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Motion Class 9 Numerical Part - 1 | Physicscatalyst
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May 22, 2025
In this video on motion numerical learn to solve motion numericals for class 9 science chapter 8 of NCERT book. Questions discussed in this video can be found at this link Motion Numericals: https://physicscatalyst.com/Class9/motion_nm.php. Please visit this link to have a preview of these questions. Link to part 2 of this video - https://www.youtube.com/watch?v=c28ZyXFWwxc Chapters : 0:00 Intro 00:30 Formula Used 4:00 Question 1 9:36 Question 2 12:06 Question 3 16:19 Question 4 19:28 Question 5 21:30 Question 6 23:28 Question 7 25:40 Outro ------------------------------------------------------------------ Links related to study material -------------------------------------------------------------------
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0:00
hello and welcome to physicscatalyst.com
0:03
this video is about numerical questions
0:05
and answers on motion for class 9
0:08
physics all the links related to study
0:10
material are given in description box so
0:13
do check them out for more information
0:16
this is part one of our video tutorial
0:19
series on motion numericals for class
0:21
nine now in this video we will have a
0:23
look at some very easy numericals level
0:26
of these numericals would increase as we
0:28
move further in our tutorial series let
0:31
us now begin our tutorial by first
0:34
having a look at formulas used in
0:37
these numerical questions so our first
0:40
formula is that of average speed
0:43
average speed is total distance upon
0:46
total time taken to cover that distance
0:49
generally
0:50
uh total distance or distance is denoted
0:53
by a letter s
0:56
or d and time taken is denoted by
1:00
letter t
1:02
now our second formula is that of
1:04
average velocity
1:06
now average velocity is equals to total
1:08
displacement upon time taken
1:11
to cover that displacement
1:13
now we all know that there is a
1:16
difference between distance and
1:18
displacement like displacement is actual
1:21
distance covered by the object for
1:24
example
1:26
let us say
1:27
a person
1:31
start moving from this point a
1:36
and he has to travel up till this point
1:40
b
1:43
now if he takes
1:45
suppose
1:46
a path
1:48
like this
1:52
then total distance would be amount of
1:55
distance traveled by the person in
1:58
reaching from point a to point b through
2:01
this path
2:02
now if we talk about displacement
2:05
then displacement is shortest possible
2:08
distance between this point a
2:11
and this point b so displacement in
2:14
reaching from point a to point b
2:17
would be
2:19
this straight line distance between
2:22
point a and b and obviously this is the
2:26
shortest possible
2:28
path between point a and point b and one
2:31
more point of difference between
2:33
distance and displacement is that
2:36
distance is a scalar quantity means it
2:39
only has
2:40
magnitude it does not have direction
2:43
and displacement is
2:45
a
2:46
vector quantity
2:49
it has both
2:50
magnitude
2:52
and direction
2:56
similarly average speed is a
2:59
scalar quantity
3:02
and average velocity is a
3:04
vector quantity
3:06
now a second formula used would be v is
3:09
equals to u plus a t
3:11
where u is the initial velocity
3:17
v is the
3:18
final velocity
3:21
a is the acceleration
3:23
and t is the time
3:27
now second equation is s is equals to ut
3:29
plus half a t square where s is the
3:32
displacement u is the initial velocity t
3:35
is the time taken and a as you know is
3:37
the acceleration
3:39
of the moving object
3:41
our third equation is v square is equals
3:44
to u square plus 2 s
3:47
where all these quantities has their
3:49
usual meaning
3:51
let us now begin with our first question
3:54
our first question states that a train
3:56
accelerates from 36 kilometer per hour
3:59
to 54 kilometer per hour in 10 seconds
4:04
we have to find the acceleration and
4:06
distance traveled by the car
4:09
let us now first find the acceleration
4:11
we know that acceleration is given by
4:15
a is equals to
4:17
delta v upon
4:19
delta t
4:20
where this delta v
4:22
is equals to
4:24
final velocity
4:26
minus
4:27
initial velocity
4:30
divided by
4:31
time
4:33
now let us first calculate our delta v
4:36
delta v is equals to
4:38
vf what is vf here it is 54 kilometer
4:43
per hour so
4:45
delta v is equals to
4:49
54
4:51
minus
4:52
vi here is 36 kilometer per hour 36
4:58
which is equals to 18
5:02
kilometer
5:04
per
5:05
hour
5:06
now
5:07
in our question time is given in second
5:10
and
5:12
our velocity is given in kilometer per
5:14
hour either we have to change this time
5:18
that is in seconds to hour or we would
5:21
have to change our velocity from
5:23
kilometer per hour to
5:26
meter per second
5:28
so we will convert it into meter per
5:31
second which is equals to 18 times here
5:35
we will use our formula one kilometer
5:37
per hour is equals to 10 by 36 meter per
