ncert solution of class 9 chapter motion
Motion
Page No: 100
11. An object has moved through a
distance. Can it have zero displacement? If yes, support your answer with an
example.
Answer
Yes,an object can have zero displacement even when it has moved through a
distance.This happens when final position of the object coincides with its
initial position. For example,if a person moves around park and stands on place
from where he started then here displacement will be zero.
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 from his initial position?
Answer
Given, Side of the
square field= 10m
Therefore, perimeter = 10 m x 4
= 40 m
Farmer moves along the boundary
in 40s.
Displacement after 2 m 20 s = 2
x 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m
Therefore, in 1s distance
covered by farmer = 40 / 40 m = 1m
Therefore,
in 140s distance covered by farmer = 1 × 140 m = 140 m
Now, number
of rotation to cover 140 along the boundary= Total Distance / Perimeter = 140 m
/ 40 m = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.
Thus, after 2 min 20 seconds
the displacement of farmer will be equal to 14.14 m north east from intial
position.
3. Which of the following is true for
displacement?
(a) It
cannot be zero.
(b) Its
magnitude is greater than the distance travelled by the object.
Answer None of the statement is true for displacement First statement is false because displacement can be zero. Second statement is also false because displacement is less than or equal to the distance travelled by the object.
Answer None of the statement is true for displacement First statement is false because displacement can be zero. Second statement is also false because displacement is less than or equal to the distance travelled by the object.
Page No: 102
1. Distinguish between speed and
velocity.
Answer
Speed
|
Velocity
|
Speed is the distance
travelled by an object in a given interval of time.
|
Velocity is the displacement
of an object in a given interval of time.
|
Speed = distance / time
|
Velocity = displacement /
time
|
Speed is scalar quantity i.e.
it has only magnitude.
|
Velocity is vector quantity
i.e. it has both magnitude as well as direction.
|
2. Under what condition(s) is the
magnitude of average velocity of an object equal to its average speed?
Answer The
magnitude of average velocity of an object is equal to its average speed, only
when an object is moving in a straight line.
3. What does the odometer of an
automobile measure?
Answer The
odometer of an automobile measures the distance covered by an automobile.
4. What does the path of an object look
like when it is in uniform motion?
Answer An
object having uniform motion has a straight line path.
5. 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 × 108 m s−1.
Answer;
Speed= 3 × 108 m s−1
Time= 5 min = 5 x 60 = 300
secs.
Distance= Speed x Time
Distance= 3 × 108 m s−1 x 300 secs. =
9 x 1010 m Page No: 103
1. When will you say a body is
in
(i) uniform acceleration?
(ii) non-uniform acceleration?
Answer (i) A body is said to be
in uniform acceleration if it travels in a straight line and its velocity
increases or decreases by equal amounts in equal intervals of time.
(ii) A body is said to be in
nonuniform acceleration if the rate of change of its velocity is not constant.
2. A bus decreases
its speed from 80 km h−1
to 60 km h−1 in 5 s. Find the acceleration of the bus.
Answer
4.3. A train starting from a railway
station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration.
Answer
Page No: 107
1.
What is the
nature of the distance - 'time graphs for uniform and non-uniform motion of an
object?
Answer When the motion is
uniform,the distance time graph is a straight line with a slope.
When the
motion is non uniform, the distance time graph is not a straight line.It can be
any curve.
2. 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 distance
time graph is a straight line parallel to the time axis, the body is at rest.
3. 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
If speed
time graph is a straight line parallel to the time axis, the object is moving uniformly.
4. What is the quantity which
is measured by the area occupied below the velocity -time graph?
Answer The
area below velocity-time graph gives the distance covered by the object.
Page No: 109
1. A bus
starting from rest moves with a uniform acceleration of 0.1 m s−2 for 2 minutes.
Find (a) the
speed acquired, (b) the distance travelled.
