Class - 9 Science (Physics) Chapter - 8 Motion Notes, NCERT Solutions & Frequently Asked Questions

 Class - 9

Science (Physics) 

Chapter - 8

Motion 

Notes, NCERT Solutions & Frequently Asked Questions 

-- Notes --

• A Reference Point is used to describe the location of an object. An object can be referred to through many reference points.
• Origin – The reference point that is used to describe the location of an object is called Origin.
• For Example, a new restaurant is opening shortly at a distance of 5 km north of my house. Here, the house is the reference point that is used for describing where the restaurant is located.

 

What is motion?

If the location of an object changes with time the object is said to be in motion.

 

Motion in a Straight Line

⏩Distance – The total path length covered by an object between two endpoints.

➧ Distance is a numerical quantity and the direction in which an object is travelling is not mentioned. 

⏩Displacement – The shortest possible distance between the initial and final position of an object is called Displacement.

➧ Displacement depends upon the direction in which the object is travelling.

 Zero Displacement – When the first and last positions of an object are the same, the displacement is zero.

 

Negative Displacement and Positive Displacement

Figure 1 – Example for negative and positive displacement

Here, displacement of object B is negative

B = Bf − Bi = 7–12 = – 5

A negative sign indicates the opposite direction.

Also, displacement of object A is positive

A = Af − Ai = 7– 0 = 7

 

What are Scalar and Vector Quantities?

• A scalar quantity describes a magnitude or a numerical value.
• A vector quantity describes the magnitude as well as the direction.
• Hence, distance is a scalar quantity while displacement is a vector quantity.

 

How is distance different from displacement?

Distance	Displacement
Distance provides the complete details of the path taken by the object	Displacement does not provide the complete details of the path taken by the object
Distance is always positive	Displacement can be positive, negative or zero
It is a scalar quantity	It is a vector quantity
The distance between two points may not be unique	Displacement between two points is always unique

What is uniform motion?

When an object travels equal distances in equal intervals of time the object is said to have a uniform motion.

 

What is non-uniform motion?

When an object travels unequal distances in equal intervals of time the object is said to have a non-uniform motion.

• The speed of an object is defined as the distance travelled by the object per unit of time. 

SI Unit: Metre (m) Symbol of Representation: m/s or ms-¹ Speed = Distance/Time

• Average Speed – If the motion of the object is non-uniform then we calculate the average speed to signify the rate of motion of that object.

For Example, If an object travels 10m in 3 seconds and 12m in 7 seconds. Then its average speed would be:

Total distance travelled = 10 m + 12 m = 22m

Total Time taken = 3s + 7s = 10s

Average speed = 22/10 = 2.2 m/s

• To describe the rate of motion in a direction the term velocity is used. It is defined as the speed of an object in a particular direction.

Velocity = Displacement/Time SI Unit: Metre (m) Symbol of Representation: M/s or ms-¹

Average Velocity (in case of uniform motion)-

Average Velocity = (Initial Velocity + Final Velocity)/2

Average Velocity (in case of non-uniform motion)-

Average Velocity = Total Displacement / Total Time taken

 

What are instantaneous speed and instantaneous velocity?

The magnitude of speed or velocity at a particular instance of time is called Instantaneous Speed or Velocity.

Figure 2 - Instantaneous Speed / Velocity 

Uniform Motion – In the case of uniform motion the velocity of an object remains constant with change in time. Hence, the rate of change of velocity is said to be zero.

Non-uniform Motion – In the case of non-uniform motion the velocity of an object changes with time. This rate of change of velocity per unit time is called Acceleration.

Acceleration = Change in velocity/ Time taken SI Unit: m/s²

Uniform Acceleration – An object is said to have a uniform acceleration if

• It travels along a straight path 
• Its velocity changes (increases or decreases) by equal amounts in equal time intervals

Non - Uniform Acceleration – An object is said to have a non-uniform acceleration if

• Its velocity changes (increases or decreases) by unequal amounts in unequal time intervals

Acceleration is also a vector quantity. The direction of acceleration is the same if the velocity is increasing in the same direction. Such acceleration is called Positive Acceleration.

