NCERT class 9 Chapter 8- Motion
Best NCERT Solutions Class 9 Science Chapter 8 motion .comprehending how objects move requires comprehending the basic ideas of motion, which are introduced in this chapter.
Intext Questions –
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer: It is possible for an object that has travelled a distance to have zero displacement if it returns to its starting point.
An illustration would be if someone were to jog in a circular park, going around once. His starting and ending positions are identical.
His displacement is therefore 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?
Answer:
Given,
Side of the given square field = 10m
Hence, the perimeter of a square = 40 m
Time taken by the farmer to cover the boundary of 40 m = 40 s
So, in 1 s, the farmer covers a distance of 1 m
Now,
Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140 m
So,
The total number of rotations taken by the farmer to cover a distance of 140 m = total distance/perimeter = 3.5
At this point, let us say the farmer is at point B from the origin O
Therefore, from Pythagoras theorem, the displacement s = √(102+102)
s = 10√2
s = 14.14 m
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.
Answer: Neither of the statements is true.
(a) The given statement is untrue because an object that moves a specific distance and then returns to its starting position has no displacement at all.
(b) The given statement is untrue since an object’s displacement can match the distance travelled, but it can never exceed it.
Intext Questions – 2
1. Distinguish between speed and velocity.
Answer:
| Difference Between Speed and Velocity | |
| Velocity | Speed |
| Velocity can be defined as the rate at which an object changes position in a certain direction | The rate at which an object covers a certain distance is known as speed |
| The velocity of the object changes with the change in direction, therefore the object must follow one direction | The average speed will continue to count even if the object changes direction |
| Vector quantity | Scalar quantity |
| Velocity can be zero, negative, or positive | Speed can never be negative or zero |
2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer: The magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement because average speed is the total distance travelled in a time frame and velocity is the total displacement in a time frame.
3. What does the odometer of an automobile measure?
Answer: An odometer, sometimes known as an odograph, is a tool used to calculate an automobile’s mileage based on the circumference of the wheel as it revolves.
4. What does the path of an object look like when it is in uniform motion?
Answer: An object moving uniformly follows 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.
Answer: The distance between the spaceship and the ground station is equal to the entire distance covered by the signal because it travels in a straight line.
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
1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer: Uniform Acceleration: An object is considered to be in uniform acceleration when it is increasing in velocity at equal intervals while moving straight ahead.
An object falling freely is an illustration of homogeneous acceleration.
Non-Uniform Acceleration: Non-uniform acceleration is the term used to describe an item that is increasing in velocity but not at equal intervals of time.
One instance of non-uniform acceleration is a bus moving or departing from the bus stop.
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.
Answer:
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
As a result, the bus’s overall acceleration is -1.112 m/s. It should be noted that the bus’s velocity is dropping as indicated by the negative sign.
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.
Answer:
Given parameters
Initial velocity (u) = 0
Final velocity (v) = 40 km/h
v = 40 × (5/18)
v = 11.1111 m/s
Time (t) = 10 minute
t = 60 x 10
t = 600 s
Acceleration (a) =?
Consider the formula
v = u + at
11.11 = 0 + a × 600
11,11 = 600 a
a = 11.11/600
a = 0.0185 ms-2
Intext Questions – 4
1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer: The distance-time graph is a straight line for uniform motion. In contrast, a non-uniformly moving object’s distance-time graph is shaped like a curve.
The uniform motion is depicted in the first graph, while the non-uniform motion is shown in the second.
2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:
The distance-time graph can be plotted as follows.
The item remains in the same location over time when the distance-time graph’s slope is a straight line parallel to the time axis. The object is therefore 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?
Answer:
The speed-time graph can be plotted as follows.
The item is considered to be in uniform motion as its velocity (Y-Axis value) does not change at any point in time (X-axis value).
4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
Considering an object in uniform motion, its velocity-time graph can be represented as follows.
Now, The area of the rectangle OABC, which is determined by OA*OC, is located beneath the velocity-time graph. However, OA stands for the object’s velocity and OC for time. Consequently, the area that is darkened can be shown as:
Area under the velocity-time graph = velocity*time.
