NCERT Solutions for Class 9 Science, Chapter 8: Motion

Class 9 is the first stepping stone for a student in the competitive world. With the introduction of the CBSE Board Exam for class 10 a few years back, this has become an important gateway for a student. Based on the results of class 9th a student selects his future stream of Science, Commerce or Arts suiting his interest.

Takshila Learning is providing NCERT Solutions for Class 9 Science as per the latest syllabus by CBSE. Class 9 is the building block for the CBSE Class 10 Board Exams, not only for your exams but also for your higher studies and career. Science is the most essential subject and the knowledge in this field opens up wider career opportunities for the students.

Below you can find the NCERT solution for Class 9th Science. You can get a Solution for the all-important question of Class 9 Science, Chapter 8: Motion

Chapter 8 Motion

Important Formulae

(1) 1st equation formula -v=u+at

(2) 2nd equation formula -s=ut+1/2at^2

(3)3rd equation formula -v^2=v^2+at

(4)momentum=mass × velocity

(5)force=mass ×acceleration

(6) acceleration =change in velocity /time take

(7)speed =distance /time

(8)velocity =displacement /time

 

Pg 100

Q1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

 

Solution

Yes. An object that has moved through a distance can have zero displacement. Displacement is that the shortest measurable distance between the initial and therefore the final position of an object. An object which has covered a distance can have zero displacement, if it comes back to its place to begin, i.e., the initial position.

 

Consider the subsequent situation. a person is walking in an exceedingly square park of length 20 m (as shown within the following figure). He starts walking from point A and after moving along all the corners of the park (point B, C, D), he again comes back to the identical point, i.e.,

A.

 

Here, the total distance covered by the man is

20 m + 20 m + 20 m + 20 m = 80 m.

However, here displacement is zero as the shortest distance between his initial and final position is zero.

 

Q2. 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?

 

Solution

The farmer takes 40 s to cover 4 × 10 = 40 m.

In 2 min and 20 s (140 s), he will cover a distance

 

40/40 X 140 = 140 m.

 

Therefore, the farmer completes 140/3 = 3.5  rounds (3 complete rounds and a half round) of the field in 2 min and 20 s.

 

That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point.

 

Now, there can be two extreme cases.

 

Case I: Starting point is a corner point of the field.

 

In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s.

 

Therefore, the displacement will be equal to the diagonal of the field.

 

Hence, the displacement will be

 

Case II: Starting point is the middle point of any side of the field.

 

In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s.

 

Therefore, the displacement will be equal to the side of the field, i.e., 10 m.

 

For any other starting point, the displacement will be between 14.1 m and 10 m.

 

Q3.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.

 

Solution

(a) Not true

 

Displacement can become zero when the initial and final position of the object is the same.

 

(b) Not true

 

Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

 

Pg 102

 

Q1.Distinguish between speed and velocity.

 

Solution

Speed	Velocity
Speed is the distance travelled by an object in a given interval of time.	Velocity is the displacement shown by an object in a given interval of time.
Speed = Distance / Time	Velocity = Displacement / Time
Speed is scalar quantity and has only magnitude.	Velocity is vector quantity and has both magnitude and direction.
The speed of an object can never be negative. At the most, it can become zero. This is because distance travelled can never be negative.	The velocity of an object can be negative, positive, or equal to zero. This is because displacement can take any of these three values.

 

Q2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Solution

Average velocity = displacement/ total time taken

If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity

 

Q3.What does the odometer of an automobile measure

Solution

The odometer of an automobile measures the distance covered by an automobile.

 

Q4.What does the path of an object look like when it is in uniform motion?

Solution

An object having uniform motion has a straight line path.

 

Q5. 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.

Solution

Time taken by the signal to reach the ground station from the spaceship

= 5 min = 5 × 60 = 300 s

 

Speed of the signal = 3 × 10⁸ m/s

 

∴Distance travelled = Speed × Time taken = 3 × 10⁸ × 300 = 9 × 10¹⁰ m

 

Hence, the distance of the spaceship from the ground station is 9 × 10¹⁰ m.

 

Pg 103

 

Q1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Solution

(i) A body is alleged to possess uniform acceleration if it travels during a straight path in such some way that its velocity changes at a homogenous rate, i.e., the speed of a body increases or decreases by equal amounts in an equal interval of your time.

 

(ii) A body is alleged to possess non-uniform acceleration if it travels during a straight path in such some way that its velocity changes at a non-uniform rate, i.e., the speed of a body increases or decreases in unequal amounts in an equal interval of your time.

Q2. A bus decreases its speed from 80 km h−1 to 60 km h−1 in 5 s. Find the acceleration of the bus.

Solution

Initial speed of the bus, u = 80 km/h

 

= 80 X 5/18 = 22.22m/s

 

Final speed of the bus, v = 60 km/h

 

= 60 X 5/18 = 16.66 m/s

 

Time take to decrease the speed, t = 5 s

 

Acceleration = (v – u)/t = (16.66 – 22.22)/5 = – 1.112 m/s²

 

Here, the negative sign of acceleration indicates that the velocity of the car is decreasing.

