NTSE Motion Resonance

Rahul Pancholi Sir Unacademy Notes Target - NTSE
1. MOTION
INTRODUCTION :
 PHYSICS : It is the branch of science in which we observe, measure and describe nature and
natural phenomena.
 MECHANICS : It is the branch of physics which deals with the study of object in the condition of
rest or motion. It is divided into two parts
(i) Statics (ii) Kinematics (iii) Dynamics
(i) Statics : It deals with the study of objects at rest or in equilibrium, even when they are under the action
of several forces.
(ii) Kinematics
Kinematics: It deals with the study of motion of objects without considering the cause of motion .
eg. equations of motion
(iii) Dynamics : It deals with the study of motion of objects with considering the cause of motion .
Eg. Newton’s laws of motion
A. REST AND MOTION

(a) Definition of Rest and Motion :
Motion is a combined property of the object and the observer. There is no meaning of rest or
motion without the observer. Nothing is in absolute rest or in absolute motion.
An object is said to be in motion with respect to a observer, if its position changes with
respect to that observer. It may happen by both ways either observer moves or object moves.
(i) Rest : An object is said to be at rest if it does not change its position w.r.t. its surroundings with the
passage of time.
Eg. : The chair, black board, table in the class room are at rest w.r.t. the students.
(ii) Motion : A body is said to be in motion if its position changes continuously w.r.t. the surroundings (or
with respect to an observer) with the passage of time.
Eg. : A car moving on the road will be in motion w.r.t. to the person standing on the road
(iii) Rest and motion are relative terms, there is nothing like absolute motion or rest.
Eg. : A train is moving on the track, the passengers are seated, will be stationary with respect to each
other but in moving condition with respect to station.
Therefore, all the motions are relative. There is nothing like absolute motion.
(iv) To study the motion of an object, following points are essential : .
(I) Concept of a Point Object : In mechanics while studying the motion of an object, sometimes its
dimensions are not important and the object may be treated as a point object without much error. When
the size of the object is much less in comparison to the distance covered by the object then the object is
considered as a point object.
Eg. Earth can be considered as a point object for studying its motion around the sun. Because length of
the path covered by the earth in one revolution is very large in comparision to the size of earth, so earth
can be considered as a point object.
(II) Frame of Reference : A fixed point or a fixed object with respect to which the given body changes its
position is known as reference point origin. To locate the position of object we need a frame of reference.
A convenient way to set up a frame of reference is to choose three mutually perpendicular axis and name
them x-y-z axis. The co-ordinates (x, y, z) of the particle then specify the position of object w.r.t. that
frame. If any one or more co-ordinates change with time, then we say that the object is moving w.r.t. this frame
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Fig. 1
(b) Motion in one ,two and three dimensions :
As position of the object may change with time due to change in one or two or all the thee co-ordinates,
so we have classified motion as follows :
(i) Motion in 1-D : If only one of the three co-ordinates specifying the position of object changes w.r.t.
time then its motion is known as 1D motion . In such a case the object moves along a straight line. This
motion is also known as rectilinear or linear motion.
E.g. (I) Motion of train along straight railway track.
(II) An object falling freely under gravity.
(II) When a particle moves from P1 to P2 along a straight line path only the x-co-ordinate changes.
x2
x1
O P1 P2 X
Fig. 2
(ii) Motion in 2-D : If two of the three co-ordinates specifying the position of object changes w.r.t. time,
then the motion of object is called two dimensional. n such a motion the object moves in a plane.
E.g. (I) Motion of queen on carom board.
(II) An insect crawling on the floor of the room.
(III) Motion of object in horizontal and vertical circles etc.
(IV) Motion of planets around the sun.
(V) A car moving along a zig-zag path on a level road.
(iii) Motion in 3-D : If all the three co-ordinates specifying the position of object changes w.r.t. time, then
the motion of object is called 3-D. In such a motion the object moves in a space.
Eg. (I) A bird flying in the sky (also kite).
(II) Random motion of gas molecules.
(III) Motion of an aeroplane in space
(c) Types of motion :
(i) Linear motion (or translatory motion) : When an object moves in straight line than motion is called
as linear motion.
Eg. : The motion of a car moving on straight road, a running person, a stone being dropped, motion of a
train on a straight track
(ii) Rotational motion : Motion of a body around a fixed axis is called rotational motion.
Eg.The motion of an electric fan, motion of earth about its own axis.
(iii) Oscillatory motion : The to and fro periodic motion of a body around a fix point is called oscillatory
motion.
Eg. : The motion of a simple pendulum, a body suspended from a spring .
(d) Scalar and vector quantities :

Physical quantities (i.e. quantities of physics) can be divided into two types :
(i) Scalar quantity : Any physical quantity, which can be completely specified by its magnitude, is
known as scalar quantity
Eg. Charge, distance, speed, time, temperature, density, volume, work, power, energy, pressure, potential
etc.

(ii) Vector quantity : The quantity which can be determined by its magnitude and the direction and also
can be added or subracted by vector algebra, is called a vector quantity.
Eg. Displacement, velocity, acceleration, force, momentum, weight and electric field etc.
(I) Representation of a vector : A vector is represented by a directed line segment drawn in the given
direction on a certain scale.The length of line shows its magnitude and arrow shows direction.
Tail head
(symbolic representation)
(iii) Difference between scalar and vector :
Scalar Vector
1. They have magnitude only. 1. They have magnitude as well as
direction.
2. They are added or 2. They are added or subtracted by
subtracted arithmetically like the process of vector addition.
3 kg + 5 kg = 8 kg
(e) Distance and displacement :

(i) Distance : It is the actual length of path covered by a moving particle . It is a scalar quantity. Its S.I.
unit is metre (m).
(ii) Displacement : It is the shortest distance between the initial and final position of the particle. It is a
vector quantity. Its S.I. unit is metre (m).
Eg. : Consider a body moving from a point A to a point B along the path shown in figure. Then total length
path covered is called distance (path-1). While the length of straight line AB in the direction from A to B is
called displacement (path-2). path-1
A B
path-2
Fig. 3
NOTE : If a body travels in such a way that it comes back to its starting position, then the displacement
is zero. However, distance travelled is never zero in case of moving body.
(iii) Some important points :
(I) When an object moves towards right from origin, its displacement considered as positive.
(III) When an object moves towards left from origin its displacement considered as negative.
(III) When an object remains stationary or it moves first towards right and then an equal distance towards
left, its displacement is zero.
(IV) Shifting origin causes no change in displacement.
(V) If body moves along the circumference of the circle of radius r then distance travelled by it is given by
2r and displacement is given by zero, for one complete revolution.
(f) Difference between distance and displacement :

Distance Displacement
1. Distance is the length of the 1. Displacement is the shortest
path actually traveled by a path between the initial and the
body in any direction. final positions of a body in the
direction of the point of the final
position.
2. Distance between two given 2. Displacement between two
points depends upon the path points is measured by the straight
chosen. path between the points.
3. Distance is always positive. 3. Displacement may be positive
as well as negative and even
zero.
4. Distance is a scalar quantity. 4. Displacement is a vector
quantity.
5. Distance will never 5. Displacement may decrease.
decrease.

Illustration 1. A body starts from A and moves according to given figure. (body retraces the path after C then
reaches to D than find distance and displacement for each path
D A 4m B 6m C
5m
Solution.
Path Distance Displacement
AB 4m 4m
ABC 10m 10m
ABCB 16m 4m
ABCA 20m 0m
ABCAD 25m –5m
Illustration 2. There are n steps each of dimension l,b & h if a man climbs n steps what is his displacement and
distance.
Sol.
By Pythagorean theorem AB = b2  h2
Similarly for each step, Displacement = b2  h2 b
So that total displacement = n × B

b2  h2 l h
Distance = n (h+b)
A
Illustration 3. A person moves in a circular path centered at O of radius 1m. He starts from A and reaches
diametrically opposite point B. Then find :
(a) distance between A and B (b) displacement between A and B
Sol. (a) Distance = Length of actual circular path from A to B = Half the circumference
2r
i.e. Distance =  r N
2
r=1m
as r = 1m W E
B 0 A
 Distance =  m S
(b) Displacement = 2r along west.

= 2m along west
D C
Illustration 4. A body moves from A to B and come back then
goes to point C. Find distance and displacement of
the journey. b
Sol. Distance = AB + BA + AC = a + a + a2  b 2  2a  a 2  b2 A B
a
Displacement = AC = a2  b 2
Illustration 5. A body moves on three quarters of a circle of radius r. Find the displacement and distance
travelled by it.

(a) Displacement = r2  r2 = r 2
S Q
3 3r
(b) Distance =  2r =
4 2
P

Q.1_ A stone is thrown vertically upwards and after ascending a height ‘h’ it comes back to the hands of the
thrower. What is the total distance covered ?
Q.2_ What is the value of distance & displacement when body travel in the straight line path.
Q.3_ Is absolute rest possible or not ?

