Motion Class 9 NCERT Solutions – The NCERT Solutions for Class 9 Science help students prepare effectively for their examinations by providing them with a comprehensive set of solutions. These solutions serve as a vital resource for students who may find it difficult to locate the answers to the questions on their own. Students can use these solutions to build a solid foundation in Science and achieve academic success.
Motion Class 9 NCERT Solutions
Q. A distance has been traversed by an object. Can it have zero displacement? If yes, give an example to prove up your assertion.
Answer – Yes, an object can have zero displacement even after moving through a distance.
Displacement is defined as the shortest distance between the initial and final positions of an object. It takes into account both the distance traveled by the object as well as the direction of the motion.
Consider an example where a person walks from point A to point B, which are 5 meters apart, and then returns to point A along the same path. In this case, the person has moved through a distance of 10 meters (5 meters from A to B and 5 meters from B to A), but the displacement is zero because the person has ended up at the same position as the starting position.
Q. A farmer walks 40 seconds along the perimeter of a square field with a side of 10 metres. What will the farmer’s displacement from his starting location be after 2 minutes and 20 seconds?
Answer –
Side square field = 10m
perimeter of square = 40m
Time taken to cover the boundary of 40m = 40s
Thus in 1s, the farmer covers a distance of 1m
Now, Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140m
The total number of rotations taken to cover distance
meters = total distance/perimeter
= 3.5
At this point, The farmer is at point B from the origin O
Therefore, the displacement theorem s = √(102+102)
s = 10√2
s = 14.14 m
Q. Which of the following statement is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is larger than the object’s travel distance.
Answer – Neither (a) nor (b) is true for displacement.
(a) is false because an object’s displacement can be zero if it returns to its initial position.
(b) is false because the magnitude of an object’s displacement can be equal to or less than the distance traveled, but never greater than it.
Q. Distinguish between speed and velocity.
Answer –
| Speed | Velocity |
|---|---|
| Scalar quantity | Vector quantity |
| Magnitude only | Magnitude and direction |
| Always positive | Can be positive, negative, or zero |
| Distance per unit time | Displacement per unit time |
| Examples: 50 km/h, 100 m/s | Examples: 50 km/h to the north, 100 m/s east |
Q. What circumstance(s) must exist for an object’s average velocity to be equal to its average speed?
Answer –
Average speed = Total distance covered / Total time taken
Average velocity = Displacement / Total time taken
The magnitude of average velocity of an object is equal to its average speed only when the object moves in a straight line without changing its direction.
Q. What is measured by an automobile’s odometer?
Answer – An automobile’s odometer is a tool that calculates the overall distance the car has covered since it was first used. It usually sits on the dashboard and shows the distance travelled in miles or kilometres. Odometers are crucial instruments for keeping track of a car’s maintenance and figuring out how much it is worth to sell it. To intentionally lower a vehicle’s mileage reading by tampering with the odometer is prohibited in many nations.
Q. What does an object’s route look like when it is moving consistently?
Answer – The term “uniform motion” refers to a straight line motion in which the direction or speed of the item does not change.
Q. A signal from a spacecraft reached the ground station in an experiment in five minutes. How far away from the ground station was the spacecraft? The signal moves at the speed of light, or 3 × 108 metres per second.
Answer –
distance = speed x time
In this case, the speed is the speed of light, which is 3 x 108 m/s, and the time is 5 minutes, or 5 x 60 = 300 seconds.
So, the distance between the spacecraft and the ground station is:
distance = speed x time
distance = 3 x 108 m/s x 300 s
distance = 9 x 1010 meters or 90,000,000,000 meters (in scientific notation)
Therefore, the spaceship was 90,000,000,000 meters or 90 billion meters away from the ground station when the signal was received.
Q. When can you state that a body is
(i) uniform acceleration?
(ii) non-uniform acceleration?
Answer –
Uniform Acceleration : An object is considered to be experiencing uniform acceleration when it is travelling straight and growing in velocity by the same amount at regular intervals. The motion’s acceleration stays constant the entire time. An object falling free is a good illustration of uniform acceleration.
Non – Uniform Acceleration : When an object’s velocity changes over time at different rates, it is said to be suffering non-uniform acceleration. The acceleration is therefore not constant during the motion, as a result. A bus departing a bus stop exhibits non-uniform acceleration since its velocity is varying, but not continuously.
