NCERT Solutions for Class 9 Science “Motion” :
NCERT Solutions for Class 9
Science Chapter 8 “Motion”:
NCERT Solutions for Class 9 Science Chapter 8
Motion.
Introducing: Our Science experts have intelligently crafted "Motion"
to align with the updated CBSE 2024-25 syllabus for class 9. By condensing
complex textbook content into straightforward solutions, students can easily
tackle NCERT questions. You can access all the NCERT Solutions for Class 9
Science Chapter 7 on our (cbsencert21.blogspot.com)
website. Furthermore, students can also find NCERT Solutions for Class 9
Science.
NCERT SOLUTIONS FOR CLASS 9 SCIENCE
CHAPTER-8 MOTION
NCERT Solutions for class 9 Science Chapter 8
Motion have been prepared and uploaded for reference by our team of academic
experts at NCERT Solution cbsencert21.blogspot.com.
Access solutions to all chapters of NCERT class 9 Science from our website for
comprehensive guidance. Utilize the following NCERT solutions of Chapter 8, crafted
by our team, as a valuable reference. Take the time to review Chapter 8 on Motion before heading into the
exam.
What we learned?
Ø Motion is a change of
position; it can be described in terms of the distance moved or the
displacement.
Ø The motion of an object
could be uniform or non-uniform depending on whether its velocity is constant
or changing.
Ø The speed of an object
is the distance covered per unit time, and velocity is the displacement per
unit time.
Ø The acceleration of an
object is the change in velocity per unit time.
Ø Uniform and non-uniform
motions of objects can be shown through graphs.
Ø The motion of an object
moving at uniform acceleration can be described with the help of the following
equations, namely
v = u + at
s = ut + ½ at2
2as = v2 – u2
5.
THE
FUNDAMENTAL UNIT OF LIFE
6.
TISSUES
7.
MOTION
9.
GRAVITATION
10.
WORK
AND ENERGY
11.
SOUND
12.
IMPROVEMENT
IN FOOD RESOURCES
Question 1.
An object has moved through a distance. Can
it have zero displacement? If yes, support your answer with an example.
Answer:
Yes, if an object has moved through a
distance it can have zero displacement because the displacement of an object is the
actual change in its position when it moves from one position to the other. So
if an object travels from point A to B and then returns back to point A again,
the total displacement is zero.
Question 2.
A farmer moves along the boundary of a square
field of side 10 m in 40 s. What will be the magnitude of displacement of the
farmer at the end of 2 minutes 20 seconds?
Answer:
Distance covered by a farmer in 40 seconds
=4 x (10) m= 40 m
Speed of the farmer = distance/time = 40m/40s
= 1 m/s.
Total time is given in the Question = 2min
20seconds
= 60+60+20 =140 seconds
Since he completes 1 round of the field in 40
seconds, he will complete 3 rounds in 120 seconds (2 mins), or 120 m
distance is covered in 2 minutes. In another 20 seconds will cover another 20 m
so the total distance covered in 2 min 20 sec
= 120 +20 =140 m.
Displacement = √102 +
102 =√200
= 10√2 m (as per diagram)
=10 x 1.414= 14.14 m.
Question 3.
Which of the following is true for
displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance traveled by the object.
Answer:
Both (a) as well as (b) are false concerning the concept of displacement.
Question 4.
Distinguish between speed and velocity.
Answer:
Speed of a body is the distance travelled by
it per unit time while velocity is displacement per unit time of the body
during movement.
Question 5.
Under what condition(s) is the magnitude of the average
velocity of an object equal to its average speed?
Answer:
If the distance travelled by an object is equal
to its displacement then the magnitude of the average velocity of an object will be
equal to its average speed.
Question 6.
What does the odometer of an automobile
measure?
Answer:
The odometer of an automobile measures the
distance covered by that automobile.
Question 7.
What does the path of an object look like
when it is in uniform motion?
