This document presents a summary of kinematics and derives the three kinematic equations of motion graphically. Kinematics deals with motion without forces. The three kinematic equations relate the variables of velocity (v), initial velocity (u), acceleration (a), time (t), and displacement (s). The document derives each equation graphically using a velocity-time graph: (1) v=u+at from the slope of the line, (2) v^2-u^2=2as from the area under the curve as a trapezoid, and (3) s=ut+1/2at^2 from considering the area under the curve as a triangle and rectangle.

1. GOOD MORNING
EVERYONE.......😃😃
I WELCOME YOU ALL TO TODAY’S
PRESENTATION...

2. TOPIC :- DERIVATION OF KINEMATICAL EQUATIONS
GRAPHICALLY.

3. KINEMATICS:-
•The branch of physics which deals with the
motion of objects without reference to the
forces which cause the motion.

4. WHAT ARE THE KINEMATICAL EQUATIONS OF
MOTION??
When a body moves from point A (where it’s initial
velocity is ‘u’) towards a point B(where it’s final
velocity becomes ‘v’), it produces some acceleration
(a), in time (t), and makes displacement (s). The
relation between all these variables of kinematics, i.
e., “v, u, a, t and s” is given by the equations which
are known as Kinematical equations.

5. LET US LOOK AT THE DIAGRAM FOR BETTER UNDERSTANDING
• U V
• a
• . A t . B
• S
•

6. THE 3 KINEMATICAL EQUATIONS
ARE:-
1)v=u+at
2) v²-u²=2as
3) s=ut+1/2at²

7. NOW LET’S DERIVE THESE
EQUATIONS OF MOTION BY
GRAPHICAL METHOD

8. FIRST EQUATION OF MOTION (V=U+AT)
• Let us derive 1st equation of motion from this velocity time graph.
• Slope of velocity Time graph gives us acceleration
• We know that, a=(v-u) /t.
• So putting the graphical values in this equation we get,
a=(BC-AO) /OC.
a=(v-u) /t
at=v-u
at+u=v
v=u+at

9. SECOND EQUATION OF MOTION (V²-U²=2AS)
• Area enclosed under the curve of a graph gives us
displacement.
• So here area under the curve is in shape of
Trapezium.
• And we know that A(Trapezium) =1/2*sum of
parallel
sides*height.
• Here A(ABCO) =1/2*(AO+BC) *OC
• So, s=1/2 *(u+v)*t
• By a=v-u/t, we can write t=v-u/a.
• So replacing t with v-u/a we get,

10. • S=1/2*(v+u) *(v-u/a)
• 2as=(v+u) (v-u)
• 2as=v²-u²
• And ultimately we get, v²-u²=2as.

11. THIRD EQUATION OF MOTION (S=UT +1/2*AT²)
• Area enclosed under the curve gives us displacement.
• Here let’s consider the area under the curve, in shape of
a triangle and a rectangle.
• So, we get s=A(🔺ABD) + A( ADCO)
• S=(1/2*b*h) +(l*b)
• S=(1/2*BD*AD) +(AD*AO)
• S=(1/2*v-u*t) +(t*u)

12. • By a=(v-u) /t, we can write v-u=at
• So by replacing v-u with at we get,
• S=(1/2*at*t) +(t*u)
• S=1/2at²+ut
• S=ut+1/2at²

13. QUESTION TIME
• What does the slope of velocity time graph give?
• Ans:-acceleration.
• What does the area enclosed under the curve of velocity time graph give?
• Ans:-Displacement.

14. OUR
SESSION
ENDS
HERE.....