5:40
second 18 times 10
5:43
divided by 36
5:45
meter per second which is equals to 5
5:49
meter per second
5:52
we now know the change in velocity and
5:55
we also know that our time delta t
5:58
is given as
6:00
10 seconds
6:02
so our acceleration would be
6:04
a is equals to delta v
6:07
upon delta t
6:10
which is equals to
6:12
5
6:13
meter per second divided by
6:16
10 seconds and this is equals to 0.5
6:22
meter per second whole square
6:26
so this is our
6:28
required answer
6:31
let us now look at second part of our
6:35
question
6:36
now in this part we have to find the
6:39
distance traveled by the car let us now
6:41
first gather all the information we have
6:44
about the motion of the car so we know
6:46
that car moved for about 10 seconds
6:50
acceleration of the car we calculated
6:53
previously
6:54
is
6:55
0.5
6:57
meter per second whole square
7:00
okay
7:01
and we also have the knowledge about
7:04
initial velocity
7:06
which is equals to 36 kilometer per hour
7:08
and final velocity which is equals to 54
7:13
kilometer per
7:15
r now if we look at our
7:18
equations of motion we could very well
7:21
use this equation s is equals to ut plus
7:24
half a t square
7:26
in solving this problem why because we
7:29
have to find s we have knowledge about
7:32
initial velocity we have knowledge about
7:34
time for which the vehicle has traveled
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and we already have calculated the
7:40
acceleration of the car so we can
7:42
comfortably use this equation
7:44
in solving the second part of the
7:46
question
7:48
so equation we would use to solve this
7:51
part of the problem is s is equals to
7:54
u t
7:56
plus
7:57
half
7:58
a t square now as you can see
8:02
from the
8:03
given quantities or calculated
8:05
quantities that our time is in seconds
8:07
acceleration is also in meter per second
8:10
square but our initial velocity is in
8:13
kilometer per hour
8:15
so for finding distance we do not need
8:18
final velocity so we would first have to
8:21
convert our initial velocity which is in
8:23
kilometer per hour to meter per second
8:26
so
8:27
it is equals to
8:30
36
8:31
times
8:34
10 divided by 36
8:37
meter per second or
8:41
it is equals to 10
8:43
meter per second now we will plug in all
8:46
our quantities
8:49
initial velocity
8:52
acceleration and time
8:55
into this equation
8:57
so s is equals to
9:00
10
9:01
times
9:02
10
9:04
plus
9:05
1 by 2
9:07
z 0.5
9:09
which is acceleration
9:11
times 10 whole square
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which is equals to 100
9:20
plus 1 by 2
9:23
times 50
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is equals to 100
9:28
plus
9:29
25 is equals to 125
9:33
meters
9:35
now let us look at our
9:37
second question
9:40
which states that a body whose speed is
9:43
constant and the option given are
9:46
option a must be accelerated option b is
9:49
might be accelerated option c is
9:51
has a constant velocity and option d is
9:54
cannot be accelerated
9:56
now the answer to this question is
9:59
option number b
10:01
might be
10:02
accelerated now why i am saying that a
10:06
body whose speed is constant might be
10:09
accelerated so let us consider a case
10:13
where a particle p is moving on the
10:16
surface of a circle in clockwise
10:18
direction now this particle p
10:21
is moving with a constant speed but in
10:24
this case we must note that as the
10:26
particle moves on the surface of the
10:29
circle
10:30
at each point of motion its direction of
10:34
motion is changing so for this how would
10:36
you find the direction of motion of this
10:39
particle at this point it is in
10:43
this direction now if this particle
10:47
moves at
10:48
this point q then at this point its
10:51
speed is constant but its direction of
10:54
motion would be like this this is not a
10:57
perfect direction what i have drawn it
11:00
is basically tangent to the circle at
11:03
this point
11:04
at this point it would be somewhat like
11:08
this in this direction so you see in
11:11
this case
11:12
of a motion of a particle on the surface
11:16
of the circle its speed is constant node
11:18
out but as the particle moves its
11:21
direction keeps on changing and since
11:24
the direction of motion of the particle
11:27
keeps on changing its velocity keeps on
11:30
changing so we can say that a body whose
11:33
speed is constant might be accelerated
11:36
like in this case of motion but if you
11:40
take a case of a straight line motion
11:42
then if a speed of the man walking is
11:46
constant and it is moving from
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point a to point b
11:51
in this direction then direction is also
11:53
constant so in this case the motion is
11:57
not accelerated so or option d is
11:59
current it might be accelerated it might
12:02
not be accelerated
12:05
let us now look at our question three a
12:08
question states that a truck traveling
12:10
at 54 kilometer per hour is slow down to
12:14
36 kilometer per hour in 10 second find