The direction of acceleration becomes opposite to that of velocity if velocity is decreasing in a direction. Such acceleration is called Negative Acceleration.

De-acceleration or Retardation – Negative acceleration is also called De-acceleration or Retardation

 

Graphical Representation of Motion

1. Distance – Time Graph

 It represents a change in position of the object with respect to time.

The graph in case the object is stationary (means the distance is constant at all time intervals) – Straight line graph parallel to x = axis

Figure 3- Distance-time Graph in case of Stationary object

The graph in case of uniform motion – Straight line graph

Figure 4- Distance-time Graph in Uniform Motion

The graph in case of non-uniform motion – Graph has different shapes

Figure 5 - Distance-time Graph in Non-Uniform Motion 

 

2. Velocity – Time Graphs

Constant velocity – Straight line graph, velocity is always parallel to the x-axis

Uniform Velocity / Uniform Acceleration – Straight line graph

Non-Uniform Velocity / Non-Uniform Acceleration – Graph can have different shapes

Calculating Displacement from a Velocity-time Graph

Consider the graph given below. The area under the graph gives the distance travelled between a certain interval of time. Hence, if we want to find out the distance travelled between time intervals t1 and t2, we need to calculate the area enclosed by the rectangle ABCD where the area (ABCD) = AB * AC.

Similarly, to calculate distance travelled in a time interval in the case of uniform acceleration, we need to find out the area under the graph, as shown in the figure below.

To calculate the distance between time intervals t1 and t2 we need to find out the area represented by ABED.

Area of ABED = Area of the rectangle ABCD + Area of the triangle ADE = AB × BC + 1/ 2 * (AD × DE)

 

Equations of Motion

The equations of motion represent the relationship between an object's acceleration, velocity and distance covered if and only if,

• The object is moving on a straight path
• The object has a uniform acceleration

 

Three Equations of Motion

1. The Equation for Velocity – Time Relation

v = u + at

2. The Equation for Position – Time Relation 

s = ut + 1/2 at²

3. The Equation for the Position – Velocity Relation

2as = v² – u²

Where

u: initial velocity

a: uniform acceleration

t: time

v: final velocity

s: distance travelled in time t

 

Deriving the Equations of Motion Graphically

Figure 10

Study the graph above. The line segment PN shows the relation between velocity and time. 

Initial velocity, u can be derived from velocity at point P or by the line segment OP

Final velocity, v can be derived from velocity at point N or by the line segment NR

Also, NQ = NR – PO = v – u

Time interval, t is represented by OR, where OR = PQ = MN

 

1. Deriving the Equation for Velocity – Time Relation

Acceleration = Change in velocity / time taken

Acceleration = (final velocity – initial velocity) / time

a = (v – u)/t

so, at = v – u

v = u + at

 

2. Deriving Equation for Position – Time Relation

We know that, distance travelled by an object = Area under the graph

So, Distance travelled = Area of OPNR = Area of rectangle OPQR + Area of triangle PQN

s = (OP * OR) + (PQ * QN) / 2

s = (u * t) + (t * (v – u) / 2)

s = ut + 1/2 at²           [because at = v – u]

 

3. Deriving the Equation for Position – Velocity Relation

We know that, distance travelled by an object = area under the graph

So, s = Area of OPNR = (Sum of parallel sides * height) / 2

 s = ((PO + NR)* PQ)/ 2 = ( (v+u) * t)/ 2

2s / (v+u) = t [equation 1]

Also, we know that, (v – u)/ a = t [equation 2]

On equating equations 1 and 2, we get,

2s / (v + u) = (v – u)/ a

2as = (v + u) (v – u) 

2as = v² – u²

 

Uniform Circular Motion

If an object moves in a constant velocity along a circular path, the change in velocity occurs due to the change in direction. Therefore, this is an accelerated motion. Consider the figure given below and observe how the directions of an object vary at different locations on a circular path.

Uniform Circular Motion – When an object travels in a circular path at a uniform speed the object is said to have a uniform circular motion.

