time relation 3) Position – velocity relation.
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Derive the following equations for uniformly accelerated motion by graphical method. a) Velocity -time relation b) Position - time relation 3) Position – velocity relation.
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Consider The velocity - time graph of an object moving under uniform acceleration as shown in the graph.
a) Equation for velocity time relation ( v = u + at ) :
∵ v = DC + u ..............(1)
Now , Acceleration (a) = Change in velocity / Time
By substituting the value of DC from (2) in (1) we get
⇒v = at +u
⇒v = u + at
b) Equation for Distance - time relation ( s = ut + 1/2 at^2 ) :
Distance covered by the object in the given time ‘t’ is given by the area of the trapezium EBDO.
Let in the given time, t the distance covered by the moving object = s
The area of trapezium, EBDO
= Distance (s) = Area of △ABD + Area of ADOE
⇒s = 1 /2 × AB × AD + (OD × OE)
⇒s = 1 /2 × DC × AD + (u + t)
⇒s = 1 /2 × at × t + (u + t)
⇒s = 1 /2 at ^2 + ut
⇒s = ut +1 /2 at ^2
c) Equation for Distance - Velocity relation ( 2as = v^2 - u^2 ) :
The distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDO
Therefore, Area of trapezium ABDO
= 1 /2 × ( sum of parallel sides+distance between parallel sides)
⇒ Distance (s) = 1 /2( DO + BE ) × OE
⇒s = 1 /2( u + v ) × t ----(iii)
Now from equation (ii)
⇒ a = v−u /t
∴ t = v − u/ a ----(iv)
After substituting the value of t from equation (iv) in equation (iii)
⇒s = 1 /2( u + v ) × ( v − u )/a
⇒s = 1 /2a ( u + v ) ( v − u )
⇒2as = ( u + v ) ( v − u )
⇒2as = v^2 - u^2
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