This document discusses equations of motion for objects with uniform acceleration moving along a straight line. It defines key terms like initial velocity (u), final velocity (v), acceleration (a), time (t), and distance (s) traveled. It then derives and explains the equations of motion: 1) v = u + at 2) s = ut + 1/2 at^2 3) s = 1/2 (v^2 - u^2)/a It also provides 10 multiple choice questions testing understanding of concepts like displacement, velocity, acceleration, and interpreting velocity-time graphs.

2. CONSIDER AN OBJECT MOVING ALONG A STRAIGH LINE WITH UNIFORM
ACCELERATION ‘a’. LET ‘u’ BE THE INITIAL VELOCITY OF THE OBJECT IN TIME ‘t’ AND
‘v’ BE THE FINAL VELOCITY OF THE OBJECT AND ‘S’ BE THE DISTANCE
TRAVELLED.VELOCITY –TIME GRAPH OF THIS MOTION IS A STRAIGHT LINE AB AS
SHOWN OA=ED=u ,OC=EB=v,
OE=AD=t.
(I) EQUATION OF MOTION
WE KNOW THE VELOCITY TIME GRAPH OF UNIFORMLY ACCELERATED MOTION
REPRESENTS THE ACCELERATION OF THE OBJECT.
SO, ACCELERATION = SLOPE OF THE VELOCITY –TIME GRAPH AB
a =BD = DB = EB-ED = v-u
AD EB OE t
THEREFORE, v-u=at OR v= u + at

3. (2)EQUATION OF MOTION
WE KNOW THAT SLOPE OF VELOCITY-TIME GRAPH OF UNIFORMLY
ACCELERATED MOTION REPRESENTS THE ACCELERATION OF THE OBJECT.
THEREFORE , ACCELERATION = SLOPE OF VELOCITY-TIME GRAPH AB
a = DB = DB
AD T
DB = at
NOW DISTANCE TRAVELLED BY OBJECT IN TIME ‘t’ IS
S = AREA OF TRAPEZIUM OABE
=AREA OF RECTANGLE OADE + AREA OF TRIANGLE ADB
=OA x OE + 1 DB x AD = ut + 1 at x t
2 2
THEREFORE , S = ut + 1 at x T
2

4. (3)EQUATION OF MOTION
DISTANCE TRAVELLED BY OBJECT IN TIME TRAVEL ‘t’ IS
S = AREA OF TRAPEZIUM OABE
= 1 ( EB+OA ) x OE = 1 ( EB ED) x OE
2 2
ACCELERATION , a = SLOPE OF VELOCITY- TIME GRAPH AB
a = DB = EB – ED
AD OE
OE = ED – EB
a
PUTTIN THIS VALUE IN EQUATION (1), WE GET
S= 1 ( EB+ED ) x (EB -ED) = 1 (EB x EB – ED x
ED)
2 a 2 a
S = 1 (v x v– u x u )
2a
2as = (v x v– u x u )

6. ANSWER MY QUESTIONS-
Q1) DISTANCE CAN NEVER BE ____ THAN DISPLACE MENT:
1) =(EQUAL) 2) >(GREATER) 3)<(SMALLER)
Q2) DISTANCE CAN NEVER BE:
1)O 2)1000 3)10000 4)100000
Q3)DISTANCE IS A __________ QUANTITY :
1)VECTOR 2) SCALAR
Q4)DISPLACEMENT IS A ___________ QUANTITY:
1) SCALAR 2) VECTOR

7. Q5) AN OBJECT IS MOVING IN A CIRCULAR PATH OF RADIUS ‘r’. THE
DISPLACEMENT AFTER HALF A CIRCLE WOULD BE?
1) 0 2) ƛr 3)2r 4)2ƛr
Q6)OUT OF THE FOLLOWING WHICH IS NOT A VECTOR QUANTITY?
1)DISPLACEMENT 2)SPEED 3)VELOCITY 4)ACCELERATION
Q7)THE AREA UNDER VELOCITY-TIME GRAPH REPRESENTS A
PHYSICAL QUANTITY WHICH IS:
1) DISTANCE 2) SPEED 3)ACCELERATION 4)DISPLACEMENT
Q8)WHEN A CAR DRIVER TRAVELLING AT A SPEED OF 10m/s APPLIES
BRAKES AND BRINGS THE CAR TO REST IN 20s,THEN
RETARDATION WILL BE:
1) 2m/s x s 2) -2m/s x s 3)-0.5m/s x s 4)0.5m/s x s

8. Q9)WHICH OF THE FOLOWING GRAPHS REPRESENTS THE
MOVEMENT OF A CAR THAT SLOWS DOWN AND COMES TO REST?
1) 2) 3) 4)
Q10)THE SLOPE OF VELOCITY- TIME GRAPH GIVES:
1)DISTANCE 2)VELOCITY 3)ACCELERATION
4)DISPLACEMENT