Non-Uniform Circular Motion – When an object travels in a circular path at a non-uniform speed the object is said to have a non-uniform circular motion

Examples of uniform circular motion:

• The motion of a satellite in its orbit
• The motion of planets around the sun

 

Velocity of Uniform Circular Motion

Velocity = Distance/ Time = Circumference of the circle / Time

v = 2πr/ t

where

v: velocity of the object

r: radius of the circular path

t: time taken by the object

----- NCERT Solutions -----

Intext Questions (Page No.- 100)

Question 1.- An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Solution:- Yes, if an object has moved through a distance it can have zero displacement because 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?

Solution:-
Distance covered by farmer in 40 seconds =4 x(10) m= 40 m

Speed of the farmer = distance/time = 40m/40s = 1 m/s.

Total time given = 2 min 20 seconds

= 60+60+20 =140 seconds

Since he completes 1 round of the field in 40 seconds so in he will complete 3 rounds in 120 seconds (2 mins) or 120 m distance is covered in 2minutes. In another 20 seconds will cover another 20 m so total distance covered in 2 min 20 sec

= 120 +20 =140m.

Displacement = √102 + 102  =√200

= 10√ 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 travelled by the object.
Solution:- Both (a) as well as (b) are false with respect to concept of displacement.

                                                      Intext Questions (Page No.- 102)

Question 1.- Distinguish between speed and velocity.
Solution:- 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 2.- Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Solution:- If distance travelled by an object is equal to its displacement then the magnitude of average velocity of an object will be equal to its average speed.

Question 3.- What does the odometer of an automobile measure?
Solution:- The odometer of an automobile measures the distance covered by that automobile.

Question 4.- What does the path of an object look like when it is in uniform motion?
Solution:- Graphically the path of an object will be linear i.e. look like a straight line when it is in uniform motion.

Question 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 x108ms-1
Solution:-

Distance = Speed x time

=

=

                                                       Intext Questions (Page No.- 103)

Question 1.- When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?
Solution:-

(i) uniform acceleration: When an object travels in a straight line and its velocity changes by equal amount in equal intervals of time, it is said to have uniform acceleration.

(ii) non uniform 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 2.- A bus decreases its speed from 80kmh-1 to 60km h-1 in 5 s. Find the acceleration of the bus.
Solution:-
Initial speed of bus (u) =
== = 

final speed of bus (v)==
=
=time (t) = 5 s

acceleration (a) = ( v – u) /t = (16.67 – 22.22)/5 = -5.55/5 = 

Question 3.- A train starting from a railway station and moving with uniform acceleration attains a speed 40km h-1 in 10 minutes. Find its acceleration.
Solution:- Since the train starts from rest(railway station) = u = zero

Final velocity of train
=

time (t) = 10 min = 10 x 60

= 600 seconds

Since a = (v – u )/t = 

= 

                                        Intext Questions (Page No.- 107)

Question 1.- What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Solution:- If an object has a uniform motion then the nature of distance time graph will be linear i.e. it would a straight line and if it has non uniform motion then the nature of distance time graph is a curved line.

Question 2.- What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Solution:- 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 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?
Solution:- Such a graph indicates that the object is travelling with uniform velocity.

Question 4.- What is the quantity which is measured by the area occupied below the velocity-time graph?
Solution:- The area occupied below the velocity-time graph measures the distance moved by any object.

                                        Intext Questions (Page No.- 109)

Question 1.- A bus starting from rest moves with a uniform acceleration of 0.1m s-2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled.
Solution:-

(a) u=o, , t= 2min = 120 seconds.

v=u+at 

so (a) speed acquired
= 

(b) 

= 720 m.