The area under the velocity-time graph indicates the object’s total displacement when the velocity value is substituted for displacement/time in the preceding equation.
Intext Questions – 5
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.
Answer:
(a) Given, the bus 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 * 120 s) + 0 m.s-1
= 12 m.s-1 + 0 m.s-1
Therefore, terminal velocity (v) = 12 m/s
(b) As per the third motion equation, 2as = v2 – u2
Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120 s, the following value for s (distance) can be obtained.
Distance, s =(v2 – u2)/2a
=(122 – 02)/2(0.1)
Therefore, s = 720 m.
The speed acquired is 12 m.s-1 and the total distance travelled is 720 m.
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.
Answer:
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?
Answer:
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-1
Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s-1
4. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Answer:
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?
Answer:
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 travelled 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
=(0-5)/-10 s
Time taken = 0.5 seconds
Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.
Exercises
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?
Answer:
Given, diameter of the track (d) = 200m
Therefore, the 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 rounds completed by the athlete in 140 seconds = 140/40 = 3.5
As a result, the athlete’s ending location on the circular track is at the other end from their starting position. Consequently, the net displacement will match the track’s 200-meter diameter.
As a result, the athlete’s net distance travelled was 2200 meters, and their total displacement was 200 meters.
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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?
Answer:
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?
Answer:
Distance travelled to reach the school = distance travelled 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 = 2d/(d/20+d/30)kmph
= 2/[(3 + 2)/60]
= 120/5 kmh-1 = 24 kmh-1
Abdul’s average speed during the course of the journey is therefore 24 km/h.
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?
Answer:
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 travelled by boat in 8 seconds = 0 + 1/2 (3)(8)2
= 96 meters
As a result, it takes the motorboat 8 seconds to cover 96 meters.
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?
Answer:
The speed v/s time graphs for the two cars can be plotted as follows.
The area under the speed-time graph can be used to determine the overall displacement of each automobile.
Therefore, displacement of the first car = area of triangle AOB
= (1/2)*(OB)*(OA)
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
= (1/2)*(OD)*(OC)
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
Consequently, the first car is moved forward 36 meters, while the second car moves forward 4.15 meters. As a result, after applying the brakes, the first car—which was moving at 52 kmph—travelled further.
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?
Answer:
(a) Line B is moving at the fastest speed since it has the largest slope.
(b) The three objects never come together at the same spot on the road because the three lines do not cross at one location.
(c) One graph unit is equal to 4/7 km since the graph has 7 unit areas between 0 and 4 on the Y axis.
ince the initial point of an 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 B passes A, the distance between the origin and C is 8km
Therefore, total distance travelled 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 travelled 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?
Answer:
Given, initial velocity of the ball (u) = 0 (since it began at the rest position)
Distance travelled by the ball (s) = 20m
Acceleration (a) = 10 ms-2
As per the third motion equation,
v2 – u2 = 2as
Therefore,
= 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 in 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?
Answer:
(a)
The car’s displacement over a 4-second interval is shown by the darkened area. It is calculable as:
(1/2) * 4 * 6 = 12 m. Thus, in the first four seconds, the car moves a total of 12 meters.
(b) The car is considered to be in uniform motion from the sixth to the tenth second because its speed does not vary between the points (x=6) and (x=10).
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.
Answer:
(a) It is feasible; when anything is launched into the air, gravity pulls on it, causing it to accelerate continuously. But its velocity is zero when it reaches its highest point.
(b) It is feasible; uniform speed denotes that the speed does not vary over time; acceleration denotes a rise or fall in speed.
One instance of an item travelling with acceleration and maintaining a constant speed is in circular motion .Because the velocity is always changing due to changes in the direction of motion, an object moving on a circular path at a constant speed is nevertheless experiencing acceleration.
(c) It is conceivable; when an object accelerates along a circular route, the acceleration is perpendicular to the path the thing is on.
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.
Answer:
Given, the radius of the orbit = 42250 km
Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km
Time is 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.