 

Pg 107

 

Q1. What is the nature of the distance−time graphs for uniform and non-uniform motion of an object?

 

Solution:

The distance−time graph for uniform motion of an object is a straight line (as shown in the following figure).

 

 

The distance−time graph for non-uniform motion of an object is a curved line (as shown in the given figure).

 

 

 

Q2. What can you say about the motion of an object whose distance−time graph is a straight line parallel to the time axis?

 

Solution:

When an object is at rest, its distance−time graph is a straight line parallel to the time axis.

 

 

 

A straight line parallel to the x-axis in a distance−time graph indicates that with a change in time, there is no change in the position of the object. Thus, the object is at rest.

 

Q3. What can you say about the motion of an object if its speed−time graph is a straight line parallel to the time axis?

 

Solution:

Object is moving uniformly.

 

 

 

A straight line parallel to the time axis in a speed−time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object.

 

Q4. What is the quantity which is measured by the area occupied below the velocity−time graph?

 

Solution:

The graph shows the velocity−time graph of a uniformly moving body.

 

Let the velocity of the body at time (t) be v.

 

Area of the shaded region = length × breath

 

Where,

 

Length = t

 

Breath = v

 

Area = vt = velocity × time …(i)

 

We know,

 

Velocity = Displacement / Time

 

∴ Displacement = Velocity × Time…(ii)

 

From equations (i) and (ii),

 

Area = Displacement

 

Hence, the area occupied below the velocity−time graph measures the displacement covered by the body.

 

Pg 109

 

Q1. 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.

 

Solution:

Initial speed of the bus, u= 0 m/s

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

s is the distance covered by the bus

(12)² – (0)²= 2(0.1) s

s = 720 m

Speed acquired finally by the bus is 12 m/s.

Distance travelled by the bus is 720 m.

 

Q2. A train is traveling 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.

 

Solution:

Initial speed of the train, u= 90 km/h = 25 m/s (1km/hr = 5/18 m/s)

Final speed of the train, v = 0 (finally the train comes to rest and its velocity becomes 0)

Acceleration = – 0.5 m s^-2

According to third equation of motion:

v²= u² + 2as

(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 coming to rest.

 

Q3. 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?

 

Solution:

Initial velocity of the trolley, u = 0 (since the trolley was initially at rest)

 

Acceleration, a = 2 cm s⁻² = 0.02 m/s²

 

Time, t = 3 s

 

According to the first equation of motion:

 

v = u + at

 

Where, v is the velocity of the trolley after 3 s from start

 

v = 0 + 0.02 × 3 = 0.06 m/s

 

Hence, the velocity of the trolley after 3 s from start is 0.06 m/s.

 

Q4. A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start?

 

Solution:

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

 

Q5. 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?

 

Solution:

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

 

 

Pg 112

 

Q1. 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?

 

Solution:

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) = 2π r

= 2 * ( 22 / 7 ) * 100

Speed of the athlete (v) = Distance / Time

= (2 x 2200) / (7 x 40)

= 4400 / 7 × 40

So, 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 rounds in 40 s =1 round

Number of rounds in 140 s =140/40

=3 ½

After taking start from position X, the athlete will be at Y after 3 ½ rounds as shown in figure

 

Therefore, Displacement of the athlete with respect to initial position at x= xy

= Diameter of circular track

= 200 m

 

Q2. 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?

 

Solution:

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

 

Q3.  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?

 

Solution:

The distance while driving from Home to School = S

Assume time taken by Abdul to cover this distance = t1

Distance Abdul covers while driving from School to Home = S

Assume time taken by Abdul to cover 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 Time taken from Home to School t1 =S / v₁av

Similarly, Time taken from 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 (V_av) for covering total distance (2S) =

Total Distance/Total Time

= 2S / (S/20 +S/30)

= 2S / [(30S+20S)/600]

= 1200S / 50S

= 24 kmh-1

 

Q4.  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?

 

Solution

Given Initial velocity of motorboat, u = 0

Acceleration of motorboat, a = 3.0 m s-2

Time under consideration, t = 8.0 s

Distance, s = ut + (1/2)at2

Therefore, 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

 

Q5.  A driver of a car 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 cars travelled farther after the brakes were applied?

 

Solution:

As given in the figure below PR and SQ is the Speed-time graph for given two cars with initial speeds 52 km/hr and 3 km/hr respectively.

Distance Travelled by first car before coming to rest =Area of Δ 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 Δ 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

 

Q6.  Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answers the following questions:

 

(a) Which of the three is traveling 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?

 

Solution:

 

(a) Object B

(b) No

(c) 5.714 km

(d) 5.143 km

 

Therefore, Speed = slope of the graph

as 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

 

Q7.  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?

 

Solution:

Assume, the final velocity with which the 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

 

Q8.  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?

 

Solution:

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 color between time 6 s to 10 s represents uniform motion of the car.

 

Q10.  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 acceleration in the perpendicular direction.

 

Solution:

(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.

 

Q11.  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.

Solution:

Radius of circular orbit, r= 42250 km

Time taken to revolve around the earth, t= 24 h

Speed of a circular moving object, v= (2π 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

 

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