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2.1 Position.
Any object is situated at point O and three observers from three different places are looking for same
object, then all three observers will have different observations about the position of point O and no one will be
wrong. Because they are observing the object from their different positions.
Observer ‘ A’ says : Point O is 3 m away in west direction.
Observer ‘ B’ says : Point O is 4 m away in south direction.
Observer ‘ C’ says : Point O is 5 m away in east direction.
Therefore position of any point is completely expressed
by two factors: Its distance from the observer and its direction
with respect to observer.
That is why position is characterised by a vector known as position vector.
Let point P is in a xy plane and its coordinates are ( x, y). Then position vector( r )
of point will be x ˆ i^ y ˆ j
and if the point P is in a space and its coordinates are ( x, y, z) then position vector can be expressed as
r x ˆ i^ y ˆ j zk ˆ.
2.2 Rest and Motion.
If a body does not change its position as time passes with respect to frame of reference, it is said to be at
rest.
And if a body changes its position as time passes with respect to frame of reference, it is said to be in motion.
Frame of Reference : It is a system to which a set of coordinates are attached and with reference to which observer
describes any event.
C 4 m
3 m
E
W
N
5 m
O
A
B
S
STUDYGUIDE
A passenger standing on platform observes that tree on a platform is at rest. But when the same passenger
is passing away in a train through station, observes that tree is in motion. In both conditions observer is right.
But observations are different because in first situation observer stands on a platform, which is reference frame
at rest and in second situation observer moving in train, which is reference frame in motion.
So rest and motion are relative terms. It depends upon the frame of references.
2.3 Types of Motion.
One dimensional Two dimensional Three dimensional
Motion of a body in a straight line is
called one dimensional motion.
Motion of body in a plane is called
two dimensional motion.
Motion of body in a space is called
three dimensional motion.
When only one coordinate of the
position of a body changes with time
then it is said to be moving one
dimensionally.
When two coordinates of the position
of a body changes with time then it is
said to be moving two dimensionally.
When all three coordinates of the
position of a body changes with time
then it is said to be moving three
dimensionally.
e.g.. Motion of car on a straight road.
Motion of freely falling body.
e.g. Motion of car on a circular turn.
Motion of billiards ball.
e.g.. Motion of flying kite.
Motion of flying insect.
2.4 Particle or Point Mass.
The smallest part of matter with zero dimension which can be described by its mass and position is defined
as a particle.
Tree is at
rest
Platform (Frame of reference)
Moving train (Frame of reference)
Tree is in
motion
STUDYGUIDE
(ii) For a moving particle distance can never be negative or zero while displacement can be.
(zero displacement means that body after motion has came back to initial position)
i.e., Distance > 0 but Displacement > = or < 0
(iii) For motion between two points displacement is single valued while distance depends on actual path
and so can have many values.
(iv) For a moving particle distance can never decrease with time while displacement can. Decrease in
displacement with time means body is moving towards the initial position.
(v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is
along a straight line without change in direction.
(vi) If rA
and rB
are the position vectors of particle initially and finally.
Then displacement of the particle
rAB rB rA
and s is the distance travelled if the particle has gone through the path APB.
Sample problems based on distance and displacement
Problem 1. A man goes 10 m towards North, then 20 m towards east then displacement is
[KCET (Med.) 1999; JIPMER 1999; AFMC 2003]
(a) 22.5 m (b) 25 m (c) 25.5 m (d) 30 m
Solution : (a) If we take east as x axis and north as y axis, then displacement 20 ˆ i 10 ˆ j
So, magnitude of displacement 20 10 10 5
2 2
= 22.5 m.
Problem 2. A body moves over one fourth of a circular arc in a circle of radius r. The magnitude of distance travelled
and displacement will be respectively
(a) , 2 2
r
r
(b) r
r , 4
(c) 2
r
r (d) r , r
Solution : (a) Let particle start from A, its position vector r ri
O (^) A
Y
X
A
B
rAB
rB
rA
s
Y
X O A
B
STUDYGUIDE
After one quarter position vector rOB r ˆ j.