Q.4_ An object move in circular path of radius r. Find the distance and dispalacement for two compete
revolution.
Q.5_ Give an example of motion with displacement = 0 and distance travelled  0
Q.6_ Is it possible to have distance travelled = 0 but displacement  0 ? If yes, give an example.
Q.7_ A Body moves 6 m north. 8 m east and 10m vertically upwards, Find the resultant displacement of the
body from initial position . (Ans. 10 2 )
Q.8_ A man goes 10m towards North, then 20m towards east then find the displacement of the body from it
s initial position. ( Ans. 10 5 )
Q.9_ A wheel of radius 1 meter rolls forward half a revolution on a horizontal ground.Find the magnitude of the
displacement of the point of the wheel initially in contact with the ground (Ans  2  4 )
Q.10_ An athlete completes one round of a circular track of radius 2m in 40 sec. Find his displacement
at the end of 2 min. 20 sec. (Ans. 4m)

B. UNIFORM AND NON UNIFORM MOTION
(a) Uniform Motion :
A body has a uniform motion if it travels equal distances in equal intervals of time, no matter how small
these time intervals may be. For example, a car running at a constant speed say, 10 metre per second,
will cover equal distances of 10 metre every second, so its motion will be uniform. Please note that
the distance-time graph for uniform motion is a straight line (as shown in the figure).
Y
A
n
io
ot
Distance
m
rm
fo
ni
U
X
O Time

(b) Non-Uniform Motion :

A body has a non-uniform motion if it travels unequal distances in equal intervals of time. For example, if
we drop a ball from the roof of a building, we will find that it covers unequal distances in equal intervals of
time. It covers :
4.9 metre in the 1st second,
14.7 metre in the 2nd second,
24.5 metre in the 3rd second, and so on.
Thus, a freely falling ball covers smaller distance in the initial ‘1 second’ interval and larger distance in the
later ‘1 second’ interval. From this discussion we conclude that the motion of a freely falling body is an
example of non-uniform motion. The motion of a train starting from the railway station is also an example
of non-uniform motion. This is because when the train starts from a station, it moves a very small distance
in the ‘first’ second. The train moves a little more distance in the ‘2nd’ second and so on. And when the
train approaches the next station, the distance travelled by it per second decreases.
Y
A
n
tio
Distance
mo
m
or
if
- Un
n
No
X
O Time
Fig. 5
Please note that the distance-time graph for a body having non-uniform motion is a curved line (as shown
in the figure). Thus, in order to find out whether a body has uniform motion or non-uniform motion, we
should draw the distance-time graph for it. If the distance time graph is a straight line, the motion will be
uniform and if the distance-time graph is a curved line, the motion will be non-uniform. It should be noted
that non-uniform motion is also called accelerated motion.
(c) Speed and velocity :

(i) Speed : The distance travelled by a body in unit time is called its speed. Therefore,
Dis tan ce d
speed = or s = .
Time t
S.I. unit of speed is m/sec. It is a scalar quantity.
(I) Types of Speed :

(1) Uniform Speed (or Constant Speed) : When an object covers equal distance in equal intervals of
time, it is said to move with uniform speed.
Eg. A car moves 10 m in every one second so its motion is uniform.
(2) Variable Speed (or Non-Uniform Speed) : If a body covers unequal distance in equal intervals of
time, its motion is said to be non-uniform.
Eg. Vehicle starting from rest, The motion of a freely falling body etc.
(3) Average Speed : For an object moving with variable speed, it is the total distance travelled by the
object divided by the total time taken to cover that distance.
total dis tan ce travelled
Average speed = total time taken
Case -1 Let initial speed of an object is v 1, final speed is v 2 and acceleration is constant, then
v1  v 2
average speed =
2
Case -2 A body covers a distance s1 in time t1, s2 in time t2 and s3 in time t3.

s1  s2  s3
Then, average speed, Vav = t  t  t
1 2 3
Case -3 A body travels with speed v 1 for a time t1 v 2 for time t2 and v 3 for the time t3.
v1t1  v 2 t 2  v 3 t 3
Then, average speed, Vav = t1  t 2  t 3
 s1 = v 1t1, s2 = v 2t2 and s3 = v 3t3
if t1 = t2 = t3 = t
t( v 1  v 2  v 3 ) (v  v 2  v 3 )
Vav =  Vav = 1
3t 3
Case -4 A body covers a distance s1 with speed v 1 , s2 with speed v 2 and s3 with speed v 3.
(s1  s 2  s 3 )
Then, average speed, Vavg=
s1 s 2 s 3
 
v1 v 2 v 3
s1 s2 s3
 t1 = v , t2 = v , t3 = v
1 2 3
Case -5 A boy goes form home to school with speed v 1 and come back to home with speed v 2.
Here distance covered by the boy is same

s
time taken by the boy, from home to school, t1 = v
1
s
and time taken by the boy, from school to home, t2 = v
2
ss 2s
Then, average speed, Vav = =
t1  t 2 s s

v1 v 2
2v1v 2
Vav = v  v
1 2
1 1 1
Case -6 If an object covered rd distance with speed u, next rd with speed V and last rd distance
3 3 3
3uvw
with speed w then, Vavg =
uv  vw  wu
(4) Instantaneous speed : The speed of object at a particular instant is called instantaneous speed. The
limiting value of average speed when the time interval approaches zero, Thus
s ds
Instantaneous speed= lim
t 0 t

dt
(ii) Velocity : It is defined as the rate of change of displacement.
displaceme nt
Therefore, velocity = or it is the distance travelled in unit time in a given direction.
time
distance travelled in a given direction

velocity = time taken
S.I. unit of velocity is m/s. It is a vector quantity.
(Magnitude of the velocity is known as speed)

NOTE :
(i) To convert m/s into km/h we multiply given quantity by 18/5
(ii) To convert km/h into m/s we multiply given quantity by 5/18
(I) Types of Velocity :
(1) Uniform Velocity (or Constant Velocity) : If a body covers equal distance in equal intervals of time
in a given direction then it is said to be moving with constant velocity.
(2) Non-Uniform Velocity (or Variable Velocity): When a body does not cover equal distances in
equal intervals of time, in a given direction (in this case speed is not constant), then it is known as non
uniform velocity.
In uniform circular motion speed is constant but velocity is not constant.
(3) Average Velocity : It is defined as the ratio of total displacement to the total time taken for this
 
displacement. It is denoted by vav or v . It is a vector quantity..
Total displacement
 Average velocity = Total time

 s
i.e. v =
t
It is a vector quantity, and its direction is in the direction of displacement.
(4) Instantaneous velocity : The velocity of an object at a particular instant of time is called instantaneous
velocity. It is equal to the limiting value of average velocity of the object when the time interval approaches
zero , thus
 
 s ds
Instantaneous velocity v = lim t 0 t

dt
It is a vector quantity.
1
NOTE :

(i) In straight line motion if displacement is positive then v is positive.

(ii) If displacement is negative then v is also negative.
(d) Difference between Speed and Velocity :
Speed Velocity
1. It is rate of change of 1. It is rate of change of position of an object in a
position of an object. specific direction.
dis tan ce travelled displaceme nt

2. Speed= 2. Velocity =
time time
3. It is a scalar quantity. 3. It is a vector quantity
4. Speed will always be 4.It will be positive or negative depending on the

positive direction of motion
5. For moving body, it will never 5.It may be zero

be zero

Illustration 6. A body satrts from A and moves according to given figure. Time four each interval is :
(tAB = 2s, tBC= 3s, tCB=2s, tBA = 3s, tAD= 4s). Than find the distance, displacement, speed and
velocity for each path.
D A 4m B 6m C
5m
Sol.
Path Distance Displacement Speed Velocity
AB 4m 4m 4/2 m/s 4/2 m/s
ABC 10m 10m 10/5 m/s 10/5 m/s
ABCB 16m 4m 16/7 m/s 4/7 m/s
ABCA 20m 0m 20/10 m/s 0/10 m/s
ABCAD 25m –5m 25/14 m/s –5/14 m/s
Illustration 7. When the average speed of an object is equal to the magnitude of its average velocity ? Give
reason also.
Total dis tan ce
Sol. As average speed = Time time
Total displacement
also, average velocity = .
Total time
When an object moves along a straight line or in the same direction its total path length is equal to the
magnitude of its displacement. Hence average speed is equal to the magnitude of its average velocity.
Illustration 8. A car is moving along x-axis. As shown in figure it moves from O to P in 18 s and returns from P
to Q in 6 second. What is the average velocity and average speed of the car in going from (i) O to P and
(ii) from O to P and back to Q.
40 80 120 160 200 240 280 320 360 400 [Metre]
O
Q P [Sec.]
0 2 4 6 8 10 12 14 16 18 20
Displacement 360m
Sol. (i) Average velocity = time int erval = = 20 ms–1
18
path length 360m

Average speed = time interval = = 20 ms–1
18
(ii) From O to P and back to Q
path length OQ 240m
Average velocity = time interval = = =10 ms–1
18  6 24
path length OP  PQ 360  120
Average speed = time int erval = = = 20 ms –1
18  6 24
Illustration 9. A car covers the 1st half of the distance between two places at a speed of 40 km h–1 and the 2nd
half with 60 km h–1. What is the average speed of the car ?
Sol. Suppose the total distance covered is 2S.
Then time taken to cover the distance ‘S’ with speed 40 km/h,
S
t1 = h
40
Time taken to cover the nextdistance ‘S’ with speed 60 km/hrs,

S
t2 = h
60
total dis tan ce SS
Vav  
total time  S S 
  
 40 60 
2S 2S
Vav    120
 3S  2S  5S  Vav  48 km / h
 
 120 
Illustration 10. A non-stop bus goes from one station to another station with a speed of 54 km/h, the same bus
returns from the second station to the first station with a speed of 36 km/h. Find the average speed of the bus
for the entire journey.
Sol. Suppose the distance between the stations is S. Time taken in reaching from one station to another
station.
S
t1 = h
54
Time taken in returning back,
S
t2 = h
36
S S 2S  3S 5S
Total time t = t1 + t2 ; t=    h
54 36 108 108
Total distance
Average speed Vav  Total time
2S 216
Vav   108  Vav  = 43.2 km/h
5S 5
Q.1_ Can a body have a constant velocity but a varying speed?

Q.2_ Are the magnitudes of average velocity and average speed equal?
Q.3_ Under what condition is the average velocity equal to instantaneous velocity?
Q.4_ A body travels with a speed of v1 from A to B and returns with a speed of v2 from B to A. Derive
an expression for the average speed of the body. What will be the average velocity of the body?
Q.5_ Define the instantaneous speed.
Q.6_ Light travels at a speed of 3  108 m s 1 . How long does light take to reach the earth from the sun,
which is 1.5  10 11 m away? (Ans. 5  102 sec.)

Q.7_ Light takes 8½ minutes to travel from the sun to the earth.The velocity of light is 3,00,000 km s 1 .
Find the distance of the sun from the earth (Ans. 153  106 km)
Q.8_ A train travels with a speed of 60 km h 1 from station A to station B and returns with a speed of
80 km h 1 from station B to station A. Find the average speed and average velocity..
(Ans. 68.57 km/h)
Q.9_ A car travels from Bangalore to Mysore on a road that is 150 km long. If half the distance is covered
at a speed of 60 km/h and the rest at 80 km/h. Find the average speed of the car .
(Ans. 68.57 km/h)
Q.10_ The speed of a magnetic audio tape is 4.5 cm s-1. Find the length of the tape in a 60 minute
cassette? (Ans. 16200 cm.)

