NCERT Most Important Questions For Class-9 Chapter-8 Motion (Physics)
The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.
1.Give an example of a body which may appear to be moving for one person and stationary for the other.
Answer
The passengers in a moving bus observe that the trees, buildings as well as the people on the roadside appear to be moving backwards. Similarly, a person standing on the roadside observes that the bus (along with its passengers) is moving in forward direction. But, at the same time, each passenger in a moving bus or train observes, his fellow passengers sitting and not moving. Thus, we can tell that motion is relative.
2.How can we describe the location of an object?
Answer
To describe the position of an object we need to specify a reference point called the origin. For example, suppose that a library in a city is 2 km north of the railway station. We have specified the position of the library with respect to the railway station i.e., in this case, the railway station acts as the reference point.
3.What do you mean by average speed? What are its units?
Answer
Average speed is defined as the average distance travelled per unit time and is obtained by dividing the total distance travelled by the total time taken. The unit of average speed is the same as that of the speed, that is, ms-1.
4.What is the difference between uniform velocity and non-uniform velocity?
Answer
Uniform velocity: An object with uniform velocity covers equal distances in equal intervals of time in a specified direction, e.g., an object moving with speed of 40km h-1 towards west has uniform velocity.
Non-uniform velocity: When an object covers unequal distances in equal intervals of time in a specified direction, or if the direction of motion changes, it is said to be moving with a non-uniform or variable velocity, e.g., revolving fan at a constant speed has variable velocity.
5.Differentiate between distance and displacement.
Answer
|
Distance
|
Displacement
|
| It is the length of the actual path covered by an object, irrespective of its dirction of motion. | Displacement is the shortest distance between the initial and final positions of an object in a given direction. |
| Distance is a scalar quantity | Displacement is a vector quantity. Displacement may be positive negative or zero. |
| Distance between two given points may be same or different for different path chosen. | Displacement between two given points is always the same. |
| Distance covered can never be negative. It is always positive or zero. | Displacement between two given points is always the same. |
6.With the help of a graph, derive the relation v = u + at.
Answer
Consider the velocity-time graph of an object that moves under uniform acceleration as shown in the figure (u≠0).
From this graph, we can see that initial velocity of the object (at point A) is u and then it increases to v (at point B) in time t. The velocity changes at uniform rate a. As shown in the figure, the lines BC and BE are drawn from point B on the time and the velocity axes respectively.
The initial velocity is represented by OA.
The final velocity is represented by BC.
The time interval t is represented by OC.
BD = BC – CD, represents the change in velocity in time interval t.
If we draw AD parallel to OC, we observe that BC = BD + DC = BD + OA
Substituting, BC with v and OA with u, we get
v = BD + u
or, BD = v – u — (i)
Thus, from the given velocity-time graph, the acceleration of the object is given by Change in velocity
a = (Change in velocity)/(Time Taken)= BD/AD= BD/OC
Substituting OC with t, we get
a = BD/t ⇒ BD = at — (ii2)
From equations (1) and (2), we have
v-u = at or v =u + at
7.Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. What is the net displacement and distance travelled by the ball?
Answer
Displacement is zero. Distance is twice the distance between position P and Q.
8.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 × 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 × 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.
9.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. |
10.What does the odometer of an automobile measure?
Answer
The odometer of an automobile measures the distance covered by an automobile.
11. 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
Answer
When the motion is uniform,the distance time graph is a straight line with a slope.
Answer
Initial speed of the bus, u= 0
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.
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
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) = 2π r
= 2 × ( 22 / 7 ) × 100
Speed of the athlete (v) = Distance / Time
= (2 × 2200) / (7 × 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 × 40) × (2 × 60 + 20)
= 4400 / (7 × 40) × 140
= 4400 × 140 /7 × 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
= 200 m
Answer
Total Distance covered from AB = 300 m
Total time taken = 2 × 60 + 30 s
=150 s
=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 × 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
Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s-1
= 0.952 m s-1
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 Distance/Total Time
= 2S / (S/20 +S/30)
= 2S / [(30S+20S)/600]
= 1200S / 50S
= 24 kmh-1
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.
= (1/2) × OR × OP
= (1/2) × 5s × 52 kmh−1
= (1/2) × 5 × (52 × 1000) / 3600) m
= (1/2) × 5 × (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) × OQ × OS
= (1/2) × 10 s × 3 kmh−1
= (1/2) × 10 × (3 × 1000) / 3600) m
= (1/2) × 10 x (5/6) m
= 5 × (5/6) m
= 25/6 m
= 4.16 m
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 = (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
NCERT Quick Revision Notes OF Chapter-8 Motion
NCERT Solution OF Chapter-8 Motion
NCERT MCQs OF Chapter-8 Motion
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