So displacement r ˆ j^ ri ˆ
Magnitude of displacement r 2.
and distance = one fourth of circumference 4 2
2 r r
Problem 3. The displacement of the point of the wheel initially in contact with the ground, when the wheel roles
forward half a revolution will be (radius of the wheel is R)
(a) 4 2
R
(b) 4
2 R (c) 2 R (d) R
Solution : (b) Horizontal distance covered by the wheel in half revolution = R
So the displacement of the point which was initially in contact with a
ground = 2 2
( R ) ( 2 R )
2 R
2.6 Speed and Velocity.
(1) Speed : Rate of distance covered with time is called speed.
(i) It is a scalar quantity having symbol .
(ii) Dimension : [ M
0
L
1
T
]
(iii) Unit : metre/second (S.I.), cm/ second (C.G.S.)
(iv) Types of speed :
(a) Uniform speed : When a particle covers equal distances in equal intervals of time, (no matter how small
the intervals are) then it is said to be moving with uniform speed. In given illustration motorcyclist travels equal
distance (= 5 m) in each second. So we can say that particle is moving with uniform speed of 5 m/ s.
Distance
Time
Uniform Speed
5 m 5 m
1 sec 1 sec 1 sec 1 sec 1 sec
5 m (^5) m 5 m
5m/
s
5m/
s
5m/s 5m/
s
5m/
s
5 m
5 m/s
1 m/s
2 R
Pnew
R
Pinitial
STUDYGUIDE
Distance averaged speed : When a particle describes different distances d 1 , d 2 , d 3 , ...... with different
time intervals t 1 , t 2 , t 3 , ...... with speeds v 1 , v 2 , v 3 ...... respectively then the speed of particle averaged over the
total distance can be given as
Total timeelapsed
Totaldistancecovered
av
1 2 3
1 2 3
t t t
d d d
3
3
2
2
1
1
1 2 3
d d d
d d d
When particle moves the first half of a distance at a speed of v 1 and second half of the distance at speed v 2
then
1 2
v v
vv
vav
When particle covers one-third distance at speed v 1 , next one third at speed v 2 and last one third at speed v 3 ,
then
1 2 2 3 31
vv vv v v
vv v
vav
(d) Instantaneous speed : It is the speed of a particle at particular instant. When we say “speed”, it usually
means instantaneous speed.
The instantaneous speed is average speed for infinitesimally small time interval ( i. e., t 0 ). Thus
Instantaneous speed
t
s
v
t
0
lim
dt
ds
(2) Velocity : Rate of change of position i.e. rate of displacement with time is called velocity.
(i) It is a scalar quantity having symbol v.
(ii) Dimension : [ M
0
L
1
T
]
(iii) Unit : metre/second (S.I.), cm/ second (C.G.S.)
(iv) Types
(a) Uniform velocity : A particle is said to have uniform velocity, if magnitudes as well as direction of its
velocity remains same and this is possible only when the particles moves in same straight line without reversing
its direction.
STUDYGUIDE
(b) Non-uniform velocity : A particle is said to have non-uniform velocity, if either of magnitude or
direction of velocity changes (or both changes).
(c) Average velocity : It is defined as the ratio of displacement to time taken by the body
Timetaken
Displacement
Average velocity ;
t
r
vav
(d) Instantaneous velocity : Instantaneous velocity is defined as rate of change of position vector of
particles with time at a certain instant of time.
Instantaneous velocity
t
r
v
t
0
lim
dt
dr
(v) Comparison between instantaneous speed and instantaneous velocity
(a) instantaneous velocity is always tangential to the path followed by
the particle.
When a stone is thrown from point O then at point of projection the
instantaneous velocity of stone is v 1 , at point A the instantaneous velocity
of stone is v 2 , similarly at point B and C are v 3 and v 4 respectively.
Direction of these velocities can be found out by drawing a tangent on the trajectory at a given point.
(b) A particle may have constant instantaneous speed but variable instantaneous velocity.
Example : When a particle is performing uniform circular motion then for every instant of its circular motion
its speed remains constant but velocity changes at every instant.
(c) The magnitude of instantaneous velocity is equal to the instantaneous speed.
(d) If a particle is moving with constant velocity then its average velocity and instantaneous velocity are
always equal.
(e) If displacement is given as a function of time, then time derivative of displacement will give velocity.
