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Access Answers of Science NCERT class 9 Chapter 8: Motion (All intext and exercise questions solved)
Intext Questions – 1 Page: 100
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Yes, an object moving a certain distance can have zero total displacement. Displacement refers to the shortest distance between the initial and the final positions of the object. Even if an object moves through a considerable distance, if it eventually comes back to its initial position, the corresponding displacement of the object would be zero.
2. A farmer moves along the boundary of a square field of side 10m 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?
Given that the farmer covers the entire boundary of the square field in 40 seconds, the total distance traveled by the farmer in 40 seconds is 4*(10) = 40 meters.
Therefore, the average distance covered by the farmer in one second is: 40m/40 = 1m
Two minutes and 20 seconds can be written as 140 seconds. The total distance traveled by the farmer in this timeframe is: 1 m * 140 = 140m
Since the farmer is moving along the boundary of the square field, the total number of laps completed by the farmer will be: 140m/40 = 3.5 laps
Now, the total displacement of the farmer depends on the initial position. If the initial position of the farmer is at one corner of the field, the terminal position would be at the opposite corner (since the field is square).
In this case, the total displacement of the farmer will be equal to the length of the diagonal line across the opposite corners of the square.
Applying the Pythagoras theorem, the length of the diagonal can be obtained as follows: √(102+102)= √200= 14.14m.
This is the maximum possible displacement of the farmer.
If the initial position of the farmer is at the mid-point between two adjacent corners of the square, the net displacement of the farmer would be equal to the side of the square, which is 10m. This is the minimum displacement.
If the farmer starts at a random point around the perimeter of the square, his net displacement after traveling 140m will lie between 10m and 14.14m.
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.
Neither of the statements are true. Statement (a) is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero. Statement (b) is false because the displacement of an object can be equal to, but never greater than the distance traveled.
Intext Questions – 2 Page: 102
1. Distinguish between speed and velocity.
|Difference Between Speed and Velocity|
|It refers to the displacement of a given object over a time interval.||It refers to the distance moved by an object over a time interval.|
|It has a specific direction||It does not have any direction.|
|Velocity = displacement/time||Speed = distance / time|
|Velocity can hold a negative value||Speed cannot hold a negative value.|
2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Since average speed is the total distance traveled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance traveled is equal to the displacement.
3. What does the odometer of an automobile measure?
The odometer measures the total distance traveled by the automobile.
4. What does the path of an object look like when it is in uniform motion?
The path of an object in uniform motion is a straight line.
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.
Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance traveled by the signal.
5 minutes = 5*60 seconds = 300 seconds.
Speed of the signal = 3 × 108 m/s.
Therefore, total distance = (3 × 108 m/s) * 300s
= 9*1010 meters.
Intext Questions – 3 Page: 103
1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Uniform Acceleration: In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a uniform rate (it changes at a constant rate in any constant time interval).
Non-Uniform Acceleration: In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a rate that is not uniform (it changes at a different rate for a given constant time interval).
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.
Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s-1
The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s-1
Time frame, t = 5 seconds.
Therefore, acceleration (a) =(v-u)/t = (16.66 m.s-1 – 22.22 m.s-1)/5s
= -1.112 m.s-2
Therefore, the total acceleration of the bus is -1.112m.s-2. It can be noted that the negative sign indicates that the velocity of the bus is decreasing.
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.
Given, the initial velocity (u) of the train = 0m.s-1 (at rest)
Terminal velocity (v) of the train = 40km/hour = 11.11 m.s-1
Time interval, t = 10 minutes = 600 s.
The acceleration of the train is given by =(v-u)/t = (11.11 m.s-1 – 0 m.s-1)/600s = 0.0185 m.s-2
Intext Questions – 4 Page: 107
1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.
The first graph describes uniform motion and the second one describes non-uniform motion.
2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
This distance-time graph can be plotted as follows.
Since there is no change in the distance traveled by the object (or the Y-Axis value) at any point in the X-Axis (time), the object 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?
This speed-time graph can be plotted as follows.
Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.
4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Considering an object in uniform motion, its velocity-time graph can be represented as follows.
Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:
Area under the velocity-time graph = velocity*time.
Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.
Intext Questions – 5 Page: 109,110
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.
(a) Given, the bust starts from rest. Therefore, initial velocity (u) = 0 m/s
Acceleration (a) = 0.1 m.s-2
Time = 2 minutes = 120 s
Acceleration is given by the equation a=(v-u)/t
Therefore, terminal velocity (v) = (at)+u
= (0.1 m.s-2 * 120s) + 0 m.s-1
= 12m.s-1 + 0 m.s-1
Therefore, terminal velocity (v) = 12m/s
(b) As per the third motion equation,
Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120s, the following value for s (distance) can be obtained.
Distance, s =(v2 – u2)/2a
=(122 – 02)/2(0.1)
Therefore, s = 720m.
The speed acquired is 12m.s-1 and the total distance traveled is 720m.
2. 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.
Given, initial velocity (u) = 90 km/hour = 25 m.s-1
Terminal velocity (v) = 0 m.s-1
Acceleration (a) = -0.5 m.s-2
As per the third motion equation, v2-u2=2as
Therefore, distance traveled by the train (s) =(v2-u2)/2a
s = (02-252)/2(-0.5) meters = 625 meters
The train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.
3. 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?
Given, initial velocity (u) = 0 (the trolley begins from the rest position)
Acceleration (a) = 0.02 ms-2
Time (t) = 3s
As per the first motion equation, v=u+at
Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms-2)(3s)= 0.06 ms-2
Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s-2
4. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Given, the car is initially at rest; initial velocity (u) = 0 ms-1
Acceleration (a) = 4 ms-2
Time period (t) = 10 s
As per the second motion equation, s = ut+1/2 at2
Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms-2)(10s)2
= 200 meters
Therefore, the car will cover a distance of 200 meters after 10 seconds.
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?
Given, initial velocity (u) = 5 m/s
Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)
Acceleration = 10 ms-2 in the direction opposite to the trajectory of the stone = -10 ms-2
As per the third motion equation, v2 – u2 = 2as
Therefore, the distance traveled by the stone (s) = (02 – 52)/ 2(10)
Distance (s) = 1.25 meters
As per the first motion equation, v = u + at
Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a
Time taken = 0.5 seconds
Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.
Exercises Page: 112,113
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?
Given, diameter of the track (d) = 200m
Therefore, circumference of the track (π*d) = 200π meters.
Distance covered in 40 seconds = 200π meters
Distance covered in 1 second = 200π/40
Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters
= (140*200*22)/(40* 7) meters = 2200 meters
Number of laps completed by the athlete in 140 seconds = 140/40 = 3.5
Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.
Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.
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?
Given, distance covered from point A to point B = 300 meters
Distance covered from point A to point C = 300m + 100m = 400 meters
Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds
Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds
Displacement from A to B = 300 meters
Displacement from A to C = 300m – 100m = 200 meters
Average speed = total distance travelled/ total time taken
Average velocity = total displacement/ total time taken
Therefore, the average speed while traveling from A to B = 300/150 ms-1 = 2 m/s
Average speed while traveling from A to C = 400/210 ms-1= 1.9 m/s
Average velocity while traveling from A to B =300/150 ms-1= 2 m/s
Average velocity while traveling from A to C =200/210 ms-1= 0.95 m/s
3. 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 30 km.h–1. What is the average speed for Abdul’s trip?
Distance traveled to reach the school = distance traveled to reach home = d (say)
Time taken to reach school = t1
Time taken to reach home = t2
therefore, average speed while going to school = total distance travelled/ total time taken = d/t1 = 20 kmph
Average speed while going home = total distance travelled/ total time taken = d/t2= 30 kmph
Therefore, t1 = d/20 and t2 = d/30
Now, the average speed for the entire trip is given by total distance travelled/ total time taken
= (d+d)/(t1+t2)kmph = (d+d)/(d/20+d/30)kmph
= 120/5 kmh-1 = 24 kmh-1
Therefore, Abduls average speed for the entire trip is 24 kilometers per hour.
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?