Question 2.- A train is travelling at a speed of 90 kmh-1. Brakes are applied so as to produce a uniform acceleration of -0.5m s-2. Find how far the train will go before it is brought to rest.
Solution:-

,

v =0(train is brought to rest)

v= u+at 

0 =25 – 0.5 x

0.5t = 25, or t
= 25/0.5
= 50 seconds

= 1250 – 625 = 625 m

Question 3.- A trolley, while going down an inclined plane, has an acceleration of 2 cms-2. What will be its velocity 3 s after the start?
Solution:-

, t= 3 s

v= u + at = 0 + 2 x 3 = 6 cm/s

Question 4.- A racing car has a uniform acceleration of 4 cms-2. What distance will it cover in 10 s after start?
Solution:-

u = 0, , t= 10 s

= 200 m

Question 5.- A stone is thrown in a vertically upward direction with a velocity of 5 cms-2. If the acceleration of the stone during its motion is 10 cms-2 the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:-

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 =0.5second.

= 2.5 – 1.25 = 1.25 m

----- Exercise -----

Question 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?

Solution:- circumference of circular track = 2πr

= 4400/7 m

rounds completed by athlete in
2 min 20 sec = s= 140/40 = 3.5

therefore, total distance covered =4400 / 7 x 3.5= 2200 m

Since one complete round of circular track needs 40s so he will complete 3 rounds in 2mins and in next 20s he can complete half round therefore displacement = diameter = 200m.

Question 2-. 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?
Solution:-
(a) distance = 300m

time = 2 min 30 seconds = 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

= 400m/210s = 1.90m/s

displacement from A to C

= (300 – 100)m =200m

time =2min30sec + 1min = 210s

velocity = displacement/time
= 200m/210s = 0.95m/s

Question 3.- Abdul, while driving to school, computes the average speed for his trip to be 20 kmh-1. On his return trip along the same route, there is less traffic and the average speed is 40 kmh-1. What is the average speed for Abdul’s trip?
Solution:-

If we suppose that distance from Abdul’s home to school = x kms

while driving to school :-

,

velocity = displacement/time

20 = x/t, or, t=x/20 hr

on his return trip :-

speed = 40 kmh-1,

40= x /t’

or, t’ =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 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?
Solution:- since the motorboat starts from rest so u= 0

time (t) = 8s, 

distance

= 96m

Question 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?
Solution:-

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 respectively.

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

= 325/9 m

= 36.11 m

Distance Travelled by second car before coming to rest =Area of  

= 25/6 m

= 4.16 m

∴ Clearly the first car will travel farther (36.11 m) than the first car(4.16 m).

Question 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?

Solution:-

(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 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 when passes to C.

Question 7.- A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10m s-2 , with what velocity will it strike the ground? After what time will it strike the ground?
Solution:-
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=20m

Downward acceleration a

As we know,

Or, 2as = 

∴ Final velocity of ball, v 

t= (v-u)/a

∴ Time taken by the ball to strike= (20-0)/10

= 20/10

= 2 seconds

Question 8.- The speed-time graph for a car is shown is Fig. 8.12.

(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?

Solution:-

(a) Distance travelled by car in the 4 second

The area under the slope of the speed – time graph gives the distance travelled by an object.

In the given graph

56 full squares and 12 half squares come under the area slope for the time of 4 second.

Total number of squares = 56 + 12/2 = 62 squares

The total area of the squares will give the distance travelled by the car in 4 second. on the time axis,

5 squares = 2seconds, therefore 1 square = 2/5 seconds

on speed axis there are 3 squares = 2m/s

therefore, area of one square

= = 4/15 m

so area of 62 squares= 

= 248/15 m = 16.53 m

Hence the car travels 16.53m in the first 4 seconds.

(b) The straight line part of graph, from point A to point B represents a uniform motion of car.

Question 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.

Solution:-

a) An object with a constant acceleration can still have the zero velocity. For example an object which 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 contant 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 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.
Solution:- 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.

= 42250 km

Radius 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 =Circumference of circular orbit

= 2πr

∴ Speed of satellite v = (2πr) / t

= 3.073 km/s

Give Us Your Feedback/Suggestions

- By Durgesh Pandey

(Eklavya Coaching Institute)

📞 8376976688, 9310533915

H-2/25, Gali No-23, Kunwar Singh Nagar, Nangloi, New Delhi -110041