10

C. ACCELERATION AND EQUATION OF MOTION

(a) Acceleration :
Mostly the velocity of a moving object changes either in magnitude or in direction or in both when the
object moves. The body is then said to have acceleration. So it is the rate of change of velocity i.e.
change in velocity in unit time is said to be acceleration.It is a vector quantity and
change in velocity
Acceleration =
time
v–u final velocity – initial velocity

a= =
t time
Its S.I unit is m/s and c.g.s unit is cm/s.2
2
(i) Uniform Acceleration (Uniformly Accelerated Motion) : If a body travels in a straight line and its
velocity increases in equal amounts in equal intervals of time. Its motion is known as uniformly accelerated
motion.
Eg.I Motion of a freely falling body is an example of uniformly accelerated motion (or motion of a body
under the gravitational pull of the earth).
II. Motion of a bicycle going down the slope of a road when the rider is not pedaling and wind resistance is
negligible.
(ii) Non-Uniform Acceleration : If during motion of a body its velocity increases by unequal amounts in
equal intervals of time, then its motion is known as non uniform accelerated motion.
Eg.I Car moving in a crowded street.
II. Motion of a train leaving or entering the platform.
(b) Types of Acceleration :

(i) Positive acceleration :If the velocity of an object increases with respect to time in the same direction,
the object has a positive acceleration.
(ii) Negative acceleration (retardation) : If the velocity of a body decreases with respect to time in the
same direction, the body has a negative acceleration or it is said to be retarding.
Eg. A train slows down, then its acceleration will be negative.
(c) Equations of uniformly accelerated motion :

There are three equation of uniformly accelerated motion. They show the relation between initial velocity
u, final velocity v, acceleration a, time t and displacement s
(i) 1st Equation of Motion : Consider a body moving with initial velocity u and its velocity changes from
u to v in time t. Then
Final velocity  Initial velocity v u
acceleration = time taken
 a=
t
So at = v – u and v = u + at
(ii) 2nd Equation of Motion :

We know
Distance covered = (Average velocity) × (Time)
uv
or s t
2
But v = u + at
Substituting the value of v in the equation above, we have
11

u  u  at 
s t
2
 2u  at   at 
or s t = u   t
 2   2
1 2
or s = ut at
2
(iii) 3rd Equation of Motion :
We know that v = u + at
v u
or t
a
Distance travelled = (Average velocity) × (time)
vu
S= t
2
u v v u
S= .
2 a
v 2  u2
or S=
2a
or v 2 – u2 = 2 as
(iv) Distance covered in nth second :
Distance travelled in nth second = Distance travelled in n sec – Distance travelled in (n –1) sec.
So, Snth = Sn – S(n–1) ........................(i)
 1 2  1 2
=  un  an  – un  1  an  1 
 2   2 
[Putting t = n and t = (n – 1) respectively in equation (i)]
1 2 1
= un + an – un + u – a(n2 – 2n + 1)
2 2
a
We have, Snth = u + (2n – 1)
2
(d) Motion under gravity (uniformly accelerated motion)

The acceleration with which a body travels under gravity is called acceleration due to gravity ‘g’. Its value
is 9.8 m/s2 ( or  10 m /s 2 ).
(i) If a body moves upwards (or thrown up ) g is taken negative (i.e. motion is against gravitation of earth).
So equation of motion becomes.
1 2 2
v = u – gt, h = u t – gt , v = u2– 2 gh.
2
(ii) If a body travels downwards (towards earth) then g is taken + ve. So equations of motion becomes
1 2 2
v = u + gt , h = ut + gt , v = u2 + 2 gh.
2
(iii) If a body is projected vertically upwards with certain velocity then it returns to the same point of
projection with the same velocity in the opposite direction.
(iv) The time for upward motion is the same as for the downward motion.
(e) Sign Conventions :
(i) g is taken as positive when it is acting in the same direction as that of motion and g is taken as
negative when it is opposing the motion.
12

(ii) Distance measured upward from the point of projection is taken as positive, while distance measured
downward from the point of projection is taken as negative.
(iii) Velocity measured away from the surface of earth (i.e. in upward direction) is taken as positive, while
velocity measured towards the surface of the earth is taken as negative.
( f) To solve numerical problems :
(i)If a body is dropped from a height then its initial velocity u = 0 but has acceleration (acting). If a body
starts from rest its initial velocity u = 0 .
(ii) If a body comes to rest ,its final velocity v = 0 or, if a body reaches the highest point after being thrown
upwards its final velocity v = 0 but has acceleration ( acting).
(iii) If a body moves with uniform velocity, its acceleration is zero i.e. a = 0.
(iv) Motion of a body is called free fall if only force acting on it is gravity (i.e. earth’s attraction).
Illustration 11. A stone drops from the edge of a roof. It passes a window 2 metre high in 0.1 second. How far is
the roof above the top of the window ?
Sol. Let the distance between the top of the window and the roof be h. This problem can be solved in two
stages.
(a) For the journey across the window i.e., from B to C
Let, Velocity at B = u m/s
Distance travelled, s = 2 m
Time taken, t = 0.1s
Acceleration, a = g = 9.8 m/s2
Using the relationship,
1 2
s = ut + gt A
2 h
B
1
2 = u × 0.1 + × 9.8 × (0.1)2 2m
2 C
2 = 0.1 u + 4.9 × 10–2
(2 – 0.049)
or, u=  19.51 m/s
0.1
The velocity of the stone at the top of the window is 19.51 m/s.
For journey from roof to the top of the window i.e., from A to B
The velocity at the top of the window is the velocity of the stone at the end of falling through ‘h’. So, for this
part of the journey,
Initial velocity, u = 0 m/s
Final velocity, v = 19.51 m/s
Acceleration due to gravity, g = 9.8 m/s2
Then by using the equation, v 2 – u2 = 2 gh, one gets
(19.51)2 – 0 = 2 × 9.8 × h
Thus, the body falls back to the earth with the same velocity with which it was thrown vertically upwards.
(19.51) 2
h= m = 19.42 m
19.6
Thus, the roof is 19.42 m from the upper end (top) of the window.
Illustration 12. A ball is thrown vertically upwards with a velocity ‘u’. Calculate the velocity with which it falls to
the earth again.
Sol. For a ball thrown vertically upwards,
Initial velocity = u ; Final velocity = v = 0
For the vertically upward motion, the equation of motion is
v =u–gt
13

u
So, 0=u–gt or t = ....(i)
g
For the return journey, when the body falls vertically downwards, the equation of motion is
v=u+gt
Since, u = 0
v
Hence v = 0 + gt or t = .... (ii)
g
From (i) and (ii),
Thus, the body falls back to the earth with the same velocity with which it was thrown vertically upwards.
Illustration 13. A car is moving at a speed of 50 km/h after two seconds it is moving at 60 km/h. Calculate the
acceleration of the car.
5 250 5 300
Sol. Here u = 50 km/h = 50  m/s = m/s and v = 60 km/h = 60  = m/s
18 18 18 18
300 250
– 50
v–u 18 18 50
Since a = = = 18 = = 1.39 m/s2
t 2 36
2
Illustration 14. A car attains 54km/h in 20 s after it starts. Find the acceleration of the car.
Sol. u = 0 (as car starts from rest)
5
v = 54 km/h = 54  = 15 m/s
18
v–u 15 – 0
As, a = a= = 0.75 m/s2
t 20
Illustration 15. A ball is thrown vertically upwards with a velocity of 20 m/s. How high did the ball go ? (take g =
9.8 m/s2).
Sol. u = 20 m/s , g = – 9.8 m/s2 (moving against gravity)
s = ? v = 0 ( at highest point)
v 2 – u2 = 2gh
(0)2 – (20)2 = 2 (–g) h
– 400 = 2 ( – 9.8) h – 400 = –19.6 h
400
19.6 = h  h = 20.4 m.
Q.1_ Give an example of motion with acceleration a = 0 and velocity v  0
Q.2_ Is it possible to have velocity v = 0 and acceleration a  0 ? If yes, give an example.
Q.3_ Define acceleration. Give its S.I. unit ?

Q.4_ Derive the first equation of motion .
Q.5_ An object is thrown in upward direction with velocity u.Derive the expression for the maxmium height
attained by the object
Q.6_ Abody moving towards positive direction has an instantaneous velocity of 5 m s –1 and
3 seconds later it is 14 m s–1.Find the acceleration of the body. (Ans. 3 ms–2 )
Q.7_ A bullet moving with an initial velocity of 20 m s 1 , strikes a target and comes to rest after penetrating
the target to a distance of 10 cm. Calculate the retardation caused by the target. (Ans. 2000 ms–2)
14

Q.8_ A train starts from rest and accelerates uniformly at a rate of 2 m s 2 for 10 s. It then maintains a
constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and it comes
to rest in 50 s. Find the maximum velocity reached. (Ans. 20 ms–1 )
Q._9 A particle moves along a straight line AB with constant acceleration. Its velocities are u and v at
u2  v2
A and B respectively. Show that its velocity at the mid-point of AB is .
2
Q._10 A ball thrown vertically upward reaches a height of 80 m. Calculate :

(a) the time to reach the highest point (Ans. 4 sec.)
(b) the speed of the ball upon arrival on the ground. (Ans. 40 ms– 1)
D. G R A P H IC A L R E P R E S EN TATIO N O F MOTIO N
( a) Distance–Time Graph :
A moving body changes its position continuously with time. The simplest way to describe the motion of
a moving body is to draw its distance–time graph.
The distance–time graphs of a body under the following three conditions are described below:
(i) When the body is at rest.
(ii) When the body is moving with a uniform speed
(iii) When the body is moving with a non–uniform speed
(i) Distance–time graph for a body at rest :
The body is at a
30
Distance(m)
distance of 15m from

the reference point
20
10
Time(s)
Fig. 6
The distance–time graph for a body at rest is a straight line parallel to the time axis
(ii) Distance–time graph for a body moving with a uniform speed :
When a body covers equal distances in equal intervals of time, it is said to have uniform speed.
5
Distance(m)
4
3
2
1
0
10 20 30 40 50
Time(s)
Fig. 7
15

Graph shows that the distance travelled by a body moving with uniform speed is directly proportional to
time.
(iii) Distance–time graph for a body moving with a non–uniform speed :
A body moving with a non–uniform speed covers unequal distances in equal intervals of time. Therefore,
the distance–time graph of a body moving with a non–uniform speed is a curve.
The shape of the distance–time graph for a body moving with non–uniform speed depends upon the way
speed of the body changes with time. Two typical cases are described below :
(I) When the speed increases with time : When the speed of a body increases with time, the distance
covered by it in one unit of time also increases with time. Therefore, the distance–time graph for a body
moving with an increasing non–uniform speed is a curve whose slope increases with time (Figure).
Distance(s) D Slope of D is greater