Let displacement
2
x A 0 A 1 t A 2 t
Instantaneous velocity ( )
2
A 0 A 1 t A 2 t
dt
d
dt
dx
v
v A 1 2 A 2 t
2
X O
v 1
Y
3
4 C
B A
STUDYGUIDE
Note : 1 |Averagespeed|
| Averagevelocity| | Av.speed||Av.velocity|
Problem 6. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/ h. Finding
the market closed, he instantly turns and walks back home with a speed of 7.5 km/ h. The average speed of
the man over the interval of time 0 to 40 min. is equal to
(a) 5 km/ h (b)
km/ h (c)
km/ h (d)
km/ h
Solution : (d) Time taken in going to market 30 min.
hr
As we are told to find average speed for the interval 40 min., so remaining time for consideration of motion is 10
min.
So distance travelled in remaining 10 min = 1. 25. 60
- 5 km
Hence, average speed = Total time
Totaldistance = km hr hr
km / 8
Problem 7. The relation 3 t 3 x 6 describes the displacement of a particle in one direction where x is in metres
and t in sec. The displacement, when velocity is zero, is
(a) 24 metres (b) 12 metres (c) 5 metres (d) Zero
Solution : (d) 3 t 3 x 6 3 x ( 3 t 6 )
2 3 x ( 3 t 6 ) 3 12 12 2 x t t
v = ( 3 12 12 ) 6 12
2 t t t dt
d
dt
dx
If velocity = 0 then,^6 t ^12 ^0 t^ ^2 sec
Hence at t = 2, x = 3(2)
2
Problem 8. The motion of a particle is described by the equation 2
x a bt where a 15 cm and b 3 cm. Its
instantaneous velocity at time 3 sec will be [AMU (Med.) 2000]
(a) 36 cm/sec (b) 18 cm/sec (c) 16 cm/sec (d) 32 cm/sec
Solution : (b)
2
x a bt v = bt
dt
dx 0 2
At t = 3 sec, v = 2 3 3 = 18 cm / sec (As b 3 cm )
STUDYGUIDE
Problem 9. A train has a speed of 60 km/h for the first one hour and 40 km/h for the next half hour. Its average speed
in km/h is [JIPMER 1999]
(a) 50 (b) 53.33 (c) 48 (d) 70
Solution : (b) Total distance travelled = 80 km
60 1 40 and Total time taken = hr hr hr 2
Average speed 53. 33 32
km/h
Problem 10. A person completes half of its his journey with speed 1 and rest half with speed 2. The average speed
of the person is [RPET 1993; MP PMT 2001]
(a) 2
(b) 1 2
(^212)
(c) 1 2
1 2
(d) 1 2
Solution : (b) In this problem total distance is divided into two equal parts. So
2
2
1
1
1 2
d d
d d av
1 2
d d
d d
1 2
(^) av = 1 2
Problem 11. A car moving on a straight road covers one third of the distance with 20 km/hr and the rest with 60 km/hr.
The average speed is [MP PMT 1999; CPMT 2002]
(a) 40 km/hr (b) 80 km/hr (c) km / hr
46 (d) 36 km/hr
Solution : (d) Let total distance travelled = x and total time taken t 1 + t 2
x / 3 x
Average speed
180
2
60
1
1
60
( 2 / 3 )
20
( 1 / 3 )
x x
x = 36 km / hr
2.7 Acceleration.
The time rate of change of velocity of an object is called acceleration of the object.
(1) It is a vector quantity. It’s direction is same as that of change in velocity (Not of the velocity)
(2) There are three possible ways by which change in velocity may occur
When only direction of velocity
changes
When only magnitude of velocity
changes
When both magnitude and direction
of velocity changes
Acceleration perpendicular to Acceleration parallel or anti-parallel Acceleration has two components one
STUDYGUIDE
e.g. (a) In uniform circular motion = 90º always
(b) In a projectile motion is variable for every point of trajectory.
(vi) If a force F acts on a particle of mass m, by Newton’s 2
nd
law, acceleration
m
F
a
(vii) By definition
2
2
dt
d x
dt
dv
a
dt
dx
v
As
i.e., if x is given as a function of time, second time derivative of displacement gives acceleration
(viii) If velocity is given as a function of position, then by chain rule
dt
dx
v
dx
d
v
dt
dx
dx
dv
dt
dv
a. as
(ix) If a particle is accelerated for a time t 1 by acceleration a 1 and for time t 2 by acceleration a 2 then average
acceleration is
1 2
11 22
t t
at at
aa
(x) If same force is applied on two bodies of different masses
m 1 and m 2 separately then it produces accelerations a 1 and a 2
respectively. Now these bodies are attached together and form a
combined system and same force is applied on that system so that
a be the acceleration of the combined system, then
F m 1 m 2 a
1 a 2
F
a
F
a
F
So,
1 2
a a a
1 2
1 2
a a
aa
a
(xi) Acceleration can be positive, zero or negative. Positive acceleration means velocity increasing with time,
zero acceleration means velocity is uniform constant while negative acceleration (retardation) means velocity is
decreasing with time.
