# Equation of motion by graphical method

Question 1 What is meant by uniform circular motion?

Question 2 A body goes around the sun with constant speed in a circular orbit.Is the motion uniform or accelerated?

Question 3 A satellite goes around the earth in a circular orbit with constant speed.Is the motion uniform or accelerated?

Question 4 Give examples of circular motion?

Question 5 What is a centripetal force.Give example?

Question 6 A cyclist goes around a circular track once every 5 minute.If the radius of the circular track is 110 meters,calculate his speed?

Question 7 Derive first equation of motion by graphical method?

Question 8 Derive second equation of motion by graphical method?

Question 9 Derive third equation of motion by graphical method?

Uniform circular motion
When a body moves in a circle,it is called circular motion.
When a body moves along a circular path,then its direction of motion keeps on changing continuously.
Sine the velocity changes(due to continuos change in direction)therefore the motion along a circular path is said to be accelerated.
When a body moves in a circular path with uniform speed its motion is called uniform circular motion.
The velocity of the body moving in a circle with uniform speed is not uniform because the direction of motion is constantly changing.

For Ex:A stone tied to a thread is rotated in circular path with uniform speed in clockwise direction.
A ———>Speed is directed towards east.
B ———>Speed is directed towards south.
Since there is change in direction of speed,the velocity is not uniform.
The motion in a circle with constant speed is an example of accelerated motion.
The force which is needed to make an object travel in circular path is called centripetal force.
For Ex: 1)Movement of artificial satellite around earth .
2)Motion of moon around earth.
3)Motion of earth around sun.
4)Tip of seconds hand of a watch.
5)Athlete moving on a circular path.

The speed of a body moving along a circular path is given by:
2 π r
v=———
t

Equation of motion by graphical method.

1)Derivation of v=u +at
Initial velocity u at A =OA
Velocity changes from A To B in time t(uniform acceleration a)
Final Velocity v=BC
BC=BD+DC
v=BD+Ao
v=BD+u
Slope of velocity time graph is equal to acceleration a.

BD

a = ——–

BD
a = ——-
t

BD = at
v=u+at

1

2)Derivation of  S = ut + —- a t2

2

The distance travelled by the body is given by area of the space between velocity time graph AB and time axis OC , which is equal to area of figure OABC.
Distance travelled=Area of figure OABC
=Area of rectangle OADC+Area of triangle ABD
1
=(OA *OC) + (—- *AD * BD)
2

1
=(u*t) + (—-  *  t * at)

2

1

S = ut + —- a t2

2
3)Derivation of v2  = u2  + 2as

The distance travelled by body in time t is given by area of figure OABC(which is a trapezium)
s = Area of trapezium OABC

Sum of parallel sides * height
s=———————————————————————-
2
(OA+OB) *OC
s=——————–
2
(u+v) * t
s=—————–

2

v=u + at

at = v – u

v – u
t = ————
a

(u+v) * (v- u)
s=———————
2a

2as = v2  – u

v = u2  + 2as