C than that of B
B
A
0
10 20 30 40 50
Time(min)
Fig. 8
(II) When the speed decreases with time : When the speed of a body decreases with time, the
distance covered by it in one unit of time also decreases with time. Therefore, the distance–time graph for
a body moving with a decreasing non–uniform speed is a curve whose slope decreases with time (Figure).
B
Distance(s)
Slope at A is greater
than that at B
0 1 2 3 4 5
Time(min)
Fig. 9
(b) Displacement – time graph :
(i) Displacement–time graph for a body at rest : The position of a body at rest remains unchanged
with time. Let us consider a body at a distance d from a reference point in a particular direction. Then from
figure.
A B
Displacement
d d
A' B'
o
t1 t2
Time
Fig. 10
Graph shows that position of the body does not change w.r.t. time, so that body is said to be in rest.
Thus the velocity of a body at rest is zero.
16

(ii) Displacement–time graph for a body moving with uniform velocity :

The displacement–time graph of a body moving with uniform (constant) velocity is a straight line inclined
to the time–axis at certain angle.
Displacement/m
2
0 10 20 30 40 50
Time/s
Fig. 11
The slope of the displacement–time graph for a body moving with uniform velocity is equal to the velocity of the body.
(iii) Displacement–time graph for a body moving with increasing non uniform velocity :
(I) When the velocity increases with time : The displacement–time graph of a body moving with an
increasing non–uniform velocity is a curve (Figure). Here, the slope of the curve increases with time. So,
the velocity of the body increases with time
i.e., velocity at B > velocity at A
velocity at B > velocity at A
B
Displacement
A B
A
Time
Fig. 12
Displacement–time graph of a body moving with decreasing non–uniform velocity.
(II) When the velocity decreases with time :The displacement–time graph of a body moving with a
decreasing non–uniform velocity is a curve (Figure). Here, the slope of the curve decreases with time. So,
the velocity of the body decreases with time.
i.e., velocity at B < velocity at A
The displacement–time graph of a body moving with a decreasing non–uniform velocity is a curve (Figure).
Here, the slope of the curve decreases with time. So, the velocity of the body decreases with time.
i.e., velocity at B < velocity at A
b
Displacement
velocity at B < velocity at A

Fig. 13
A

Time
17

(c) Speed – time graph :

(i) Speed–time graph for a body moving with constant speed :
speed in cm/s
15
10
5
10
Time in s
(a)
Fig. 14
Figure (14) shows that for any time, the speed has the same value (10 cm/s). Thus it represents an
object moving with a constant speed. Whenever an object moves with a constant speed, its speed–time
graph is a straight line, parallel to the time–axis.
(ii)Speed–time graph for a body moving with increase speed at constatnt rate :
speed in cm/s
10
10
Time in s
(b)
Fig. 15
Figure (15) shows that the speed continuously increases with time. At time t = 0, the speed is 0. At t =
10 s, it becomes 10 cm/s. The straight–line nature of the graph indicates that the speed increases at a
constant rate.
(iii) Speed–time graph for a body moving with decrease speed at constatnt rate :
speed in cm/s
10
Time in s
(c)
` Fig. 16
Figure (16) shows that the speed is 10 cm/s at t = 0 and gradually decreases as time passes. Thus it
represents a decelerating object. Here also the speed changes at constant rate. At t = 10s, the speed
becomes zero.
speed in cm/s
10
Fig. 17
5
5 10
Time in s
18

Figure (17) represents the motion of an object which speeds up from t = 0 to t = 5s, then moves at a
constant speed from t = 5s to 10s and then decelerates to stop at t = 15 s.
 NOTE : Distance covered by the body is shown by the area under speed–time graph.
(d) Velocity – time graph :

If a graph is plotted taking the velocity of an object moving along a straight line on the vertical axis and
time on the horizontal axis, we get a velocity–time graph.
Suppose an object moves along a straight line in a fixed direction. That means the object does not turn
around during its motion. Taking the direction of motion as the positive direction, the velocity of the object
is given by the same value as its speed. Thus the speed–time graph for such an object is also its
velocity–time graph.
The area under the speed–time graph gives the distance covered. But for a particle moving in a fixed
direction, the distance covered in a time interval has the same value as its displacement in that time
interval. So, the area under the velocity–time graph of an object gives its displacement.
(i) Velocity–time graph for a body dropped from a height: A ball is dropped from a height. We take
the downward direction as positive. As the ball falls, its velocity increases. The velocity of the ball at
different instants are given in Table-1. The velocity versus time graph is shown in figure.
Table-1 : Velocity of the falling ball at different instants :
Time in s Velocity in m/s

0 0
0.1 1
0.2 2
0.3 3
0.4 4
0.5 5
0.6 6
6 A
Velocity (m/s)
1
B
O 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Time(s)
FIg. 18
What is the displacement of the ball in the time interval 0 to 0.6s ? It is equal to the area under the
velocity–times graph from t = 0 to t = 0.6 s. This area is in the shape of a triangle. The area is
1 1
× base × height = (OB) × (AB)
2 2
1
= × (0.6s) × (6 m/s) = 1.8 m.
2
The ball has fallen through 1.8 m in 0.6 s.
19

(ii) Velocity–time graph of an object thrown upwards : Suppose a ball is thrown upwards. We take
the upward direction as the positive direction. The velocity decreases as the ball goes up. Table-2 gives
the velocity of the ball at different instants.
Table-2 : Velocity of the rising ball at different instants :
Velocity (m/s)
10
8
Time in (s) Velocity in m/s
0 10 6
0.2 8
0.4 6 4
0.6 4
0.8 2 2
1.0 0 B
0 0 0.2 0.4 0.6 0.8 1.0 1.2 Time(s)
Fig. 19
Figure shows the velocity–time graph. The plotted points fall on a straight line, AB. At t = 1.0 s, the
velocity becomes zero. This means that the ball reaches the highest point at t = 1.0 s.
(e) Acceleration from velocity–time graph :

Suppose a particle moves with a uniform acceleration of 2 m/s2 along a straight line. This means that the
velocity increases by 2 m/s in one second. Also suppose its speed at t = 0 is 10 m/s.
Let us plot the velocity–time graph for this situation. We first find the values of the velocity at certain
instants. At t = 0, the velocity is 10 m/s, at t = 1s it will become 10 m/s + 2m/s = 12 m/s, at t = 2s, it will
become 14 m/s and so on. These values are given in Table-3 and the velocity–time graph is shown in
Figure.
Table-3 :
20
Velocity (m/s)
Time (s) Velocity (m/s)

0 10 15
1 12
10
2 14
3 16 5
4 18
5 20 0 1 2 3 4 5
Time (s)
Fig. 20
We see that the graph is a straight line. Whenever the acceleration is uniform the velocity–time
graph is a straight line. We will now show that the slope of the velocity–time graph gives the acceleration.
Suppose the velocity–time graph of a particle moving along a straight line is as shown in figure. The graph
is a straight line. At time t1, the velocity is v 1, and at time t2, it is v 2. These values are represented by the
points A and B on the graph.
The acceleration of the object is
v 2  v1 OE  OD DE BC
a = t  t  OG  OF  FG  AC = tan ,
2 1
where  is the angle made by the graph with the time–axis.
20

Fig. 21
BC
As defined earlier, the ratio = tan  is called the slope of the line. Thus, we have the following :
AC
The slope of the velocity–time graphs gives the acceleration for an object moving along a straight line.
( f) Non–uniform acceleration :
If the acceleration of an object moving along a straight line is not constant, the veolcity–time graph is not
a straight line. Consider the velocity–time graph shown in figure. To find the acceleration at time t1 we
should treat the small part AB as a straight line and find its slope. Similarly, the slope of the small part CD
gives the acceleration at time t2. It is clear from the figure that the slope of CD is greater than that of AB.
Thus, the acceleration at t2 is greater than that at t1. The graph in figure represents a motion in
which acceleration increases with time.
Velocity
D
C
A B
t1 t2 Time
Fig. 22
The area under the velocity–time curve along time axis gives the displacement of the particle
(g) Graphical derivation of Equations of motion :

(i) First Equation :
v = u + at
It can be derived from v – t graph , as shown in figure v Q
S
From line PQ, the slope of the line = acceleration velocity
QR SP
a= =
RP RP
 SP = v – u u R
P
C
and RP = t O t time
v u
So a = Fig. 23
t
or v = u + at
(ii) Second Equation :

1 2
S = ut + at
2
21

It can also be derived from v - t graph as shown in figure.

From relation, Velocity
Distance covered = Area under v - t graph v Q
s = Area of trapezium OPQS
= Area of rectangle OPRS + Area of triangle PQR
RQ  PR
= OP × PR + Putting values, 2
2 ½ at
1 P R
s=u×t+ (v – u) × t ( RQ = v – u & PR = OS = t)
2 u ut
S
O t Time
1
=u×t+ at × t ( v – u = at) Fig. 24
2
1 2
or s = ut + at
2
(iii) Third Equation :
v 2 = u2 + 2aS
From above graph
OP = u, SQ = v, OP + SQ = u + v
QR
a=
PR
QR v – u
Or PR = 
a a
S = Area of trapezium
OP  SQ
OPQS =  PR
2
On putting the values,
u  v v – u v 2 – u2
S= × = or v2 = u2 + 2as
2 a 2a
Illustration 16. Find the distance covered by a particle during the time interval t = 0 to t = 20s for which the
speed–time graph is shown in figure 4.13.
Speed in m/s
20
15
10
5
5 10 15 20 25 0
Time in seconds
Sol. The distance covered in the time interval 0 to 20s is equal to the area of the shaded triangle. It is
1
× base × height
2
1
= × (20s) × (20 m/s) = 200 m.
2
Illustration 17.A ball is thrown vertically upwards with a velocity of 10m/sec. It strikes the ground after 2 sec. Its
velocity–time graph is as shown in figure below. Find the displacement travelled by the ball in 2 second.
22

Velocity
10m/sec A
1 sec 2 sec
O
B D time
–10m/sec
C
Sol. Displacement = area under velocity–time curve along time axis