Given, initial velocity of the boat = 0 m/s
Acceleration of the boat = 3 ms-2
Time period = 8s
As per the second motion equation, s = ut + 1/2 at2
Therefore, the total distance traveled by boat in 8 seconds = 0 + 1/2 (3)(8)2
= 96 meters
Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.
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?
The speed v/s time graphs for the two cars can be plotted as follows.
The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.
Therefore, displacement of the first car = area of triangle AOB
But OB = 5 seconds and OA = 52 km.h-1 = 14.44 m/s
Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms-1) = 36 meters
Now, the displacement of the second car is given by the area of the triangle COD
But OC = 10 seconds and OC = 3km.h-1 = 0.83 m/s
Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms-1) = 4.15 meters
Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 kmph) traveled farther post the application of brakes.
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?
(a) since the slope of line B is the greatest, B is traveling at the fastest speed.
(b) since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.
(c) since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.
Since the initial point of object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km
When A passes B, the distance between the origin and C is 8km
Therefore, total distance traveled by C in this time = 8 – (16/7) km = 5.71 km
(d) the distance that object B has covered at the point where it passes C is equal to 9 graph units.
Therefore, total distance traveled by B when it crosses C = 9*(4/7) = 5.14 km
7. 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?
Given, initial velocity of the ball (u) = 0 (since it began at the rest position)
Distance traveled by the ball (s) = 20m
Acceleration (a) = 10 ms-2
As per the third motion equation,
= 2*(10ms-2)*(20m) + 0
v2 = 400m2s-2
Therefore, v= 20ms-1
The ball hits the ground with a velocity of 20 meters per second.
As per the first motion equation,
Therefore, t = (v-u)/a
= (20-0)ms-1 / 10ms-2
= 2 seconds
Therefore, the ball reaches the ground after 2 seconds.
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?
The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:
(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.
(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6th to the 10th second.
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 with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.
(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.
(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time
(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.
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.
Given, radius of the orbit = 42250 km
Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km
Time taken for the orbit = 24 hours
Therefore, speed of the satellite = 11065.4 km.h-1
The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.
|NCERT Exemplar Solutions for Class 9 Science Chapter 8|
|CBSE Notes for Class 9 Science Chapter 8|
NCERT Solutions for Class 9 Science Chapter 8: Motion
NCERT Class 9 Science Chapter 8 explains the concept of motion, types of motion with relevant day today examples for the clear understanding of the concept. It explains the cause of the happenings of sunrise, sunset, and changing of the seasons. It helps students understand uniform and non-uniform motion. Distance-time graph and velocity-time graph, which are considered as the important concepts for examination are explained in an easy way in NCERT Solutions. It describes how the acceleration of an object is the change in velocity per unit time. NCERT Class 9 Science Chapter 8 is covered under Unit III: Motion, Force and Work and can get you maximum marks.
- NCERT Solutions for Class 9 explain motion in terms of distance moved or the displacement.
- Uniform and non-uniform motions of objects are explained through the graph and examples.
- Uniform circular motion concept is made understandable in a simple way.
- Problems on acceleration, velocity, and average velocity are also solved.
Key Features of NCERT Solutions for Class 9 Science Chapter 8: Motion;
- The simple and easily understandable approach is followed in NCERT solutions to make students aware of topics.
- Provides completely solved solutions to all the questions present in the respective NCERT textbooks.
- NCERT Solutions offer detailed answers to all the questions to help students in their preparations.
- These solutions will be useful for CBSE board exams, Science Olympiads, and other competitive exams.
Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 8
What will I learn from the Chapter 8 of NCERT Solutions for Class 9 Science?
2. Examples and graphs are provided for uniform and non uniform motions of objects.
3. The concept of uniform circular motion is explained in an understandable way.
4. Numericals based on the average velocity, velocity and acceleration are solved in a stepwise manner.
How will the NCERT Solutions for Class 9 Science Chapter 8 help us to score well in the exam?
2. These solutions contain answers to all the questions present in the NCERT textbook.
3. Detailed answers improve logical and analytical thinking abilities among students.
4. Students can also use these solutions to prepare for various other competitive exams along with the annual exams.
How can I solve the problems from the NCERT Solutions for Class 9 Science Chapter 8 effortlessly?