Q. In 5 seconds, a bus may go from 80 km h–1 to 60 km h–1. Determine the bus’s acceleration.
Answer –
To find the acceleration of the bus, we can use the formula:
acceleration = (final velocity – initial velocity) / time
Here, the initial velocity (u) of the bus is 80 km/h, the final velocity (v) is 60 km/h, and the time taken (t) for the bus to change its velocity from u to v is 5 seconds.
First, we need to convert the velocities from km/h to m/s, since acceleration is measured in meters per second squared (m/s2).
So, initial velocity (u) = 80 km/h = 80 x (1000/3600) m/s = 22.22 m/s (rounded to two decimal places)
Final velocity (v) = 60 km/h = 60 x (1000/3600) m/s = 16.67 m/s (rounded to two decimal places)
We can now enter the values into the formula as follows:
acceleration = (final velocity – initial velocity) / time
acceleration = (16.67 m/s – 22.22 m/s) / 5 s
acceleration = -1.112 m/s2 (rounded to two decimal places)
Therefore, the acceleration of the bus is -1.112 m/s2, which means that it is decelerating or slowing down. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity of the bus.
Q. A train can reach a speed of 40 km h–1 in 10 minutes when it departs from a railway station and moves with uniform acceleration. the acceleration of it.
Answer –
First, we need to convert the final velocity from km/h to m/s, since acceleration is measured in meters per second squared (m/s2).
Final velocity (v) = 40 km/h = 40 x (1000/3600) m/s = 11.11 m/s (rounded to two decimal places)
The time taken (t) for the train to reach this velocity is 10 minutes, which is 10 x 60 = 600 seconds.
Now, we can use the formula:
acceleration = (final velocity – initial velocity) / time
Here, the initial velocity (u) of the train is 0 m/s, since it starts from rest.
Substituting the values, we get:
acceleration = (11.11 m/s – 0 m/s) / 600 s
acceleration = 0.0185 m/s2 (rounded to four decimal places)
Therefore, the acceleration of the train is 0.0185 m/s2.
Q. What kind of graphs do uniform and non-uniform motion of an item have in terms of distance and time?
Answer – Non-uniform motion has a curving distance-time graph with a fluctuating slope, whereas uniform motion has a straight line with a constant slope. The object’s acceleration can be calculated from the graph’s curvature. The object is at rest when the graph shows a horizontal line.
Q. What may be said about an object’s motion when the distance-time graph is a straight line that runs parallel to the time axis?
Answer – The distance travelled by an item does not change over time if the object’s distance-time graph is a straight line parallel to the time axis. In other words, the object is not moving, and it is at rest.
Q. If an object’s speed-time graph is a straight line parallel to the time axis, what can you infer about its motion?
Answer – A straight line parallel to the time axis on an object’s speed-time graph indicates that the object is travelling at a constant speed. This is the case because the acceleration of the item is represented by the slope of the speed-time graph, and if the slope is zero (i.e., a horizontal line), then the acceleration is also zero.
Q. What is the value that the space below the velocity-time graph represents?
Answer –
Using the equation Area under the velocity-time graph = velocity*time, where velocity is the speed of the object and time is the duration of its travel, we can calculate the area under the velocity-time graph.
However, since velocity is the rate of change of displacement, we can also express velocity as displacement/time. Substituting this value into the previous equation, we get:
Area under the velocity-time graph = displacement
This means that the area under the velocity-time graph represents the total displacement or distance traveled by the object. Therefore, if we know the velocity-time graph of an object, we can find its displacement by calculating the area under the graph.
Q. A bus that starts off at rest travels for two minutes at a constant acceleration of 0.1 m s-2. Find (a) the speed acquired, (b) the distance travelled.
Answer –
Initial velocity, u = 0 (as the bus is starting from rest)
Acceleration, a = 0.1 m/s²
Time, t = 2 minutes = 120 seconds
We need to find:
(a) The speed acquired by the bus.
(b) The distance traveled by the bus.
Solution:
(a) The final velocity (v) acquired by the bus can be calculated using the formula:
v = u + at
where v is the final velocity, u denotes the starting velocity, a denotes the acceleration, and t denotes the time required.
Substituting the given values, we get:
v = 0 + (0.1 m/s²) x (120 s)
v = 12 m/s
Therefore, the speed acquired by the bus is 12 m/s.
(b) The distance traveled (s) by the bus can be calculated using the formula:
s = ut + (1/2)at²
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.