= area of triangle AOB + area of triangle BDC
1 1 1 1
= × OB × AO + × BD × CD = × 1 sec × 10 m/sec + × 1 sec × (–10m/sec)
2 2 2 2
= 5m – 5m = 0m
Illustration 18.Velocity–time curve for a body moving with constant accelaration is shown in the figure. Calculate
the displacement travelled by the body in 10 sec.
V
A
10 m/sec
B
O 10 sec
t
Sol. : Displacement = area under the velocity time curve along time axis = area AOB
Now AOB is a triangle with base = 10 sec and height = 10 m/sec.
1
So area = × Base × height
2
1
= × 10 sec × 10 m/sec = 5 × 10 = 50 m
2
Illustration 19.The velocity versus time graph of a linear motion is shown in figure. Find the distance from the origin in 8
second.
V
(m/sec)
+4
4 5 6 7 8
t (insec)
1 2 3
–2
Sol. Distance in 8 second
S = Area of OABC + Area of CDMN
(2  4)  4 (2  4)  2
S= 
2 2
S = 18 m
Illustration 20. A graph between the square of the velocity of a particle and the distance is moved by the particle
is shown in the figure. Find the acceleration of the particle in km/h2.
23

3600
2
v
2
(km/hr) 900
0 0.5
s (km)
Sol. Given : u2 = 3600 (km/h)2, v 2 = 900 (km/h)2,
s = 0.6 km
From III equation of motion,
v 2  u2 (900 )  (3600 ) 2700
v 2 = u2 + 2as a =  a =  = –2700 km/h2
2s 2  0.5 2  0.5

Q.1_ If the displacement-time graph of a body is a straight line parallel to the time axis, what is the nature of
motion of the body?
Q.2_ What does the slope of a displacement–time graph represent? Can displacement-time graph be parallel
to the displacement axis? Give reason for your answer.
Q.3_ A ball starts from rest, rolls down an inclined plane and then moves on a horizontal ground. Neglecting
friction, draw the v-t graph.
Q.4_ From the position time graph for two particles A and B is shown below. Graph A and graph B are making
angles 60º and 30º with the time axis. The ratio of velocities v A : v B will be : (Ans. VA : VB = 3 : 1)
Y
A
Displacement
B
60º
30º
X
O Time
Q.5_ A ball is dropped from a height ‘h’ above the ground. It rebounds to a height (h/2). Draw the v-t and s-t
graph for the motion neglecting air resistance.
Q.6_ The variation of velocity of a particle with time moving along a straight line is illustrated in the following
figure. Than find the distance travelled by the particle in four seconds. (Ans. 65 m)
30
Velocity (m/s)
20
10
0
1 2 3 4
Time in second
Q.7_ From the following displacement-time graph find out the velocity of a moving body (Ans. 3 m/s)
Time (sec)
30o
O
Displacement (meter)
24

Q.8_ The velocity time graph shows the motion of a cyclist. Its acceleration and the distance covered by the
cyclist in 15 second will be : (Ans. 300 m)
Velocity (ms –1 )
25
20
15
10
5
5 10 15 25 20
Time
Q.9_ A particle moves according to given velocity time graph. Then, the ratio of distance travelled in last 2
seconds to the total distance travelled will be : (Ans. 1/4)
10
v(m/s)
1 3 5 7
t(s)
Q.10_ The velocity-time graph is a straight line parallel to the time axis for a body moving with uniform velocity
as shown in figure than find the acceleration and distance and displacement in 6sec (Ans. 24 m)
y
5
A
4 B
Velocity (m s1)
3
2
1
O C x
0 1 2 3 4 5 6 7
Time (s)
Fig. 5.12. Velocity-time graph

TYPE (I) : VERY SHORT ANSWER TYPE QUESTIONS : [01 MARK EACH]
1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance
covered and the displacement at the end of 2 minutes 20 s?
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then
turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and
velocities in jogging (a) from A to B and (b) from A to C?
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip
along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed
for Abdul’s trip?
25

TYPE (II) : SHORT ANSWER TYPE QUESTIONS : [02 MARKS EACH]

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s.
How far does the boat travel during this time?
5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite
direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly
and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which
of the two cars travelled farther after the brakes were applied?
TYPE (III) : LONG ANSWER TYPE QUESTIONS: [03 MARK EACH]
6. Shows the distance-time graph of three objects A,B and C. Study the graph and answer the following
questions:
(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How
far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with
what velocity will it strike the ground? After what time will it strike the ground?
TYPE (IV): VERY LONG ANSWER TYPE QUESTIONS [05 MARK EACH]
8. The speed-time graph for a car is shown is Figure

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the
distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
9. State which of the following situations are possible and give an example for each of these: (a) an object
with a constant acceleration but with zero velocity (b) an object moving in a certain direction with an
acceleration in the perpendicular direction.
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24
hours to revolve around the earth.
26

 Marked Questions can be used as Revision Questions.
SUBJECTIVE QUESTIONS
SUBJECTIVE EASY, ONLY LEARNING VALUE PROBLEMS
SECTION (A) : REST AND MOTION
A-1. Define rest and motion and give two examples of each.
A-2. Can a body be at rest and motion at the same time ? Explain.
A-3. Give three examples to explain that motion is relative
A-4. Is distance a vector quantity ?
A-5. What is vector ?
A-6. Define scalar quantity and give two examples.
A-7. What is the S.I. unit of displacement ?
A-8. Given an example when displacement can be zero ?
A-9. Are distance and displacement equal in magnitude ?
A-10. A runner running along a circle, runs the circle completely. What is his distance and displacement ?
A-11. Write five difference between distance and displacement.
SECTION (B) : UNIFORM AND NON UNIFORM MOTION
B-1. What is the S.I. unit of velocity ?

B-2. Define speed
B-3. What is velocity ?
B-4. When is a body said to move with uniform acceleration?
B-5. State S.I. unit of speed
B-6. Give the name of the physical quantity that corresponds to the rate of change of displacement ?
B-7. Apart from velocity name two other quantities which are vector ?
B-8. When is a body said to have uniform velocity ?
B-9. A particle is moving with uniform velocity. Is it necessary moving with uniform speed ? Is it necessary that
it is moving along a straight line ?
B-10. Write differences between speed and velocity .
B-11. A train covers 80 km in 2 hours. Find its average speed in kmh-1, m min-1 and ms-1.
B-12. Which one of the following have maximum and the least average speed ?
(a) Sanjeev moving with 12 kmh-1
(b) Rajeev running with 5 ms-1
(c) Kabir moving with 150 m min-1
B-13. Write the short notes on :
(a) Uniform motion (b) Non uniform motion
(c) Average speed (d) Velocity
27

SECTION (C) : ACCELERATION AND EQUATION OF MOTION
C-1. Derive the formula for the distance covered by a body in nth sec.
C-2. How is the position of a moving particle along a straight line described by a number ? How is the direction
of motion specified by the number describing position ?
C-3. What do you mean by term retardation?
C-4. A car accelerated from 6 m/s to 16 m/s in 10 second. Calculate.
(a) the acceleration and
(b) the distance covered by the car in that time.
C-5. A ball is thrown vertically upward from the ground with a velocity 39.2 ms–1. Calculate :
(a) the maximum height to which the ball rises and
(b) the time taken by the ball to reach the highest point.
C-6. A boy standing near the edge of a cliff 125 m above a river throws a stone downward with a speed of
10 ms –1. Find :
(a) with what speed will the stone hit water and
(b) how long will it take to descend ?
C-7. A stone is dropped from the top of a building 200 m high and at the same time another stone is projected
vertically upward from the ground with a velocity of 50 ms–1. Find where and when the two stones will
meet.
C-8. A ball thrown vertically upward reaches a height of 80 m. Calculate :
(a) the time to reach the highest point
(b) the speed of the ball upon arrival on the ground.
SECTION (D) : GRAPHICAL REPRESENTATION OF MOTION
D-1. What does the slope of a speed-time graph indicate?

D-2. What quanitity is given by the area under the velocity -time graphy ?
D-3. Draw V-t graphs in the following cases : (a) uniform retardation (b) non uniform acceleration
D-4. The speed-time graph is a line parallel to time axis, what conclusion can you draw from the graph about
the speed of the body.
D-5. A stone is thrown vertically upward which takes time ’t’ to reach the maximum height ‘h’. After next ‘t’
seconds it reaches the ground from the maximum height. Draw (i) distance-time graph and (ii) displacement
time graph for the motion of the stone.
D-6. From the following (V-t) graph find :
V(m/s)
20 —
10 —
t(sec)
5 10 15
–5—
(a) Distance and displacement in 10 second.

(b) Distance and displacement in 15 second.
D-7. Draw velocity-time graph for an uniformly accelerated object. Using velocity-time graph, derive v = u+at.
D-8. Derive the equation of motion v = u + at using graphical method.
28

OBJECTIVE QUESTIONS
SINGLE CHOICE OBJECTIVE, STRAIGHT CONCEPT/FORMULA ORIENTED
A-1. A body whose position with respect to surrounding does not change, is said to be in a state of :
(A) Rest (B) Motion (C) Vibration (D) Oscillation
A-2. In case of a moving body :
(A) Displacement > Distance (B) Displacement < Distance
(C) Displacement > Distance (D) Displacement < Distance
A-3. A body is said to be in motion if :
(A) Its position with respect to surrounding objects remains same
(B) Its position with respect to surrounding objects keep on changing
(C) Both (A) and (B)
(D) Neither (A) nor (B)
A-4. Distance is always :
(A) shortest length between two points (B) path covered by an object between two points
(C) product of length and time (D) none of the above
A-5. Displacement :
(A) is always positive (B) is always negative
(C) may be positive as well as negative (D) is neither positive nor negative
A-6. Which of the following is not characteristic of displacement ?
(A) It is always positive.
(B) It has both magnitude and direction.
(C) It can be zero.
(D) Its magnitude is greater than or equal to the actual path length of the object.
A-7. S.I. unit of displacement is :
(A) m (B) ms-1 (C) ms-2 (D) none of these
A-8. In five minutes, distance between a pole and a car changes progressively. What is true about the car ?
(A) Car is at rest (B) Car is in motion
(C) Nothing can be said with this information (D) None of the above
A-9. Distance :
(A) Is always positive (B) Is always negative
(C) May be positive as well as negative (D) Is neither positive nor negative
A-10. Motion is a change of :
(A) mass (B) position (C) time (D) shape of an object
A-11. Vector quantities are those which have :
(A) Only direction (B) Only Magnitude
(C) Magnitude and direction both (D) None of these
A-12. What is true about scalar quantities ?

(A) Scalars quantities have direction also. (B) Scalars can be added arithmetically.
(C) There are special laws for scalar addition. (D) Scalars have special method to represent.
A-13. Examples of vector quantities are :
(A) velocity, length and mass (B) speed, length and mass
(C) time, displacement and mass (D) velocity, displacement and force
A-14. Which of the following is not a vector ?

