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Physics
Session
Kinematics - 2
Session Opener
or !
REST
MOTION
Session Objectives
Session Objective
1. Rest and motion
2. Distance and displacement
3. Uniform and non-uniform motion
4. Velocity
5. Acceleration
6. Equations of motion
Rest and Motion
Body at rest : Position constant
w.r.t. fixed point as time
increases
Fixed Point : Origin of a coordinate
system
Y
r   ox  oy
2
1
2 2

P (position)
y
r
oy
tan 
ox
ox  r cos 
oy  r sin 
900
O
(origin)

x
X
Motion
What is motion ?
Change in position of an object
with time, with respect to a given
co ordinate system.
Actual distance traveled :
Curve P0P1P2P3P4.
Displacement : Straight line
P0P4 directed from P0 to P4
y
P4(t=t4)
P3(t=t3)
P2(t=t2)
P1(t=t1)
P0(t=0)
x
Distance and Displacement
1. Distance  Displacement.
2. Distance = Displacement
(If direction remains same.)
3.Distance is always 0 or+ve.
Displacement can be +ve,0
or –ve.
x
2
4. Distance always increases
with motion.
o
1
t
Average Speed
y
P4(t=t4)
P3(t=t3)
P2(t=t2)
P1(t=t1)
P0(t=0)
x
actual dis tan ce cov ered
Average speed 
time taken
Curve lengthP0P1P2P3P4

time int erval (t 4  t 0 )
Courtesy:www.physicsclassroom.com
Instantaneous Speed
When time is infinitesimal (=t  0),
distance is infinitesimal(=s  0)
Instantaneous speed =
Speed : Scalar
Unit
Lim
t 0
s
t
m/s
s
Dimension LT-1
P1
s
Po
s
t
t
t
Average Velocity
Average velocity is defined as
displacement divided by time
taken.
 


r  r r
v av  2 1 
t 2  t1 t
Nature : vector

vav
Unit : m/s
Dimension : [LT-1]
r
r2
r1
t
t1
t2
Displacement and average velocity are in same direction
Instantaneous Velocity
If t= (t2 – t1)is extremely
small (t  0)
v
lim
t 0
r
t
v is instantaneous velocity
v is a vector.
Unit of v : m/s
r
r2
r2
r1
t1 t 2
t2
t
Uniform Motion (One dimension)
Equal displacement (x) traveled
in equal time interval (t)
x1  x0 x2  x1
x  x0

... 
v
t1  0
t2  t1
t 0
v is constant.
x = x0 + vt
If x0= 0 at t = 0 , x = vt
x0
x1
x2
x3
x4
x
0
t1
t2
t3
t4
t
Class Exercise
Class Exercise - 3
Graph in the figure below shows the
variation of displacement with respect to
time for a particle in one-dimensional
motion. Which of the following represents
the velocity-time graph of the particle in
motion?
x
0
5
10
t
15
20
Options is in next slide
Class Exercise - 3
(a)
v
(d)
0
5
10
t
15
20
v
(c)
0
5
t
v
(b)
0
5
10
t
v
0
5
10
t
15
20
10 15
15
20
20
Solution - 3
For t = 0 to 5
v
dx
(ve and cons tan t)
dt
t  5 to 15
v
dx
( 0)
dt
t  15 to 20
v
dx
(ve and constant)
dt
Hence answer is (c)
Class Exercise - 8
An object travels half the distance
with v1, with v2 for half of the
remaining time and with v3 for the
remaining half of the time. If the
object never reverses the direction
of motion, find the average velocity
during the motion.
Solution :
2v1  v2  v3
4
Average Acceleration
Change in velocity divided by the
time interval during which the
change occurs.

aav



v2  v1 v


t 2  t1
t
Nature : vector

aav
unit : m/s²
v
v2
v1
Dimension: [LT-2]
t1
t2
t
Non uniform motion (constant
acceleration)
Instantaneous acceleration
a lim
t 0
v
t
For constant acceleration
v (final velocity)  u(initial velocity)
a
t (time taken)
v = u + at Equation of motion (1)
t=t2-t1
v
v
u
t1
t2
t
Class Exercise
Class Exercise - 1
Which graph represents increasing
acceleration?
C
A
B
v
t
(a) A
(b) B
(c) C
(d) None of these
Solution - 1
dv
a
dt
Increasing a  Increasing slope of
v-t curve. By observation, we find that the
velocity is increasing at an increasing rate. So
acceleration is increasing.
Hence answer is (b).
Class Exercise - 9
An object starts from rest. It
accelerates at 2 m/s2 till it reaches
its maximum velocity. Then it
retards at 4 m/s2 and finally comes
to rest. If the total time taken is 6s,
find vmax and the displacement of
the object.
Solution - 9
Vmax = 2t1
Vmax = 4t2
4t2  2t1
t2 1