Answer:
Graphically the path of an object will be
linear i.e. look like a straight line when it is in uniform motion.
Question 8.
During an experiment, a signal from a
spaceship reached the ground station in five minutes. What was the distance of
the spaceship from the ground station? The signal travels at the speed of
light, that is 3 x108ms-1
Answer:
Distance = Speed x time
= 5 min x (3 x 108 m s-1
)
= (5 x 60) seconds x (3 x 108 m s-1
)
= 300 sec x (3 x 103 m s-1
)
= 900 sec x (3 x 108 m s-1
)
= 9 x 1010 m
Question 9.
When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?
Answer:
(i) Uniform acceleration: When an object
travels in a straight line and its velocity changes by an equal amount in equal
intervals of time, it is said to have uniform acceleration.
(ii) nonuniform acceleration: It is also
called variable acceleration. When the velocity of an object changes by unequal
amounts in equal intervals of time, it is said to have non-uniform
acceleration.
Question
10.
A bus
decreases its speed from 80 kmh-1 to 60 km h-1 in
5 s.
Find the acceleration of the bus.
Answer:
The initial speed of bus (u) = 80 km-1
= 80 x 1000 / 60 x 60 sec
= 200 / 9 ms-1
= 22.22 ms-1
final speed of
bus (v)=60 km-1 =60 x 1000 / 60 x 60 seconds
=50 / 3 ms-1
=16.67 ms-1 time (t) = 5 s
acceleration (a)
= ( v – u) /t
= (16.67 –
22.22)/5
= -5.55/5
= -1.11 s
Question 11.
A train starting from a railway station and
moving with uniform acceleration attains a speed of 40 km h-1 in
10 minutes. Find its acceleration.
Answer:
Since the train starts from rest(railway
station) = u = zero
The final velocity of the train:
V= 40 km h-1
=(40 x 1000) / 60 x 60 ms-1
= 100/9
ms-1
= 11.11 ms-1
time (t) = 10 min = 10 x 60
= 600 seconds
Since a = (v – u )/t
= 11.11 ms-1 / 600 seconds
= 0.018 m/s2
Question 12.
What is the nature of the distance-time
graphs for uniform and non-uniform motion of an object?
Answer:
If an object has a uniform motion then the
nature of the distance-time graph will be linear i.e. it would be a straight line and
if it non-non-uniformform motion then the nature of the distance time graph is a curved
line.
Question 13.
What can you say about the motion of an
object whose distance-time graph is a straight line parallel to the time axis?
Answer:
If the object’s distance-time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest i.e. not moving.
Question 14.
What can you say about the motion of an
object if its speed-time graph is a straight line parallel to the time axis?
Answer:
Such a graph indicates that the object is traveling with uniform velocity.
Question 15.
What is the quantity which is measured by the
area occupied below the velocity-time graph?
Answer:
The area occupied below the velocity-time
graph measures the distance moved by any object.
Question 16.
A bus starting from rest moves with a uniform
acceleration of 0.1m s-2 for 2 minutes. Find (a) the speed
acquired, and (b) the distance traveled.
Answer:
(a) u=0, a=0.1
ms-1, t= 2 min = 6-+60=120 seconds.
v=u+at =
0 + .1 x 120
= 12 ms-1
so (a) speed
acquired
v= 12 ms-1
(b) s = ut + ½ at2
=0 x 120 + ½ x 0.1 x 1202
= 720 m.
Question 17.
A train is traveling at a speed of 90km h-1.
Brakes are applied to produce a uniform acceleration of -0.5m s-2.
Find how far the train will go before it is brought to rest.
Answer:
= 90 km h-1
= (90 x 1000) / 60 x 60
= 25 ms-1
a = -0.5 ms-2
v =0(train is brought to rest)
v= u+at = 25 + (-0.5) x t
0 =25 – 0.5 x t
0.5t = 25,
or t = 25/0.5
= 50 seconds
S = ut + ½ at2
= 25 x 50 +1/2 x (-0.5) x 502
= 1250 – 625 = 625m
Question 18.