(A) Speed (B) Velocity (C) Weight (D) Acceleration
29

A-15. Time is an example of :

(A) Scalar (B) Vector (C) Scalar or vector (D) Neither scalar nor vector
A-16. Out of energy and acceleration which is vector ?

(A) Acceleration (B) Energy (C) Both (D) None of these

B-1. When a body covers equal distance in equal intervals of time, its motion is said to be :
(A) Non-uniform (B) Uniform (C) Accelerated (D) Back and forth
B-2. The motion along a straight line is called :

(A) Vibratory (B) Stationary (C) Circular (D) Linear
B-3. A particle is travelling with a constant speed. This means:

(A) Its position remains constant as time passes
(B) It covers equal distances in equal interval of time
(C) Its acceleration is zero
(D) It does not change its direction of motion
B-4. The rate of change of displacement is :

(A) Speed (B) Velocity (C) Acceleration (D) Retardation
B-5. Speed is never :

(A) Zero (B) Fraction (C) Negative (D) Positive
B-6. The motion of a body covering different distances in same intervals of time is said to be :
(A) Zig-Zag (B) Fast (C) Slow (D) Variable
B-7. Unit of velocity is :

(A) ms (B) ms-1 (C) ms2 (D) none of these
B-8. Metre per second is not the unit of :

(A) Displacement (B) Velocity (C) Speed (D) None of them
B-9. A particle moves with a uniform velocity then :

(A) The particle must be at rest (B) The particle moves along a curved path
(C) The particle moves along a circle (D) The particle moves along a straight line
B-10. A quantity has value of –6.0 ms-1. It may be the :

(A) Speed of a particle (B) Velocity of a particle
(C) Position of a particle (D) Displacement of a particle
B-11. In 10 minute, a car with speed of 60 kmh-1 travels a distance of :
(A) 6 km (B) 600 km (C) 10 km (D) 7 km
B-12. A particle covers equal distances in equal intervals of time, it is said to be moving with uniform :
(A) Speed (B) Velocity (C)Acceleration (D)Retardation
B-13. The SI unit of the average velocity is :
(A) m/s (B) km/s (C) cm/s (D) mm/s
B-14. S.I. Unit of speed is :
(A) km h–1 (B) cm s–1 (C) h Km–1 (D) m s–1
B-15. Average speed of a car is given as 50 km /h . In S.I. units, it can be expressed as :
(A) 13.9 m/s (B) 5 m/s (C) 50 m/s (D) 139 m /s
B-16. The value on converting Km/h into m/s is:

(A) 5/18 (B) 5/36 (C) 5/54 (D) 5/324
30

B-17. A body is moving with uniform velocity of 10 ms–1. The velocity of the body after 10 s is :
(A) 100 ms–1 (B) 50 ms–1 (C) 10 ms–1 (D) 5 ms–1
B-18. In 12 minutes a car whose speed is 35 kmh–1 travels a distance of :

(A) 7 km (B) 3.5 km (C) 14 km (D) 28 km
B-19. A bus travelled the first one-third distance at the speed of 10 km/h, the next one-third at 20 km/h and the
last one-third at 60 km/h. The average speed of the bus is
(A) 9 km/h (B) 16 km/h (C) 18 km/h (D) 48 km/h
B-20. When the distance travelled by an object is directly proportional to the time, it is said to travel with :
(A) zero velocity (B) constant speed
(C) constant acceleration (D) uniform velocity
C-1. A car accelerates uniformly from 18 km/h to 36km/h in 5 s. The acceleration in ms–2 is :
(A) 1 (B) 2 (C) 3 (D) 4
C-2. C.G.S. unit of acceleration is :

(A) ms–2 (B) cm s–2 (C) ms2 (D) cm s2
C-3. A train starting from a railway station and moving with uniform acceleration, attains a speed of 40 kmh–1
in 10 minutes. Its acceleration is :
(A) 18.5 ms–2 (B) 1.85 cm s–2 (C) 18.5 cms–2 (D) 1.85 m s–2
C-4. The brakes applied to a car produce a negative acceleration of 6ms–2. If the car stops after 2 seconds, the
initial velocity of the car is :
(A) 6 ms–1 (B) 12 ms–1 (C) 24 ms–1 (D) Zero
C-5. A body is moving along a straight line at 20 ms–1 undergoes an acceleration of 4 ms–2. After 2 s, its speed
will be :
(A) 8 ms–1 (B) 12 ms–1 (C) 16 ms–1 (D) 28 ms–1
C-6. A car increase its speed from 20 kmh–1 to 50 kmh–1 in 10 s., its acceleration is :
(A) 30 ms–2 (B) 3 ms–1 (C) 18 ms–2 (D) 0.83 ms–2
C-7. A body freely falling from rest has a velocity V after it falls through a height h. The distance it has to fall
further for its velocity to become double is :
(A) 3 h (B) 6 h (C) 8 h (D) 10 h
C-8. The velocity of a bullet is reduced from 200 m/s to 100 m/s while travelling through a wooden block of
thickness 10 cm. The retardation, assuming it to be uniform, will be :
(A) 10 × 104 m/s2 (B) 1.2 × 104 m/s2 (C) 13.5 × 104 m/s2 (D) 15 × 104 m/s2
C-9. A body starts falling from height ‘h’ and travels distance h/2 during the last second of motion. The time of
travel (in sec.) is :
(A) 2 – 1 (B) 2  2 (C) 2  3 (D) 3  2
C-10. A stone is dropped from the top of a tower. Its velocity after it has fallen 20 m is [Take g = 10 ms–2] :
(A) 5 ms–1 (B) 10 m s–1 (C) 15 m s–1 (D) 20 m s–1
C-11. An object undergoes an acceleration of 8 ms–2 starting from rest. Distance travelled in 1 sec is :
(A) 2m (B) 4m (C) 6m (D) 8m
31

D-1. Area between speed- time graph and time axis gives :
(A) Distance (B) Velocity (C) Speed (D) None of these
D-2 The velocity-time graph of a body moving in a straight line is shown in figure. The displacement and
distance travelled by the body in 6 seconds are respectively :
(A) 8 m, 16 m (B) 16 m, 8 m (C) 16 m, 16 m (D) 8 m, 8 m
D-3. For the velocity time graph shown in figure,what fraction is the distance covered by the body in the last
two seconds of the total distance covered in all the seven seconds ?
10
Velocity (m/s)
8
6
4
2
0 1 2 3 4 5 6 7
Time (s)
(A) 1/2 (B) 1/4 (C) 1/3 (D) 2/3
D-4. Velocity-time graph AB (Figure) shows that the body has :

(A) A uniform acceleration
(B) A non-uniform retardation
(C) Uniform speed
(D) Initial velocity OA and is moving with uniform retardation
D-5. In figure BC represents a body moving :

(A) Backward with uniform velocity
(B) Forward with uniform velocity
(C) Backward with non-uniform velocity
(D) Forward with non-uniform velocity
D-6. Area of acceleration time graph gives :

(A) Rate of change of velocity with the time (B) Rate of change of acceleration with time
(C) Change in velocity (D) Change in accelration
D-7. Which of the acceleration - time graph is not possible?
a a a a
(A) (B) (C) (D)
t t t t
32

D-8. Which of the following is correct for uniformly accelerated motion ?
displacement
displacement
displacement
(A) (B) (C) (D) All are correct
t t t
D-9. Slope of the acceleration time graph gives :

(A) Rate of change of velocity with the time (B) Rate of change of acceleration with time
(C) Change in velocity (D) Change in acceleration
D-10. If the displacement - time graph of a particle is perpendicular to the time axis, then velocity of the particle
is :
(A) infinity (B) equal to the acceleration of the body
(C) unity (D) zero
D-11. Velocity - time graph shows that the body has constant velocity for part :
A B
D
v C
t
(A) AB (B) BC (C) CD (D) Both (A) and (C)
D-12. In the above question acceleration is non-zero for which part of the graph :
(A) AB (B) BC
(C) CD (D) Both (A) and (C)
D-13. Which of the figure corresponds to a case when body is moving with uniform velocity ?
displacement
displacement
displacement
displacement
(A) (B) (C) (D)
time time time time
D-14. Which of the figure corresponds to a cast when body travels for a certain time with uniform acceleration
and then with a uniform velocity for the rest of the time ?
v
(A) (B) v
(C) v (D) v
t t t t
33

OBJECTIVE QUESTIONS
1. If the distance travelled by an object is zero, then the displacement of the object is
(A) zero (B) not zero (C) negative (D) may or may not be zero
2. Ram reached Rahul‘s house walking 10 km in 20 minutes towards south. Displacement will be
(A) 0.5 km in south direction. (B) 10 km in south direction.
(C) 0.5 km in north direction. (D) 10 km in north direction.
3. The velocity of an object can be changed by

(A) changing the speed
(B) changing the direction of motion
(C) changing both the speed and direction of motion
(D) all (a), (b) and (c) are true.
4. An object is moving with a velocity of 2 ms–1. Its velocity changes at a uniform rate to 5 ms–1. The average
velocity of the object is
(A) 3 ms–1 (B) 3.5 ms–1 (C) 4 ms–1 (D) 5.5 ms–1
1
5. A boy running at an average speed of 4 km h–1 reaches school from his home in hour. The distance of
2
the school from his home is :
(A) 2 km (B) 8 km (C) 4 km (D) 6 km
6. A girl swims in a swimming pool of length 100 m. She swims from one end to another end and reaches the
starting point again in 2 minutes. The average velocity of the swimmer is
(A) 100 ms–1 (B) 0.83 ms–1 (C) 1.67 ms–1 (D) zero
7. A car is moving with a uniform velocity of 40 km h–1. Its acceleration after 1 hour is
(A) 40 km h–1 (B) 20 km h–1 (C) 30 km h–1 (D) zero
8. A body will have uniform acceleration if its
(A) speed changes at uniform rate (B) velocity changes at uniform rate
(C) speed changes at non-uniform rate (D) velocity remains constant
9. A car travels from one town to the other with average speed 20 km/hr .If the first half is travelled at
average speed 30 km/hr , then the average speed of the car in the other half will be:
(A) 30 km/hr (B) 40 km/hr (C) 50 km/hr (D) 15 km/hr
10. A bird is sitting on train A moving towards East with a velocity 300 km/hr. Another train B of same
speed is moving in West direction on the same track. When the trains are 6 km apart , the bird
starts flying with a velocity 30 km/hr with respect to ground towards B . After touching B, it returns
back to A and continue repeating this process until the trains collide. In this process, the total
distance travelled by the bird is :
(A) 100 m (B) 200 m (C) 300 m (D) 6 km
34

