SAMAGAM INITIATIVE BY : PARMAR OFFICERS PHYSICS NOTES
What is Motion?
&
Relative change in position
↳ What is needed?
>
-
Reference point
· ·
A B
RS
Point of origin
CE
Two physical quantities
Shortest part between two points (is a straight line)
L -
Distance Displacement
I
↳
N
Case 1
-----------------
50 m
FF Case 2
50 m
O
L
↓
Distance covered Distance = 100 m
Displacement = 0
R
A
Distance = 50 m
Displacement = 50 m
Displacement: vector quantity
RM
>
- Distance: scalar quantity
4
3
- -
PA
↓
1
Case 1: shortest distance
Displacement
Types of Motion
5 sec 5 sec 5 sec -
> Equal distance covered in equal
intervals: Uniform Motion
-
A l
5m 5m 5m
2 cases
↳ 5 sec 6 sec 3 sec Equal distance in unequal intervals:
RS
>
-
↑ ↑
Non-uniform motion
5m 5m 5m
W
CE
~
Wind speed is decreasing after increasing: Retardation
Rate of change of motion -
> Called as speed
I
↓
Time involved
>
Speed = Distance
Time
FF
- m/km
O
↳ s/hr
When motion is non-
uniform
S.I Unit: m/s >
- Represented
↑
R
Average speed = Total distance
A
Km/hr >
- m/sec Total time
↳x 5
RM
18
m/s > Km/hr
x 18
PA
>
-
Avg. speed = 16m + 16m
4+2
= 32 = 16 m/s
6 3
Speed + Direction = Velocity
Velocity = Displacement
Time
Unit: m/s
RS
Scalar quantity Vector quantity
Only represents numerical value Numerical value + Direction
CE
Distance Displacement
I
Speed Velocity
Time
FF
O
R
Avg. speed = 180 = 3 m/s
60
A
Avg. velocity = 0 = 0 m/s
60
RM
PA
Rate of change in velocity Is called Acceleration
5 m/s
5 m/s 5 m/s
Velocity changing with
direction
Change in velocity = Final speed - Initial speed
Acceleration = m/s = m/s
s
a
RS
CE
Case 1
Initial speed = 0 30 s a = 6 - 0 = 1 = 0.2 m/s
I
Final speed = 6 m/s 30 5
Case 2
Initial speed = 6 m/s
5s
FF
a = 4 - 6 = -2 = -0.4 m/s
O
Final speed = 4 m/s
5 5
R
Graph Representation
A
RM
Slope = speed
Slope = Acceleration
Distance Velocity
PA
Time Time
Area = Motion Area = Displacement
Numerical of Motion in straight line: The ‘UTSAV’ Concept
3 equations of motions
1 v = u + at Horizontal motion u = initial velocity
2 s = ut + 1 at t = time
2 s = distance
3 v - u = 2 as a = acceleration
RS
v = final velocity
CE
Straight line motion Rectilinear motion
I
Vertical
Horizontal
FF
O
Vertical motion (a = -g)
Against the gravity
a=g Acceleration due to gravity
R
Free fall u=0 v = u - gt
h = ut - 1 gt
A
Initial speed
v = u + gt 2
h = ut + 1 gt v - u = -2gh
RM
2
v - u = 2gh
Final point = 0
PA
Final velocity = 0
RS
5 x 60 = 300 s
u=0 5 min
v = 72 km/hr
CE
72 x 5 = 20 m/s
18
a = 20-0 = 20 = 1 m/s
I
300 300 15
s = ut + 1 at
2
FF
O
s = 0 + 1 x 1 x (300)
2 15
R
= 3000 m
s = 3 km
A
RM
PA
u = 18 km/h x 5 a = 10-5 = 5 = 1 m/s
18 5 5
= 5 m/s
s = ut + at
v = 36 x 5 = 10 m/s = 5 (5) + 1 x (5)
18 2
= 37.5 m
Uniform Circular Motion motion of a body moving with speed along the circular path
Uniform = speed a =v
r
Centripetal acceleration
RS
Acceleration towards the
centre in circular path
CE
Change in velocity at every point
I
FF
O
A Displacement = 0
R
100 m
A
10 s = 100 m
RM
1 min = 60 sec
B
One Liners (MCQs)
PA
Distance in a particular direction is called velocity
Displacement = velocity
Time