Substituting the given values, we get:
s = 0 x 120 + (1/2) x (0.1 m/s²) x (120 s)²
s = 720 m
Therefore, the distance traveled by the bus is 720 meters.
Q. 90 kilometres per hour is the speed of a train. The use of brakes results in a uniform acceleration of -–0.5 m s-2. Find out how far the train will go before coming to a stop.
Answer –
Initial velocity, u = 90 km/h = 25 m/s (as 1 km/h = 1000 m/3600 s = 5/18 m/s) Acceleration, a = -0.5 m/s² (negative sign indicates deceleration) We need to find the distance traveled (s) by the train before it comes to rest.
We can use the following formula to find the distance traveled by the train:
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
As the train is brought to rest, the final velocity v = 0.
Substituting the given values, we get:
0 = (25 m/s)² + 2(-0.5 m/s²) s
Simplifying the equation, we get:
s = (25 m/s)² / (2 x 0.5 m/s²) s = 625 m
Therefore, the train will travel a distance of 625 meters before it is brought to rest.
Q. A trolley accelerates by 2 cm s-2 when moving down an inclined plane. What will its speed be three seconds after the start?
Answer –
Acceleration, a = 2 cm/s² Time, t = 3 s Initial velocity, u = 0 (as the trolley starts from rest)
We need to find the final velocity (v) of the trolley after 3 seconds.
We can use the following formula to find the final velocity of the trolley:
v = u + at
where u is the starting speed, v is the end speed, an is the acceleration, and t is the total duration.
Substituting the given values, we get:
v = 0 + (2 cm/s²) x (3 s) v = 6 cm/s
Therefore, the velocity of the trolley after 3 seconds is 6 cm/s.
Q. A race car accelerates uniformly at 4 m s-2. How far will it travel in 10 seconds after starting?
Answer –
Acceleration, a = 4 m/s² Time, t = 10 s Initial velocity, u = 0 (as the racing car starts from rest)
We need to find the distance traveled (s) by the racing car in 10 seconds.
We can use the following formula to find the distance traveled by the racing car:
s = ut + (1/2)at²
where s denotes the distance covered, u the speed at which the journey began, a the acceleration, and t the amount of time required.
Substituting the given values, we get:
s = 0 x 10 + (1/2) x (4 m/s²) x (10 s)² s = 200 m
Therefore, after starting, the racing car will travel 200 metres in 10 seconds.
Q. A stone is thrown with a 5 m s-1 velocity in a vertical upward direction. What height will the stone reach and how long will it take to get there if the acceleration of the stone during its motion is 10 m s–2 in the downward direction?
Answer –
Initial velocity, u = 5 m/s (upward) Acceleration, a = -10 m/s² (downward) We need to find the maximum height attained by the stone (h) and the time taken to reach there (t).
When the stone reaches its maximum height, its final velocity (v) becomes zero. We can use the following equation to find the maximum height reached by the stone:
v² = u² + 2as
where s is the distance travelled, an is the acceleration, v is the final velocity, and u is the initial velocity.
At the maximum height, v = 0, so we can write:
0² = (5 m/s)² + 2(-10 m/s²) s
Simplifying the equation, we get:
s = (5 m/s)² / (2 x 10 m/s²) s = 1.25 m
Therefore, the maximum height reached by the stone is 1.25 meters.
We can use the following equation to find the time taken to reach the maximum height:
v = u + at
where v is the final velocity, u denotes the starting velocity, a denotes the acceleration, and t denotes the time required.
Substituting the given values, we get:
0 = 5 m/s + (-10 m/s²) t
Simplifying the equation, we get:
t = 5/10 s
t = 0.5 s
Therefore, the stone takes 0.5 seconds to reach the maximum height.
Q. A 200 m-diameter circular track is traversed by an athlete in 40 seconds. What will the displacement and distance be after two minutes and twenty seconds?
Answer –
Diameter of circular track, d = 200 m
Radius of circular track, r = d/2 = 100 m
Time taken to complete one round, t = 40 s
We need to find the distance covered and the displacement of the athlete at the end of 2 minutes 20 seconds.
First, let’s calculate the distance covered by the athlete in one round of the circular track. Distance covered in one round = Circumference of the circle
Circumference of the circular track = 2πr
= 2 x 3.14 x 100 m
= 628 m
Therefore, the distance covered by the athlete in one round of the circular track is 628 meters.
