Motion

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08 Equation of Motion by Graphical Representation

EQUATION OF MOTION BY GRAPHICAL REPRESENTATION

The relation between quantities like velocity, time, acceleration and displacement provided the acceleration remains constant are collectively known as equation of motion.

FIRST EQUATION OF MOTION

The first equation of motion is:

v = u + a t

Derivation of v= u + at by Graphical Method :

Initial velocity of the body u = OA , … (1)

And, Final velocity of the body v = BC … (2)

But from the graph BC = BD + DC and

Therefore, v = BD + DC … (3)

Again DC = OA

So, v = BD + OA

Now, from equation (1), OA = u

So, v = BD + u

∴ v - u = B D . . . . . . . . . . . . . . . . . . . . (4)

We should find out the value of BD now. We know that the slope of a velocity-time graph is equal to acceleration, a.

a = \frac{C h a n g e i n v e l o c i t y}{t} = \frac{B D}{t}

a = \frac{B D}{t}

BD = at

Now, putting this value of BD in equation (4) we get:

v = at + u

This equation can be rearranged to give:

v = u + at

SECOND EQUATION OF MOTION

Derivation of

s = u t + \frac{1}{2} a t^{2}

by Graphical Method

= A r e a o f r e c \tan g l e O A D C + A r e a o f t r i a n g l e A B D

(i) A r e a o f r e c \tan g l e O A D C = O A \times O C

= u \times t

= u t

……………………..(5)

(i i) A r e a o f t r i a n g l e A B D = \frac{1}{2} \times A r e a o f r e c \tan g l e A E B D

= \frac{1}{2} \times A D \times B D

= \frac{1}{2} \times t \times a t (b e c a u s e A D = t a n d B D = a t)

= \frac{1}{2} a t^{2}

……………………(6)

Distance travelled,

s = A r e a o f r e c \tan g l e O A D C + A r e a o f t r i a n g l e A B D

O r, s = u t + \frac{1}{2} a t^{2}

THIRD EQUATION OF MOTION

v^{2} = u^{2} + 2 a s

D e r i v a t i o n o f v^{2} = u^{2} + 2 a s b y g r a p h i c a l m e t h o d

Distance travelled, s = Area of trapezium OABC

s = \frac{(S u m o f p a r a l l e l s i d e s) \times h e i g h t}{2}

s = \frac{(O A + C B) \times O C}{2}

Now, OA + CB = u + v and OC = t. Putting these values in the above relation, we get:

s = \frac{(u + v) \times t}{2} . . . . . . . . . . . . . . . . . . . (7)

v = u + at (first equation of motion)

And, at = v – u

So,

t = \frac{(v - u)}{a}

s = \frac{(u + v) \times (v - u)}{2 a}

2 a s = v^{2} - u^{2} [B e c a u s e (v + u) (v - u) = v^{2} - u^{2}]

v^{2} = u^{2} + 2 a s

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