A trolley, while going down an inclined
plane, has an acceleration of 2cm s-2. What will be its velocity 3 s
after the start?
Answer:
u=0, a=2 cm/s2, t= 3s
v= u +at
= 0 + 2 x 3 = 6 cm/s
Question 19.
A racing car has a uniform acceleration of 4 cm
s-2. What distance will it cover in 10 seconds after start?
Answer:
u = 0, a=5m/s2, t= 10 s
s= ut + ½ at2
= 0x10 +(4 x 10 x 10)/2
= 0 + 200
= 200 m
Question 20.
A stone is thrown in a vertically upward
direction with a velocity of 5 cm s-2. If the acceleration of the stone during
its motion is 10 cm s-2n in the downward direction, what will be the height
attained by the stone and how much time will it take to reach there?
Answer:
v = 0 (since at maximum height its velocity
will be zero)
v = u + at
= 5 + (-10) x t
0 = 5 – 10t
10t = 5, or, t = 5/10 =0.5 second.
= 2.5 – 1.25 = 1.25m
Question 21.
An
athlete completes one round of a circular track of diameter 200 m in 40 seconds. What
will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer:
circumference of circular track = 2πr
= 4400/7 m
rounds completed by athletes in
2 min 20 sec = s= 140/40 = 3.5m
therefore, total distance covered =4400 / 7 x
3.5
= 2200 m
Since one complete round of circular track
needs 40s he will complete 3 rounds in 2mins and in next 20s he can complete
half round therefore displacement = diameter = 200m.
Question 22.
Joseph jogs from one end A to the other end B
of a straight 300 m road in 2 minutes 50 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?
Answer:
(a) distance = 300m
time = 2min30seconds = 150 seconds
average speed from A to B = average velocity
from A to B
= 300m/150s = 2m/s
(b) average speed from A to C =
(300+100)m/(150+60)sec
= 400 m/210 s = 1.90 m/s
displacement from A to C
= (300 – 100) m =200 m
time =2 min 30 sec + 1 min = 210 s
velocity = displacement/time
= 200 m/210 s = 0.95 m/s
Question 23.
Abdul, while driving to school, computes the
average speed for his trip to be 20km h-1. On his return trip along the same
route, there is less traffic and the average speed is 40km h-1. What is the average
speed for Abdul’s trip?
Answer:
If we suppose that distance from Abdul’s home
to school = x km
while driving to school:-
velocity = displacement/time
20 = x/t, or, t=x/20 hr
on his return rip:-
speed = 40 km–1,
40= x /t’
or, to =x/40 hr
total distance travelled = x + x = 2x
total time = t + t’ = x/20 + x/40
=(2x + x)/40 = 3x/40 hr
average speed for Abdul’s trip
= 2x/(3x/40) = 80x/3x = 26.67km/hr
Question 24.
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?
Answer:
since the motorboat starts from r u= 0
time (t) = 8s,
s= ut + ½ at2
= 0 + ½ x 3 x 82
distance = 96 m
Question 25.
A driver traveling 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 traveled led her after the brakes were applied?
Answer:
As given in the figure below AB (in red line)
and CD(in red line) are the Speed-time graph for given two cars with initial
speeds 52 kmh-1 and 3 kmh-1respectively.
Distance Travelledthe by the first car before coming
to rest =Area of DOAB
= (1/2) x OB x OA
= (½) x 5 s x 52 kmh-1
= (1/2) x 5 x(52 x 1000)/3600 m
= (1/2) x 5 x (130/9) m
= 325/9 m
= 36.11 m
Distance Travelledthe by the second car before
coming to rest =Area of DOCD
=(1/2) x OD x OA
=(1/2) x 10 s x 3 kmh-1
=(1/2)
x 10 s x(3 x 1000) / 3600 m
= 5 x 5 /6 m
= 25/6 m
= 4.16 m
∴ Clearly the first car will travel farther
(36.11 m) than the first car(4.16 m).