(xii) For motion of a body under gravity, acceleration will be equal to “ g”, where g is the acceleration due to
gravity. Its normal value is
2
9. 8 m/s or
2
980 cm/s or
2
32 feet/s.
Sample problems based on acceleration
a 1
F m 1
a 2
F m 2
m 1 m 2
a
F
STUDYGUIDE
Problem 12. The displacement of a particle, moving in a straight line, is given by 2 2 4
2
s t t where s is in metres
and t in seconds. The acceleration of the particle is [CPMT 2001]
(a) 2 m/ s^2 (b) 4 m/ s^2 (c) 6 m/ s^2 (d) 8 m/ s^2
Solution : (b) Given 2 2 4
2
s t t velocity ( v) 4 t 2
dt
ds
and acceleration ( a)
2 4 ( 1 ) 0 4 m / s dt
dv
Problem 13. The position x of a particle varies with time t as. 2 3 x at bt The acceleration of the particle will be zero
at time t equal to [CBSE PMT 1997; BHU 1999; DPMT 2000; KCET (Med.) 2000]
(a) b
a (b) b
a
3
(c) b
a
3
(d) Zero
Solution : (c) Given
2 3 x at bt velocity
2 ( ) 2 at 3 bt dt
dx
v and acceleration ( a) = 2 a 6 bt.
dt
dv
When acceleration = 0 2 a 6 bt 0 b
a
b
a t 6 3
Problem 14. The displacement of the particle is given by. 2 4 y a bt ct dt The initial velocity and acceleration are
respectively [CPMT 1999, 2003]
(a) b , 4 d (b) b , 2 c (c) b , 2 c (d) 2 c , 4 d
Solution : (c) Given
2 4
y^ a bt ct dt v =
3 0 b 2 ct 4 dt dt
dy
Putting t 0 , vinitial = b
So initial velocity = b
Now, acceleration ( a)
2 0 2 c 12 dt dt
dv
Putting t = 0, ainitial = 2 c
Problem 15. The relation between time t and distance x is , 2
t x x where and are constants. The retardation
is ( v is the velocity) [NCERT 1982]
(a)
3
2 v (b)
3
2 v (c)
3
2 v (d)
2 3
2 v
Solution : (a) differentiating time with respect to distance x
dx
dt 2
dt x
dx v 2
So, acceleration ( a) =
dt
dx
dx
dv
dt
dv . =
2 3 2
vv v x
v
dx
dv v
Problem 16. If displacement of a particle is directly proportional to the square of time. Then particle is moving with
[RPET 1999]
(a) Uniform acceleration (b) Variable acceleration
(c) Uniform velocity (d) Variable acceleration but uniform velocity
Solution : (a) Given that
2 x t or
2
x Kt (where K= constant)
STUDYGUIDE
Problem 19. A body of mass 10 kg is moving with a constant velocity of 10 m/ s. When a constant force acts for 4 sec on
it, it moves with a velocity 2 m/ sec in the opposite direction. The acceleration produced in it is [MP PET 1997]
(a) 2 3 m/s (b) 2 3 m/s (c) 2
- 3 m/s (d) 2 0. 3 m/s
Solution : (b) Let particle moves towards east and by the application of constant force it moves towards west
1 10 m / s
and 2 2 m / s
. Acceleration Time
Changeinvelocity t
a 4
2 3 m / s
2.8 Position Time Graph.
During motion of the particle its parameters of kinematical analysis ( u, v, a, r) changes with time. This can
be represented on the graph.
Position time graph is plotted by taking time t along x-axis and position of the particle on y-axis.
Let AB is a position-time graph for any moving particle
As Velocity =
2 1
2 1
Time taken
Changeinposition
t t
y y
…(i)
From triangle ABC
2 1
2 1
tan
t t
y y
AC
AD
AC
BC
….(ii)
By comparing (i) and (ii) Velocity = tan
v = tan
It is clear that slope of position-time graph represents the velocity of the particle.