12:17
the retardation now let us first write
12:20
down the information given to us in the
12:23
question so we are given the initial
12:25
velocity as
12:27
vi
12:29
is equals to
12:31
54
12:33
kilometer
12:34
per hour
12:36
our final velocity would be
12:38
vf
12:40
is equals to
12:43
36 kilometer
12:47
per
12:48
r and time t in reaching from initial
12:51
velocity v i to final velocity is
12:55
t is equals to 10 second
12:58
now both our initial velocities and
13:00
final velocities are given in kilometer
13:02
per hour so we would have to first
13:05
convert them into meter per second now
13:07
we know that one kilometer
13:10
per hour
13:12
is equals to
13:14
10 by 36
13:17
meter per second
13:20
so
13:21
vi is equals to
13:24
54
13:26
times
13:28
10 by 36
13:30
meter per second
13:32
which is equals to 15
13:35
meter per second similarly we will uh
13:38
convert our final velocity
13:41
which is vf is equal to 36 kilometer per
13:43
hour into meter per second so
13:46
36 times
13:49
10 by 36
13:52
meter per second which is equals to 10
13:56
meter per second
14:02
now let us
14:03
before going any further let us first
14:05
look at our term retardation
14:14
so retardation is nothing but a
14:17
negative
14:20
acceleration
14:28
so uh we know that
14:31
change in velocity of a body may either
14:34
be positive
14:41
or
14:42
negative
14:44
now if the change in velocity
14:47
of the moving body is positive
14:50
then acceleration is said to be
14:54
positive
14:56
and if the velocity decreases or change
15:00
in velocity is negative then it is
15:03
called a deceleration
15:07
or retardation
15:11
so retardation is simply nothing but
15:14
negative acceleration so now we would
15:17
calculate retardation
15:19
from the given data
15:22
so we know that
15:23
acceleration is given by
15:27
a is equals to delta v
15:29
upon
15:31
delta t
15:33
which is equals to
15:35
final velocity minus
15:37
initial velocity upon change in time
15:41
which in this case is
15:43
10 seconds
15:44
so now putting in the values
15:46
we get
15:47
10
15:48
which is our final velocity
15:51
minus 15
15:53
which is our initial velocity divided by
15:58
10
15:59
and the answer is minus
16:02
0.5
16:04
meter per second whole square so here
16:08
the negative sign
16:10
implies that
16:12
change in velocity is negative and
16:16
it is retardation or deceleration
16:20
let us now look at our fourth question
16:22
now it states that a particle is moving
16:25
in a circle
16:26
of diameter 20 meter what is its
16:30
distance as per the table given below so
16:33
let us solve this problem now here we
16:36
have to fill the table for the particle
16:39
moving on the surface of a circle now
16:42
the circle has a diameter of 20 meter
16:45
now if the particle completes one round
16:48
of the circle and return to this
16:51
position o then what would be its
16:53
displacement and what would be the
16:55
distance covered by this particle so
16:58
displacement of the particle in this
17:01
case would be zero because the particle
17:03
has returned to its original position to
17:06
the position from which it has started
17:09
to move and distance
17:12
in this case would be equal to
17:14
circumference of this whole circle so
17:16
distance is equals to 2 pi r
17:20
for one round
17:21
it is equals to
17:23
20
17:25
pi
17:26
because diameter is equals to 2r so in
17:30
this case distance is equals to 20
17:33
pi
17:34
now if the particle completes 1.5 round
17:39
means particle starts from this point o
17:42
then cover the entire circumference of
17:45
the circle and then half round of the
17:47
circle let's say it is at this point a
17:51
so now in this case
17:53
displacement would be 20 meter
17:57
okay displacement would be 20 meter
18:00
which is equals to diameter of the
18:02
circle and
18:05
our distance is equals to
18:08
2 pi r plus pi
18:11
r which is equals to
18:13
20 pi
18:15
plus 10 pi so it would be
18:18
35
18:20
now if it completes two rounds means it
18:23
starts from position o
18:25
then make one complete round and then
18:27
again make one complete round it is
18:30
again at its initial position so
18:32
displacement would be zero
18:34
and
18:35
distance covered would be equal to
18:38
two times
18:40
two pi r
18:42
which is equals to 40 pi
18:45
and if it completes 2.5 rounds
18:49
then again displacement would be
18:53
20 meter
18:55
and distance would be
19:00
d is equals to
19:02
2 times
19:04
2 pi r
19:05
plus
19:06
pi r because again after completing
19:10
two rounds
19:11
complete and one half round it is at
19:14
point a
19:16
so
19:17
our distance in this case would be equal
19:19
to 40 pi
19:21
plus 10 pi which is equals to
19:25
50 pi now let us look at our question 5
19:28
it states that a scooter traveling at 10
19:31
meter per second speed up to 20 meter
19:34
per second in 4 second find the
19:36
acceleration of the scooter now the
19:39
solving this question involves a simple
19:41
formula of acceleration
19:44
a is equals to delta v upon delta t
19:48