Now, let’s calculate the total number of rounds completed by the athlete in 2 minutes 20 seconds.
Total time taken = 2 minutes 20 seconds = 140 seconds
Number of rounds completed = Total time taken / Time taken to complete one round
= 140 seconds / 40 seconds
= 3.5 rounds (rounded to one decimal place)
Therefore, the athlete completes 3.5 rounds of the circular track in 2 minutes 20 seconds.
Now, let’s calculate the distance covered by the athlete in 3.5 rounds of the circular track:
Distance covered = Distance covered in one round x Number of rounds completed
= 628 m x 3.5
= 2198 m
Therefore, the athlete has travelled 2198 metres after 2 minutes and 20 seconds.
Finally, let’s calculate the displacement of the athlete at the end of 2 minutes 20 seconds. Since the athlete starts and ends at the same point, the displacement is zero.
Therefore, the displacement of the athlete at the end of 2 minutes 20 seconds is zero.
Q. In 2 minutes and 30 seconds, Joseph runs a straight 300-meter route from point A to point B. He then turns around and runs the final 100 metres to point C. What are Joseph’s typical running distances and velocities (a) from A to B and (b) from A to C?
Answer –
We can start by using the formulas for speed and velocity:
Speed = distance / time
Velocity = displacement / time
where displacement refers to a change in an object’s position.
(a) From A to B:
Distance = 300 m
Time = 2 minutes 30 seconds = 150 seconds
Speed = Distance / Time
Speed = 300 m / 150 s
Speed = 2 m/s
Note that since Joseph returns to point A, his displacement is zero. Therefore, his velocity from A to B is also zero.
(b) From A to C:
Distance = 300 m + 100 m = 400 m
Time = 2 minutes 30 seconds + 1 minute = 210 seconds
Speed = Distance / Time
Speed = 400 m / 210 s
Speed = 1.9 m/s
Joseph’s displacement from A to C is 100 m (he ends up at point C). Therefore, his velocity is:
Velocity = Displacement / Time
Velocity = 100 m / 210 s
Velocity = 0.48 m/s
So Joseph’s average speed from A to B is 2 m/s and his average velocity is zero. His average speed from A to C is 1.9 m/s and his average velocity is 0.48 m/s.
Q. Abdul calculates his average journey speed while driving to school to be 20 km.h–1. The amount of traffic is lower and the average speed is 30 km.h–1 on his return trip over the same route. What is Abdul’s journey’s average speed?
Answer –
The entire distance travelled divided by the total time taken is the formula we may use to determine the average speed:
Average speed = Total distance / Total time
Let’s assume the distance from Abdul’s home to school is d km. Then, the time taken for Abdul to travel from home to school at an average speed of 20 km/h would be:
Time taken = distance / speed
Time taken = d / 20
Similarly, the time taken for Abdul to travel from school to home at an average speed of 30 km/h would be:
Time taken = distance / speed
Time taken = d / 30
The total distance covered by Abdul is 2d (since he travels from home to school and then back from school to home). The total time taken is the sum of the time taken for the forward and return journeys:
Total time taken = d / 20 + d / 30
We can simplify this expression by finding a common denominator:
Total time taken = (3d / 60) + (2d / 60)
Total time taken = 5d / 60
Total time taken = d / 12
Now we can substitute the values for total distance and total time taken into the formula for average speed:
Average speed = Total distance / Total time taken
Average speed = 2d / (d / 12)
Average speed = 24 km/h
Abdul’s journey will therefore have an average speed of 24 km/h.
Q. A speedboat on a lake begins at rest and accelerates in a straight line for 8.0 seconds while maintaining a constant speed of 3.0 m s–2. During this time, how far does the boat travel?
Answer –
The distance traveled with constant acceleration is:
Distance = initial velocity * time + (1/2) * acceleration * time2
where the initial velocity is the velocity of the object before acceleration begins.
In this case, the boat starts from rest, so its initial velocity is 0 m/s. The boat accelerates for 8.0 s at a steady speed of 3.0 m/s2. Hence, we may add these values to the formula as follows:
Distance = 0 m/s * 8.0 s + (1/2) * 3.0 m/s2 * (8.0 s)2
Distance = 0 + 96.0 m
Distance = 96.0 meters
Therefore, the boat travels 96.0 meters during this time.