11. Distance -time graph of a body is a straight line parallel to time axis. The body is
(A) moving with constant speed (B) moving with constant velocity
(C) at rest (D) moving in a straight line
12. Slope of distance -time graph of a moving body is equal to

(A) velocity of the body (B) speed of the body
(C) acceleration of the body (D) none of these
13. The speed of a body is 1 ms–1. The angle between the distance-time graph of the body and the time axis is
(A) 0º (B) 30º (C) 45º (D) 60º
14. Area under velocity -time graph is equal to the

(A) speed of the body (B) distance travelled by the body
(C) magnitude of the displacement of the body (D) none of these
15. Area under speed-time graph is equal to the

(A) velocity of the body (B) magnitude of the displacemnt
(C) distance travelled by the body (D) none of these
16. Velocity  Time graph of a particle is given. Which of the following statement(s) is/are correct .
(A) net displacement is zero (B) total distance travelled is 80 metre

(C) acceleration of the particle is constant (D) direction of velocity is constant
NTSE PROBLEMS (PREVIOUS YEARS)
1. A student starts with a velocity 40 km/hr for school at 4 km away from his house. Due to closing of school
he returns soon to his house with a velocity of 60km/hr.His average velocity will be:
(Raj./ NTSE Stage-I/2007)
(A) zero (B) 10 km/hr (C) 48 km/hr (D) 50 km/hr
2. A boy sitting on the top most berth in the compartment of a train which is just going to stop on the railway
station, drops an apple aiming at the open hand of his brother situated vertically below his hands at a
distance of about 2m. The apple will fall : (Punjab/ NTSE Stage-I/2013)
(A) In the hand of his brother
(B) Slightly away from the hand of his brother in the direction of the motion of the train.
(C) Slightly away from the hands of his brother in the direction opposite to the direction of the motion of
the train.
(D) None of these
35

3. A velocity-time graph for a moving object is shown below. What would be the total displacement
during time t = 0 to t = 6s? (Orissa/ NTSE Stage-I/2013)
(A) 10 m
(B) 20 m .
(C) 2.5 m
(D) 0.0 m
4. The velocity-time graph of a body falling from rest under gravity and rebounding from a solid surface is
represented by: (Raj./ NTSE Stage-I/2014)
(A) (B) (C) (D)

t t t t
O O O O
5. A bullet of mass 10 g traveling horizontally with a velocity of 160 ms–1 strikes a stationary wooden block
and comes to rest in 0.02 s. The distance of penetration of the bullet into the block will be :
(Raj./NTSE Stage-I/2014)
(A) 1.20 m (B) 1.60 m (C) 2.00 m (D) 2.40 m
6. The acceleration versus time graph of an object is as shown in figure. The corresponding velocity-time
graph of the objects is : (Raj./NTSE Stage-I/2014)
v
(A) (B) (C) (D)
7. The graph below describe the motion of a ball rebounding from a horizontal surface being released from a
point above the surface. (Haryana./ NTSE Stage-I/2014)
The quantity represented in the y-axis is the ball's :

(A) displacement (B) velocity (C) acceleration (D) momentum
8. Value of one Fermi is : (M.P../ NTSE Stage-I/2014)

(A) 10–13 metre (B) 10–14 metre (C) 10–15 metre (D) 10–16 metre
9. Unit of Impulse is : (M.P./ NTSE Stage-I/2014)

2
(A) Newton (B) Newton  second (C) Newton  (second) (D) Newton per second
36

10. A person takes time t to go once around a circular path of diameter 2R. The speed () of this person would
be: (Raj./ NTSE Stage-I/2015)
t 2R R 2
(A) (B) (C) (D) 2R.t
2R t t
11. A body of mass 2 kg is moving on a smooth floor in straight line with a uniform velocity of 10 m/s.
Resultant force acting on the body is: (Raj./ NTSE Stage-I/2015)
(A) 20 N (B) 10 N (C) 2 N (D) zero
12. A body falling from rest describes distances S1, S2 and S3 is the first, second and third seconds of its fall.
Then the ratio of S1 : S2 : S3 is : (Delhi/ NTSE Stage-I/2014)
(A) 1 : 1 : 1 (B) 1 : 3 : 5 (C) 1 : 2 : 3 (D) 1 : 4 : 9
13. A body starts from rest at time t = 0, the acceleration time graph is shown in figure. The maximum
velocity attained by the body will be : (Delhi/ NTSE Stage-I/2014)
(A)1110 m/s
10
(B) 55 m/s 2
a(m/s )
(C) 650 m/s
O 11 t(s)
(D) 550 m/s
14. A body covers half the distance with a speed of 20m/s and the other half with 30m/s. The average speed
of the body during the whole journey is : (West Bengal/ NTSE Stage-I/2014)
(A) Zero (B) 24m/s (C) 25 m/s (D) None of the above
15. Correct relation is............... (Madhya Pradesh/ NTSE Stage-I / 2015)
(A) v 2 = u2 + 2a2s2 (B) v 2 = u2 – 2a2s2 (C) v 2 = u2 + 2as (D) v 2 = u2 + 2a2s
16. The figure given below shows the displacement plotted time for a particle. In which regions is the force
acting on the particle zero ? E (Bihar/ NTSE Stage-I/2014)
Displacement
B C
(A) AB
(B) BC
D
(C) CD
A
(D) DE
Time
17. Two cars of unequal masses use similar tyres. If they are moving with same initial speed, the minimum
stopping distance : (Jharkhand/ NTSE Stage-I/2014)
(A) is smaller for the heavier car. (B) is same for both the cars
(C) is smaller for the lighter car. (D) depends on the volume of the car
18. A ball hits a wall horizontally with a velocity of 6.0 ms–1. After hitting wall it rebounds horizontally with a
velocity of 4.4 ms–1 . If the balls remains in the contact of all for 0.040 sec. the acceleration of ball would
be - (Uttrakhand/ NTSE Stage-I/2014)
(A) –260 m/s2 (B) +260 m/s2 (C) –26 m/s2 (D) +26 m/s2
g
19. If the length of pendulum executing simple harmonic motion is metre then the time period of the
4 2
pendulum is : (Uttrakhand/ NTSE Stage-I/2015)
(A) 2.5 sec. (B) 1.5 sec. (C) 1 sec. (D) 2 sec.
20. The speed of a train decreases from 80 km/hour to 60 Km/hour in 5 seconds. In this process, find out the
acceleration of the train : (Uttrakhand/ NTSE Stage-I/2015)
(A) 2.22 m/sec2 (B) –2.22 m/sec2 (C) –1.11 m/sec2 (D) 1.11 m/sec2
21. A ball thrown vertically upward returns to the thrower after 6s. The ball is 5m below the highest
point at t=2s. The time at which the body will be at same position, (take g=10 m/s2)
(Delhi/ NTSE Stage-I/2015)
(A) 2.5s (B) 3s (C) 4s (D) 5s
37

22. A particle starts its motion from rest under the action of a constant force. If the distance covered
in first 10s is S1 and that covered in first 20s is S2 then : (Delhi/ NTSE Stage-I/2015)
(A) S2 = S1 (B) S2=2S1 (B) S2 = 3S1 (D) S2 = 4S1
23. A car travels 40 kms at an average speed of 80 km/h and then travels 40 kms at an average speed of 40
km/h. The average speed of the car for this 80 km trip is : (Raj./ NTSE Stage-I/2015)
(A) 40 km/h (B) 45 km/h (C) 48 km/h (D) 53 km/h
24. Which motion does the graph of distance and time shows for accelerated motion?
(Gujrat/ NTSE Stage-I/2015)
(A) non uniformly accelerated
(B) constant velocity
(C) uniformly accelerated
(D) uniformly retarded motion
25. A body is dropped from certain height from a uniformly ascending balloon. The correct graph showing
variation of velocity with time for body is : (Haryana/ NTSE Stage-I/2015)
v v v v
(A) (B) (C) (D)

t t t t
26. A thin uniform metal rod of mass 'm' and length '' standing vertically is now allowed to fall due to its own
weight. The velocity with which its mid-point will strike ground will be : (Haryana/ NTSE Stage-I/2015)
5 3
(A) g (B) 5g (C) 3g  (D) g
2 4
27. A stone is dropped from the top of a tower. Its velocity after it has fallen 20m is (take g = 10m/s2):
(Bihar/ NTSE Stage-I/2015)
(A) – 10 m/s (B) 10 m/s (C) – 20 m/s (D) 20 m/s
28. If the length of a simple pendulum is increased to 2 times its value then its time period will be-
(A) halved (B) doubled (C) becomes 2 times (D) reduces by 2
29. In the adjacent V – T diagram what is the relation between acceleration A1 and A2?
(West Bengal/ NTSE Stage-I/2015)
V(Velocity)
T(Time)
(A) A2 = A1 (B) A2 > A1 (C) A2 < A1 (D) Cannot be predicted
38

30. A balloon is moving up from the ground in such a way that its acceleration is linearly decreasing with its
height above the ground. It starts from the ground with acceleration 4 m/ s2 and with zero initial velocity.
Its acceleration becomes zero at a height 3 m. The speed of the balloon at a height 1.5 m is :
(Andra Pradesh/ NTSE Stage-I/2015)
(A) 4 m/s (B) 8 m/s (C) 6 m/s (D) 3 m/s
31. A train can accelerate at 20 cm/s2 and decelerate at 100 cm/s2. Then the minimum time for the train to
travel between the stations 2 km apart is..... (The train should start at one station and stop at another
station) (Andra Pradesh/ NTSE Stage-I/2015)
(A) 125 s (B) 100 s (C) 155 s (D) 200 s
32. The brakes applied to a car produce an acceleration of 8 m/s2 in the opposite direction to the motion. If the
car takes 3 seconds to stop after the application of brakes, the distance it travels during the time will be
(A) 30 m (B) 36 m (C) 25 m (D) 40 m
33. A car covers 30 km at a uniform speed of 60 km/hr. and the next 30 km at a uniform speed of 40 km/hr.
The total time taken is: (M.P./ NTSE Stage-I/2017)
(A) 30 min (B) 45 min (C) 75 min (D) 120 min
34. A stone is thrown upwards with a speed "u" from the top of a tower. It reaches the ground with a velocity
"3u". The height of the tower is: (M.P./ NTSE Stage-I/2017)
u2 2u 2 3u 2 4u2
(A) (B) (C) (D)
g g g g
35. The velocity time graph of the prticl in motion is parallel to time axis shows
(Chhatt./ NTSE Stage-I/2017)
(A) Uniform motion of particle (B) Particle is in rest
(C) Non uniform motion of particle (D) Accelerated motion of particle
36. Trippling the speed of a motor car multiplies the distance needed for stopping it by :
(Haryana/ NTSE Stage-I/2017)
(A) 3 (B) 6 (C) 9 (D) 12
37. Two bodies of masses ma and mb are dropped from different heights ‘a’ and ‘b’. The ratio of time taken by
them to reach the ground is: (Haryana/ NTSE Stage-I/2017)
 