Question 26.
Fig 8.11 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 was traveled when B passes A?
(d) How far has B travelled by the time it
passes C?
Answer:
(a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.
(b) All of them never come at the same point
at the same time.
(c) According to graph; each small division
shows about 0.57 km.
A is passing B at point S which is in line
with point P (on the distance axis) and shows about 9.14 km
Thus, at this p,point C travels about
9.14 - (0.57 x 3.75)km
= 9.14 km – 2.1375 km
= 7.0025 km
Thus, when A passes B, C travels about 7 km.
(d) B passes C at point Q at the distance
axis which is
=5.28 km
Therefore, B travelled about 5.28 km wpassedsses to C.
Question 27.
A ball is gently dropped from a height of 20
m. If its velocity increases uniformly at the rate of 10ms-2,
with what velocity will it strike the ground? After what time will it strike
the ground?
Answer:
Let us assume, the final velocity with the ich ball will strike the ground is ‘v’ the time it takes to strike the ground be ‘t’
Initial Velocity of ball u=0
Distance or height of fall s=20 m
Downward acceleration a= 10 ms-2
As we know,
V2=u2–2as
Or, 2as = v2-u2
= v2=2as + u2
= 2 x 10 x 20 + 0
V=400 ms-1The financial velocity of the ball, v
t= (v-u)/a
∴ Time taken by the ball to strike= (20-0)/10
= 20/10
= 2 seconds
Question 28.
The speed-time graph for a car is shown is Fig. 8.12.
Fig. 8.12
(a) Find how long does the travels are 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 nts
uniform motion of the car?
Answer:
(a) Distatraveledlled by car in the 4 second
The area under the slope of the speed – time
graph gives the distance traveled by an object.
In the given graph
56 full squares and 12 half squares come
under the area slope for the time of second.
Total number of squares = 56 + 12/2 = 62
squares
The total area of the squares will give the distance traveled by the car a second. On the time axis,
5 squares = 2 seconds, therefore 1 square =
2/5 seconds
on the ed, axis there are 3 squares = 2m/s
therefore reread one square
= 4/15 the
so area of 62 squares=4/15 x 62
= 248/15 m = 16.53 m
Hence the car travels 16.53m in the first 4
seconds.
(b) The straight line part of the graph, from
point A to point B represents the car.
Question 29.
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.
Answer:
(a) An object with a constant acceleration
can still have the zero velocity. For exa, simply an object that is at rest on the
surface of earth will have zero velocity but still being acted upon by the
gravitational force of earth with an acceleration of 9.81 ms-2 towards the
center of earth. Hence when an object starts falling freely can have constant acceleration but with zero velocity.
(b) When an athlete moves with a velocity of
constant magnitude along the circular path, the only change in his velocity is
due to the change in the direction of motion. Here, the motion of the athlete
moving along a circular path is, therefore, an example of an accelerated motion
where acceleration is always perpendicular to direction of motion of an object
at a given instance. Hence it is possible when an object moves on a circular
path.
Question 30.
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.
Answer:
Let us assume an artificial satellite, which
is moving in a circular orbit of radius 42250 km covers a distance ‘s’ as it
revolve around earth with speed ‘v’ in given time ‘t’ of 24 hours.
= 42250The radiusdius of circular orbit r =42250 x 1000 m
Time taken by artificial satellite t=24 hours
=24 x 60 x 60 s
Distance covered by satellite s
=Circumferencea of a circular orbit
= 2πr
∴ Speed of satellite v = (2πr) / t
= [2 x (22/7) x 42250 x 1000]/(24 x 60 x 60)
= (2 x 22 x 42250 x 1000)/(7 x 24 x 60 x 60)
ms-1
= 3073.74 ms-2
= 3.073 km/s
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