Q. A 52 km h–1 vehicle’s driver evenly decelerates in one direction before accelerating in the opposite direction. In 5 seconds, the automobile stops. Another driver in another automobile slows down and stops in 10 seconds while travelling at a speed of 3 km h–1. Plot the speed versus time graphs for the two cars on the same sheet of graph paper. Which of the two cars made a longer distance after using the brakes?
Answer – The equations for the two cars are:
First car: 0 m/s = 14.44 m/s + acceleration x 5 s
Second car: 0 m/s = 0.83 m/s + acceleration x 10 s
The speed versus time graphs for the two cars are:
To determine the distance traveled by each car, use the formula:
distance = (initial speed x time) + (0.5 x acceleration x time2)
For the first car:
distance = (14.44 m/s x 5 s) + (0.5 x -2.888 m/s2 x (5 s)2) = 36.1 m
For the second car:
distance = (0.83 m/s x 10 s) + (0.5 x -0.083 m/s2 x (10 s)2) = 4.17 m
Therefore, the first car traveled farther after the brakes were applied.
Q. From a distance of 20 metres, a ball is gently dropped. What speed will it hit the ground at if its velocity rises uniformly at a rate of 10 m s-2? How long before it hits the ground?
Answer –
Using the kinematic equation:
v2 = u2 + 2as
where:
v = final velocity of the ball just before it strikes the ground
u = initial velocity of the ball (which is 0 m/s since it is dropped gently)
a = acceleration of the ball due to gravity (which is -9.8 m/s2 since it is directed downwards)
s = distance travelled by the ball (which is 20 m since it is dropped from a height of 20 m)
We can rearrange this equation to solve for v:
v2 = u2 + 2as
v2 = 0 + 2(-9.8 m/s2)(20 m)
v2 = -392 m2/s2
Since we can’t have a negative velocity, we need to take the square root of both sides of the equation and include a negative sign to indicate the direction of the velocity (downwards):
v = – √392 m/s ≈ -19.8 m/s
Therefore, the ball will strike the ground with a velocity of approximately 19.8 m/s downwards.
To find the time it takes for the ball to strike the ground, we can use the kinematic equation:
s = ut + (1/2)at2
where:
t = time taken for the ball to strike the ground (which is what we’re solving for)
Rearranging this equation, we get:
t = √(2s/a)
Substituting the values, we get:
t = √(2 x 20 m / 9.8 m/s2) ≈ 2.02 s
Therefore, the ball will take approximately 2.02 seconds to strike the ground.
Q. The speed-time graph for a car is shown below
(A) Measure the distance the car covers in the first four seconds. Cover the portion of the graph that corresponds to the distance the car travelled during the time period. (b) Which area of the graph best captures the car’s constant motion?
Answer –
(a) The shaded area on the speed-time graph represents the displacement of the car over a time period of 4 seconds, and the value of the area is 12 meters. Therefore, the car travels a total distance of 12 meters in the first four seconds.
(b) From the 6th to the 10th second, the speed of the car remains constant, which means the car is said to be in uniform motion during that time interval. This uniform motion is represented by the horizontal section of the speed-time graph between x=6 and x=10.
Q. Provide examples for each of the following circumstances and indicate which are possible:
(A) a stationary object with continuous acceleration but no motion
(b) A moving item that exhibits an acceleration yet maintains a constant speed.
(c) A thing that is travelling in one direction while accelerating in the opposite direction.
Answer –
a) It is possible for an object to have constant acceleration but zero velocity, such as a ball thrown straight up into the air at its highest point.
b) It is not possible for an object to have an acceleration and also move with uniform speed.
c) It is possible for an object to have an acceleration perpendicular to its direction of motion, such as a car turning a corner, which experiences sideways acceleration directed towards the centre of the turning circle.
Q. An artificial satellite is travelling in a circular orbit with a 42250 km radius. If it takes 24 hours to complete one orbit of the earth, then determine its speed.
Answer –
We can use the formula for the speed of an object in circular motion:
v = 2πr/T
where v is the speed of the object, r is the radius of the circular orbit, and T is the time taken to complete one revolution.
In this case, the radius of the orbit is 42,250 km, and the time taken for one revolution is 24 hours or 86,400 seconds (since there are 24 hours in a day and 60 seconds in a minute).
Plugging these values into the formula, we get:
v = 2πr/T = 2π(42,250 km) = 265571.42 km
Therefore, the speed of the satellite in its circular orbit is approximately 265571.42 km.