Answer
Initial speed of the bus, u= 0
Acceleration,
a = 0.1 m/s2
Time taken,
t = 2 minutes = 120 s
(a) v= u + at
v= 0 + 0×1 × 120
v= 12 ms–1
(b) According to the third equation of
motion:
v2 - u2= 2as
Where, s is
the distance covered by the bus
(12)2 - (0)2= 2(0.1) s
s = 720 m
Speed
acquired by the bus is 12 m/s.
Distance
travelled by the bus is 720 m.
Page No: 110
11.A train is travelling at a speed of
90 km h−1. Brakes are applied so as to produce
a uniform acceleration of −0.5 m s−2.
Find how far the train will go before it is brought to rest.
Answer
Initial speed of the train, u= 90 km/h = 25
m/s
Final speed
of the train, v = 0 (finally the train comes to rest) Acceleration = - 0.5 m s-2
According to
third equation of motion:
v2= u2+
2 as
(0)2= (25)2+
2 ( - 0.5) s
Where, s is
the distance covered by the train
The train
will cover a distance of 625 m before it comes to rest.
22.A trolley, while going down an
inclined plane, has an acceleration of 2 cm s−2. What will be its velocity 3 s after
the start?
Answer
Initial
Velocity of trolley, u= 0 cms-1
Acceleration,
a= 2 cm s-2
Time, t= 3 s
We know that
final velocity, v= u + at = 0 + 2 x 3 cms-1
Therefore,
The velocity of train after 3 seconds = 6 cms-1
4. A racing car has a uniform
acceleration of 4 m s - '2. What distance will it cover in 10 s after start?
Answer
Initial
Velocity of the car, u=0 ms-1
Acceleration,
a= 4 m s-2
Time, t= 10 s
We know
Distance, s= ut + (1/2)at2
Therefore,
Distance covered by car in 10 second
= 0 × 10 +
(1/2) × 4 × 102
= 0 + (1/2)
× 4× 10 × 10 m
= (1/2)× 400
m = 200 m
5. A stone is thrown in a vertically
upward direction with a velocity of 5 m s−1. If the acceleration of the stone during its motion is 10 m
s−2 in the downward direction, what will be the height attained
by the stone and how much time will it take to reach there?
Answer Given
Initial velocity of stone, u=5 m s-1
Downward of negative Acceleration, a= 10 m s-2
We know that
2 as= v2- u2
Page No: 112
Excercise
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?
Answer
Diameter of circular track (D) = 200 m
Radius of
circular track (r) = 200 / 2=100 m
Time taken
by the athlete for one round (t) = 40 s
Distance
covered by athlete in one round (s) = 2Ï€ r
= 2 x ( 22 /
7 ) x 100
Speed of the
athlete (v) = Distance / Time
= (2 x 2200)
/ (7 x 40)
= 4400 / 7 ×
40
Therefore,
Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7
x 40) x (2 x 60 + 20)
= 4400 / ( 7
x 40) x 140
= 4400 x 140 /7 x 40
= 2200 m
Number of round in 40 s =1
round
Number of round in 140 s
=140/40
=3 1/2
After taking start from
position X,the athlete will be at postion Y after 3 1/2 rounds as
shown in figure
Hence,
Displacement of the athlete with respect to initial position at x= xy
= Diameter of circular track
= 200 m 2.
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?
Answer
Total Distance covered from AB
= 300 m
Total time taken = 2 x 60 + 30
s
=150 s
Therefore,
Average Speed from AB = Total Distance / Total Time
=300 / 150 m
s-1
=2 m s-1
Therefore,
Velocity from AB =Displacement AB / Time
= 300 / 150
m s-1
=2 m s-1
Total
Distance covered from AC =AB + BC =300 + 200 m
Total time taken from A to C =
Time taken for AB + Time taken for BC
= (2 x 60+30)+60 s
= 210 s
Therefore, Average Speed from
AC = Total Distance /Total Time
= 400 /210 m s-1
= 1.904 m s-1
Displacement (S) from A to C =
AB – BC
= 300-100 m
= 200 m
Time (t) taken for displacement
from AC = 210 s
Therefore,
Velocity from AC = Displacement (s) / Time(t)
= 200 / 210
m s-1 = 0.952 m s-1
2. 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 40 km h−1.