we have used this formula in previous
19:51
problems also
19:53
so
19:54
what you have to do is plug in the
19:56
values
19:58
and find your answers so delta v is
20:00
equals to vf minus vi
20:04
upon
20:05
delta t
20:06
now here vf is 20 meter per second vi is
20:09
10 meter per second so putting in the
20:12
values we get
20:14
20 minus 10 divided by 4
20:18
is equals to 2.5
20:22
meter per second whole square
20:24
[Music]
20:26
now let us look at our question five it
20:28
states that a scooter traveling at 10
20:31
meter per second speed up to 20 meter
20:33
per second in 4 second find the
20:36
acceleration of the scooter now the
20:38
solving this question involves a simple
20:41
formula of acceleration a is equals to
20:44
delta v upon delta t we have used this
20:47
formula in previous
20:49
problems also so what you have to do is
20:53
plug in the values
20:55
and find your answers so delta v is
20:58
equals to vf minus vi
21:01
upon
21:02
delta t
21:04
now here vf is 20 meter per second vi is
21:07
10 meter per second so putting in the
21:09
values we get
21:11
20 minus 10 divided by 4
21:15
is equals to 2.5
21:19
meter per second whole square
21:22
let us now look at our question 6 which
21:24
states that a train starts from rest
21:31
and accelerate uniformly at a rate of 5
21:34
meter per second square
21:36
for 5 second
21:38
calculate the velocity of train in 5
21:41
seconds
21:43
we will first gather the information
21:45
given in the question
21:47
it is a very easy question simple and
21:49
straightforward question so what is the
21:51
information given
21:52
in the question it is
21:55
initial velocity of the train vi is
21:59
equals to 0
22:00
and
22:02
we are given that it is accelerating
22:04
uniformly at a rate of 5 meter per
22:07
second square so acceleration of the
22:10
train is equals to 5
22:12
meter
22:13
per second whole square now we have to
22:16
calculate the velocity of train in 5
22:18
seconds so time given is t is equals to
22:22
5 second
22:23
so what we have to calculate we have to
22:26
calculate the velocity of the train
22:29
after 5 second when it is accelerating
22:32
at a rate of 5 meter per second square
22:35
so here we are giving initial velocity
22:38
acceleration and time t so
22:41
which formula we can use we can use
22:43
formula
22:45
v is equals to
22:47
u plus a t
22:50
u also denotes initial velocity
22:54
so generally we denote initial velocity
22:57
using two symbols either vi or
23:01
u
23:02
so putting in all these values we get
23:04
final velocity v f
23:07
is equals to
23:08
0 plus
23:11
5 meter per second square times
23:14
t which is equals to 5 second on
23:16
calculating we get final velocity equals
23:19
to 25
23:21
meter per second and
23:24
this is our required answer
23:26
let's now look at our question 7 this is
23:29
a multiple choice question now the
23:31
question states that object moves with a
23:34
uniform positive acceleration its
23:37
velocity time graph will be option a is
23:39
a straight line parallel to the time
23:41
axis option b is a straight line
23:44
inclined at an obtuse angle to the time
23:47
axis option c is a straight line
23:49
inclined at an acute angle to the time
23:52
axis and option
23:54
d is none of these let us now consider
23:56
our option a where velocity time graph
23:59
is a straight line parallel to time
24:02
access as can be seen in this graph so
24:05
here in this case we can see that the
24:08
velocity of the moving object does not
24:11
change with the passage of time this
24:14
means that object this means that
24:17
object is
24:19
not accelerating so since the object is
24:22
not accelerating so this straight line
24:26
parallel to the time axis does not
24:28
represent the velocity time graph of
24:32
object moving with uniform positive
24:34
acceleration now let's consider or
24:37
option b where a straight line inclined
24:39
at an obtuse angle to the time axis so
24:43
straight line inclined at an obtuse
24:46
angle to the time axis looks like this
24:49
so this line is making an obtuse angle
24:51
with the time axis now as that can be
24:53
seen clearly from this graph that the
24:55
velocity
24:56
is decreasing as time goes on so in this
25:00
case we would get a negative
25:01
acceleration
25:04
or retardation so this option b is also
25:07
not correct now a third option is a
25:09
straight line inclined at an acute angle
25:12
to the time axis from this velocity time
25:15
graph it can be clearly seen that as the
25:18
time goes on velocity increases so in
25:22
this case acceleration of the object
25:27
would be positive
25:29
so our option c where a straight line
25:32
inclined at an acute angle to the time
25:34
axis is our correct option
25:39
this video we have solved several
25:41
numerical problems in our next video
25:43
tutorial we will cover some more such
25:45
problems
25:46
do not forget to like this video and
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subscribe to our channel for more such
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