t1 2
t1  t2  6
s = Area of triangle
Vmax
\ t1 = 4 s, t2 = 2 s
Hence, Vmax = 8 m/s
1
s   6  8  24 m
2
O
6
t1
t2
Class Exercise - 2
A particle is thrown vertically
upwards with velocity v. It returns to
the ground in time T. Which of the
following graphs correctly represents
the motion?
(b)
(a)
v
T
2
v
t
T
2
T
–v
(d)
t
(c)
v
v
T
2
t
–v
T
2
T t
Solution - 2
The acceleration is constant
(= –g). So slope has to be
negative throughout the
motion and velocity varies
between v and –v.
Hence answer is (c).
Class Exercise - 5
If a particle has an initial velocity of




3 i  4 j and an acceleration of 0.4 i  0.3 ,j
its speed after 10 s is
(a) 10 units
(b) 7 units
(c) 7 2 units
(d) 8.5 units
Solution - 5
v  u  at
 


 
Hence, v   3 i  4 j    0.4 i  3 j 10

 


 



7 i7 j
Speed  v  7 2 m/s
Hence answer is (c).
One dimensional equations of
motion
Distance s = area under v-t graph
= ½ (u+v)t
Using equation of motion (1)
s = ut + ½ at2 Equation of motion (2)
But, s = vavgt
Hence, vavg = (u+v)/2
As t 
v u
, equation of motion (2)
a
v
v
t=t2-t1
u
gives
v2  u2  2 a.s Equation of motion (3)
t1
t2
t
One dimensional equations of
motion
v = u + at
s  ut 
1 2
at
2
v² = u² + 2as
Distance traveled in nth second
 2n  1 
Sn  u  
a

 2 
Class Exercise
Class Exercise - 6
A particle moves along the X-axis as
x = u(t – 3) + a(t – 3)2, then
Which of the following are true?
(a) initial velocity of particle is u at t = 0
(b) acceleration of particle is a
(c) at t = 3 the particle was at origin
(d) the particle may have negative velocity
Solution - 6
The observation of displacement
has started at time t = 3 s, after the
object has actually started. So if it
represents the time for which the
object has traveled and s be the
displacement after the observation
has started, then general form is
1 2
s  ut  at
2
compare with s  u(t – 3)  a(t – 3)2
Acceleration  2a, initial velocity at t  3s  u
Class Exercise - 10
The magnitude of maximum
acceleration, retardation of an object
is ‘a’ m/s2. What is the minimum time
taken by the object to cover a
displacement ‘s’ if it starts from rest
and finally comes to rest?
Solution :
The minimum time would be when
the acceleration is at maximum and
deceleration is also maximum. Half
the time accelerating at a and the
rest of the time deceleration at ‘a’.
1
s
t
Hence  a    t  s  t  2
sec
2
a
2
Uniform and Non-Uniform Motion
Let us see a comparison of uniform
and non-uniform motion
Courtesy:www.physicsclassroom.com
Class Exercise
Class Exercise - 7
An object has one-dimensional motion.
If V = 6t + 4t3, then
(a) what is the distance covered from
t = 3 s to t = 5 s?
(b) when is the acceleration < 0 for the
first time?
Solution :
5
(a)
 (6t  4t)3 dt  592
3
dv
(b) Since
is never zero
dt
So acceleration is
never negative.
Class Exercise - 4
A particle moves in a straight line,
starting from rest. The acceleration of
the particle is given by a  sin t 
1
t
 1
2
What is the distance traveled by the
particle in the time interval 0 to p
seconds.
Solution - 4
dv
1
 sin t 
dt
 t  12


1
 dt
 dv   sint 
2

 t  1 
0
0
t

t

t

1 
 v  t    cos t 
t  10

 v  t   2  cost 
1
t 1
[Check that v(0) = 0]
Solution - 4
Now v(t) 

dx  t 
dt
dx
1
 2  cos t 
dt
t 1
x
p


0
0
 dx 
1 

2

cos
t


 dt
t  1

 x  [2t  sin t  log  t  1]
 x = 2p – log(p + 1)
p
0
Thank you