What is the average speed for Abdul’s trip?
Answer
The distance
Abdul commutes while driving from Home to School = S
Let us assume
time taken by Abdul to commutes this distance = t1 Distance Abdul commutes while driving from School to Home = S
Let us
assume time taken by Abdul to commutes this distance = t2 Average speed from home to school v1av = 20 km h-1
Average
speed from school to home v2av = 30 km h-1
Also we know
Time taken form Home to School t1 =S /
v1av
Similarly
Time taken form School to Home t2 =S/v2av
Total
distance from home to school and backward = 2 S
Total time
taken from home to school and backward (T) = S/20+ S/30
Therefore,
Average speed (Vav) for covering total distance (2S) =
Total Dostance/Total Time
= 2S / (S/20
+S/30)
= 2S /
[(30S+20S)/600]
= 1200S /
50S
= 24 kmh-1
34.constant rate of 3.0 m s−2 for 8.0 s. How far does the boat travel during this time?
Answer
Given
Initial velocity of motorboat, u = 0
Acceleration
of motorboat, a = 3.0 m s-2
Time under
consideration, t = 8.0 s
We know that
Distance, s = ut + (1/2)at2
Therefore,
The distance travel by motorboat
= 0 x 8 +
(1/2)3.0 x 8 2
= (1/2) x 3
x 8 x 8 m
= 96 m
4.5. 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?
Answer As
given in the figure below PR and SQ are the Speed-time graph for given two cars
with initial speeds 52 kmh-1 and 3 kmh-1 respectively.
Distance Travelled by first car
before coming to rest =Area of △ OPR
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) /
3600) m
= (1/2) x 5x (130 / 9) m = 325
/ 9 m
= 36.11 m
Distance Travelled by second
car before coming to rest =Area of △ OSQ = (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) /
3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m
5.6. 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 has C travelled
when B passes A?
(d)How far has B travelled by
the time it passes C?
Answer (a)
Object B
(b) No
(c) 5.714 km
(d) 5.143 km
Therefore,
Speed = slope of the graph Since slope of object B is greater than objects A
and C, it is travelling the fastest.
(b) All three objects A, B and C never
meet at a single point. Thus, they were never at the same point on road.
On the
distance axis:
7 small
boxes = 4 km
Therefore,1
small box = 4 / 7 Km
Initially,
object C is 4 blocks away from the origin.
Therefore,
Initial distance of object C from origin = 16 / 7 Km
Distance of
object C from origin when B passes A = 8 km
Distance covered by C
Page No: 113
6. 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?
Answer Let us assume, the final velocity with
which ball will strike the ground be 'v' and 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 m s-2
As we know, 2as =v2-u2
v2 = 2as+ u2
= 2 x 10 x 20 + 0
= 400
Final velocity of ball, v = 20
ms-1
t = (v-u)/a
∴Time
taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds
8. The speed-time graph for a
car is shown is Fig. 8.12.
(a) Find out how far the car
travels 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?
Answer (a)
The shaded
area which is equal to 1 / 2 x 4 x 6 = 12 m represents the distance travelled
by the car in the first 4 s.
(b)
The part of
the graph in red colour between time 6 s to 10 s represents uniform motion of
the car.
10. 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) Possible When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.
(b) Possible When a car is moving in a circular track, its acceleration is perpendicular to its direction.
(b) Possible When a car is moving in a circular track, its acceleration is perpendicular to its direction.
11. 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 Radius
of the circular orbit, r= 42250 km
Time taken
to revolve around the earth, t= 24 h
Speed of a
circular moving object, v= (2Ï€ r)/t
=[2×
(22/7)×42250 × 1000] / (24 × 60 × 60)
=(2×22×42250×1000)
/ (7 ×24 × 60 × 60) m s-1
=3073.74 m s -1