(A) a : b (B) a : b (C) : (D) ma : mb
a b
38. A person throws ball with a velocity ‘v’ from top of a building in vertically upward direction. The ball reaches
the ground with a speed of ‘3v’. The height of the building is : (Haryana/ NTSE Stage-I/2017)
4v 2 3v 2 6v 2 gv 2
(A) (B) (C) (D)
g g g g
39. Two particles of masses m1 and m2 are allowed to fall freely from height h1 and h2. They reach the
ground at time t1 and t2 respectively. Then (West Bengal/ NTSE Stage-I/2017)
t1 h1 t1 h2 t2 h2 t2 h1
(A) t  h2 (B) t  h1 (C) t  h (D) t  h
2 2 1 1 1 2
40. Position of a particle moving along x-axis is given by x = 3t – 4t2 + t3, where x is in metre and t is in
second. Find the average velocity of the particle in the time interval form t = 2 second to t = 4 second.
(West Bengal/ NTSE Stage-I/2017)
(A) 7 m/s (B) 1 m/s (C) 13 m/s (D) 5 m/s
39

41. Consider the following five graphs ( note the axes carefully ) Which of the following represents motion at
constant speed ? (Raj./ NTSE Stage-I/2018)
(A) D only (B) D and E (C) A, B and C (D) A and D
42. A ball is shot vertically upward with a given initial velocity. It reaches a maximum height of 100 m. If on a
second shot, the initial velocity is doubled then the ball will reach a maximum height of
(A) 70.7 m (B) 141.4 m (C) 200 m (D) 400 m
43. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically
to height d/2. Neglecting air resistance, its velocity v varies with the height h above the ground as :
(A) (B)
(C) (D)
44. Two objects moving along the same straight line are leaving point 4 with acceleration a, 2a and initial
velcoity 2u, u at time t = 0. The distance moved by objects with respects to point A when one object,
initially behind other, overtakes the other is : (Bihar/ NTSE Stage-I/2017)
6u 2 2u 2 4u2 8u 2
(A) (B) (C) (D)
a a a a
45. A body falls freely from a tower and travels a distance of 40 m in its last two seconds. The height of the
tower is (take g = 10 m/s–2) (Delhi/ NTSE Stage-I/2017)
(A) 54 m (B) 45 m (C) 80 m (D) 65 m
40

46. Velocity-time graph of a body moving with uniform acceleration is shown in the diagram. The
distance travelled by the body in 3 seconds is (Rajasthan /NTSE Stage-1/2018)
(A) 90 m (B) 45 m (C) zero (D) 10 m
47. A car accelerates uniformly from 18km/h to 36 km/h in 5 sec. Calculate the acceleration.
(Gujarat/ NTSE Stage-1/2018)
–2 –2 –2
(A) 1 ms (B) 3.6 ms (C) 2 ms (D) 2.6 ms–2
48. If the distance travelled by an object is zero, then the displacement of the object is :
(Madhya Pradesh/ NTSE Stage-1/2018)
(A) zero (B) not zero (C) negative (D) May or may not be zero
49. A body describes the first half of the total distance with velocity v1 and the second half with
velocity v2. The average velocity is : (Jharkhand/ NTSE Stage-1/2018)
v1 + v 2 1 1 v v 2v1v 2
(A) (B) + (C) 1 2 (D)
2 v1 v 2 v1 + v 2 v1 + v 2
50. A boy starts from rest is accelerated uniformly for 30s. If x1 x2 x3 are the distances travelled in
I 1
first 10s, next 10s and last 10s respectively, then x1 : x2 : x3 is (Delhi/ NTSE Stage-1/2018)
(A) 1 : 2 : 3 (B) 1 : 1 : 1 (C) 1 : 3 : 5 (D) 1 : 3 : 9
51. The velocity–time graph of a moving body is shown in the figure. (Delhi/ NTSE Stage-1/2018)
Which of the following statement is true?

(A) The acceleration is constant and positive.
(B) The acceleration is constant and negative.
(C) The acceleration is increasing and positive.
(D) The acceleration is decreasing and negative.
41
41
52. The displacement time graph of a body in motion is given as below –

(Haryana/ NTSE Stage-1/2018)
Velocity of body is (in m/s) -

1 1
(A) 3 (B) (C) 3 (D)
3 3
53. V2 – S graph of moving body in a straight line is as shown in figure. Which one among the
following is not true? (Haryana/ NTSE Stage-1/2018)
(A) Motion is uniformly accelerated. (B) Corresponding s-t graph will be parabola.
(C) Initial velocity of particle is zero (D) Velocity is time varying.
42

BOARD LEVEL EXERCISE
TYPE (I) : VERY SHORT ANSWER TYPE QUESTIONS
1. Distance=2200m and displacement=200m
2. (a) Average speed=1.765m/s and average velocity=1.765m/s

(b)Average speed=1.739m/s and average velocity=0.87m/s
3. Average speed=26.67m/s
TYPE (II) : SHORT ANSWER TYPE QUESTIONS
4. 96 m
TYPE (III) : LONG ANSWER TYPE QUESTIONS
6. (a) B (b) No (c) 5.714 Km (d) 5.143 Km 7. 20 m/s and 2s
TYPE (IV): VERY LONG ANSWER TYPE QUESTIONS
10. 3070.9m/s
EXERCISE - 1
SUBJECTIVE QUESTIONS
SUBJCTIVE EASY, ONLY LEARNING VALUE PROBLEMS
B-11. 40 kmh-1, 666.7 m min-1, 11.1 ms-1
C-4. (a) 1m/s2 (b) 110 m C-5. (i) 78.4 m (ii) 4 s
C-6. (a) 50.5 ms–1 (b) 4.13 s C-7. After 4 second, it will be at a height of 121.6 m from the ground.
C-8. (a) 4.04s (b) 39.59 ms–1
D-6. (i) 100 m, 100 m (ii) 112.5 m, 87.5 m
43

OBJECTIVE QUESTIONS
A-1 (A) A-2 (D) A-3 (B) A-4 (B) A-5 (C) A-6 (D)
A-7 (A) A-8 (B) A-9 (A) A-10 (B) A-11 (C) A-12 (B)
A-13 (D) A-14 (A) A-15 (A) A-16 (A)
B-1 (B) B-2 (D) B-3 (B) B-4 (B) B-5 (C) B-6 (D)
B-7 (B) B-8 (A) B-9 (D) B-10 (B) B-11 (C) B-12 (A)
B-13 (A) B-14 (D) B-15 (A) B-16 (A) B-17 (C) B-18 (A)
B-19 (C) B-20 (B)
C-1 (A) C-2 (B) C-3 (B) C-4 (B) C-5 (D) C-6 (B)
C-7 (A) C-8 (D) C-9 (B) C-10 (D) C-11 (B)
D-1 (A) D-2 (A) D-3 (B) D-4 (D) D-5 (A) D-6 (C)
D-7 (D) D-8 (A) D-9 (B) D-10 (A) D-11 (D) D-12 (B)
D-13 (B) D-14 (C)
EXERCISE - 2
OBJECTIVE QUESTIONS
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. A B D B A D D B D C C B C C C
Que. 16
Ans. D
44

EXERCISE - 3
NTSE PROBLEMS (PREVIOUS YEARS)
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. A B A C B D A C B B D B B B C A B A C C
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. C D D C A D D C B D C B C D A C A A A A
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53
Ans. D D A A B B A A D C C D C
45

NTSE Motion Resonance

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Rahul Pancholi Sir Unacademy Notes Target - NTSE

A. REST AND MOTION

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(d) Scalar and vector quantities :

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(e) Distance and displacement :

(f) Difference between distance and displacement :

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Similarly for each step, Displacement = b2  h2 b

So that total displacement = n × B

(b) Displacement = 2r along west.

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Q.3_ Is absolute rest possible or not ?

Q.5_ Give an example of motion with displacement = 0 and distance travelled  0

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(b) Non-Uniform Motion :

(c) Speed and velocity :

(I) Types of Speed :

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Here distance covered by the boy is same

distance travelled in a given direction

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(d) Difference between Speed and Velocity :

dis tan ce travelled displaceme nt

3. It is a scalar quantity. 3. It is a vector quantity

4. Speed will always be 4.It will be positive or negative depending on the

5. For moving body, it will never 5.It may be zero

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Path Distance Displacement Speed Velocity

AB 4m 4m 4/2 m/s 4/2 m/s

ABC 10m 10m 10/5 m/s 10/5 m/s

ABCB 16m 4m 16/7 m/s 4/7 m/s

ABCA 20m 0m 20/10 m/s 0/10 m/s

ABCAD 25m –5m 25/14 m/s –5/14 m/s

path length 360m

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Q.1_ Can a body have a constant velocity but a varying speed?

which is 1.5  10 11 m away? (Ans. 5  102 sec.)

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C. ACCELERATION AND EQUATION OF MOTION

v–u final velocity – initial velocity

(b) Types of Acceleration :

(c) Equations of uniformly accelerated motion :

(ii) 2nd Equation of Motion :

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(d) Motion under gravity (uniformly accelerated motion)

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Q.1_ Give an example of motion with acceleration a = 0 and velocity v  0

Q.2_ Is it possible to have velocity v = 0 and acceleration a  0 ? If yes, give an example.

Q.3_ Define acceleration. Give its S.I. unit ?

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Q._10 A ball thrown vertically upward reaches a height of 80 m. Calculate :

(ii) When the body is moving with a uniform speed

(iii) When the body is moving with a non–uniform speed

(i) Distance–time graph for a body at rest :

distance of 15m from

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