Various position – time graphs and their interpretation
Time
A
B
C
D
x
y
y 2
y 1
O (^) t 1 t 2
Positio
n
STUDYGUIDE
o
so v = 0
i. e., line parallel to time axis represents that the particle is at rest.
o
so v =
i. e., line perpendicular to time axis represents that particle is changing its position
but time does not changes it means the particle possesses infinite velocity.
Practically this is not possible.
= constant so v = constant, a = 0
i. e., line with constant slope represents uniform velocity of the particle.
is increasing so v is increasing, a is positive.
i. e., line bending towards position axis represents increasing velocity of particle.
It means the particle possesses acceleration.
is decreasing so v is decreasing, a is negative
i. e., line bending towards time axis represents decreasing velocity of the particle.
It means the particle possesses retardation.
constant but > 90
o
so v will be constant but negative
i. e., line with negative slope represent that particle returns towards the point of
reference. (negative displacement).
T
P
O
T
P
O
T
P
O
T
P
O
O T
P
O T
P
STUDYGUIDE
v m s v 6 m / s 1
and v 10 m / s 1
i.e. the speed is increasing at a constant
rate so motion is uniformly accelerated.
Problem 21. Which of the following graph represents uniform motion [DCE 1999]
(a) (b) (c) (d)
Solution : (a) When distance time graph is a straight line with constant slope than motion is uniform.
Problem 22. The displacement-time graph for two particles A and B are straight lines inclined at angles of 30 o and 60 o
with the time axis. The ratio of velocities of vA : vB is [CPMT 1990; MP PET 1999; MP PET 2001]
(a) 1 : 2 (b) 1 : 3 (c) 3 : 1 (d) 1 : 3
Solution : (d) v tan from displacement graph. So
tan 60
tan 30
o
o
B
A v
v
Problem 23. From the following displacement time graph find out the velocity of a moving body
(a) 3
m/ s
(b) 3 m/ s
(c) 3 m/ s
(d) 3
Solution : (c) In first instant you will apply tan and say,
tan 30
o
m/ s.
But it is wrong because formula tan is valid when angle is measured with time axis.
Here angle is taken from displacement axis. So angle from time axis
o o o 90 30 60.
Now tan 60 3 o
Problem 24. The diagram shows the displacement-time graph for a particle moving in a straight line. The average
velocity for the interval t = 0, t = 5 is
(a) 0
(b) 6 ms–^1
O
30 o
Time (
sec
)
Displacement (meter)
2 4
5 O^ t
10
20
x
t
s
t
s
t
s
t
s
STUDYGUIDE
(c) – 2 ms
(d) 2 ms
Solution : (c) Average velocity =
Totaltime
Totaldisplacement
5
= – 2 m/s
Problem 25. Figure shows the displacement time graph of a body. What is the ratio of the speed in the first second and
that in the next two seconds
(a) 1 : 2
(b) 1 : 3
(c) 3 : 1
(d) 2 : 1
Solution: (d) Speed in first second = 30 and Speed in next two seconds = 15. So that ratio 2 : 1
2.9 Velocity Time Graph.
The graph is plotted by taking time t along x-axis and velocity of the particle on y-axis.
Distance and displacement : The area covered between the velocity time graph and time axis gives the
displacement and distance travelled by the body for a given time interval.
Then Total distance|^ A 1^ || A 2 || A 3 |
= Addition of modulus of different area. i.e.
s | | dt
Total displacement A 1 A 2 A 3
= Addition of different area considering their sign. i.e.
r dt
here A 1 and A 2 are area of triangle 1 and 2 respectively and A 3 is the area of trapezium.
Acceleration : Let AB is a velocity-time graph for any moving particle
As Acceleration =
2 1
2 1
Time taken
Changeinvelocity
t t
v v
…(i)
From triangle ABC,
2 1
2 1
tan
t t
v v
AC
AD
AC
BC
….(ii)
By comparing (i) and (ii)
Acceleration ( a) =tan
Time
A
B
C
D
x
y
v 2
v 1
O (^) t 1 t 2
Velocity
t
1
2
3
1 2 3
0
10
20
30
Y
X Displacement
Time
STUDYGUIDE
