NCERT Solutions Class 9 Science Chapter 8 Motion

NCERT Solutions Class 9 Science 

The NCERT Solutions in English Language for Class 9 Science Chapter – 8 (Motion) has been provided here to help the students in solving the questions from this exercise. 

Chapter – 8 (Motion) 

Questions

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer – An object can have zero displacement even when it has moved a distance and shown some displacement. This takes place when the final position of an object coincides with its own initial position. Let’s say, if a person moves around circular area and comes back to the place from where he started then the displacement will 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?
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
Therefore,  from Pythagoras theorem, the displacement s = √(10²+10²)
s = 10√2
s = 14.14 m
Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from intial position.

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) Given statement is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.
(b) Given statement is false because the displacement of an object can be equal to, but never greater than the distance travelled.

Questions 

1. Distinguish between speed and velocity.
Answer – Difference Between Speed and Velocity

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 of an object is equal to its average speed, only when an object is moving in a straight line.

3. What does the odometer of an automobile measure?
Answer –  The odometer of an automobile measures the distance covered by a vehicle or an automobile.

4. What does the path of an object look like when it is in uniform motion?

Answer – An object having uniform motion has a straight line path.

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 – 
Speed= 3 × 108 m s^-1
Time= 5 min = 5 × 60 = 300 seconds.
Distance = Speed × Time
Distance = 3 × 10⁸ m s^-1 × 300 secs.
= 9 × 10¹⁰m

Questions 

1. When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?
Answer –
Uniform Acceleration – A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal time intervals.

Non-Uniform Acceleration – A body is said to be in non-uniform acceleration if the rate of change of its velocity is not constant, that is differs in different time intervals.

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 –  Firstly, we will change the speed from km/h to m/s, as the given time is in seconds.
Initial speed, u = 80 km/h
=

= 22.22 m/s        —————— (i)

Final speed of bus, v = 60 km/h
=

= 16.66 m/s       ——————— (ii)
Time taken, t = 5 s    —————— (iii)
As we know,
Acceleration, a =  
=

=
= – 1.11 m/s²
Here, negative sign shows the negative acceleration of retardation.

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 –
Initial speed, u = 0 (As it starts from rest)
Final speed, v = 40 km/h
=

= 11.11 m/s
And,
Time taken, t = 10 minutes
= 10 × 60 seconds
= 600 s
Now,
Acceleration, a =
=  

= 0.0185 m/s²
= 1.85 × 10^-2 ms^-2

Questions 

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer –  Distance-time graph is the plot of distance travelled by a body against time. So it will tell us about the journey made by a body and its speed.

(i) For uniform motion of an object, its distance-time graph is a straight line with constant slope.
(ii) For non-uniform motion of an object, its distance-time graph is a curved line with increasing or decreasing slope.

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.

When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means 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?
Answer – The 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?
Answer – Considering an object in uniform motion, its velocity-time graph can be represented as follows.
The slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. The magnitude of distance is measured by the area occupied below the velocity-time graph.

Questions

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 =
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,
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 + at²
Therefore, the total distance covered by the car (s) = 0 × 10m + (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 in 40 s =1 round
Number of rounds in 140 s =140/40 =3 ½
Therefore, Displacement of the athlete with respect to initial position at x = xy
= Diameter of circular track
= 200 m

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 – Firstly, we have to draw a line segment to show the movement of Joseph during his jogging:

(a) Given,
Total distance from A to B = 300 m
Total time taken = 2 minutes 30 seconds
= 2 × 60 + 30
= 120 + 30
= 150 s
Average speed (from A to B) = Total Distance / Total Time Taken
=  
= 2.0 m/s    —————(i)
The displacement and the time taken of Joseph is same ingoing from A to B i.e., 300m and 150 s respectively.
Therefore, Average velocity (from A to B) = Displacement / Total time taken
=
= 2.0 m/s     —————- (ii)
So, from (i) and (ii) it is clear that the average speed and average velocity of Joseph during his jogging is same.

(b) Given,
Total distance from A to C = 300 + 100
= 400 m
Total time = 2 minutes 20 seconds + 1 minutes
= 150 s + 60 s
= 210 s
Average Speed (from A to C) = Total Distance / Total Time Taken
=
= 1.90 m/s  ——————- (iii)
Now the average velocity of Joseph from A to C:
Displacement = 300 – 100
= 200 m
Total time taken is same as that of the time taken from A to C i.e., 210 s
Average velocity (from A to C) =Total Distance / Total Time Taken
=  
= 0.95 m/s —————— (iv)
From (iii) and (iv) it is clear that the average speed of Joseph is different from his average his velocity.

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 –
Let, the school is at a distance of x km 
Suppose the time taken while driving to school is t₁
Given,
Average speed = 20 km/h
Speed = Distance / Time taken 
20 = x/t₁
Time taken, t₁ = x/20 ———— (1)
on returning from the trip the average speed is 30 km/h
Let, the time taken for the return trip is t₂
Speed = Distance / Time taken 
30 = x/t₂
So, Time taken, t₂ = x/30  ————— (2)
Now, total distance of the whole trip from going to trip to returning back to school is:
Total distance = x + x
= 2x Km  ————— (3)
Time taken = x/20 + x/30
= (3x + 6x)/60
= 5x/60
= x/12  —————– (4)
Now, we will calculate the average speed of the whole trip:
Average speed =Total Distance covered / Total Time Taken
=2x × 12 /x
= 24 km/h
Therefore, the average speed for Abdul’s whole trip from going from to school to return back is 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 –
Initial speed, u = 0
Time, t = 8.0 s
Acceleration, a = 3.0 m/s²
We know that,
s = ut +  at²
s = 0 × 8.0 + × 3 × (8.0)²
s = 0 + × 3 × 64
s = 96 m
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?
Answer – 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
= × (OB) × (OA)
OB = 5 seconds
OA = 52 km.h^-1 = 14.44 m/s
Therefore, the area of the triangle AOB is given by:
= × (5s) × (14.44ms^-1)
= 36 meters
Now, the displacement of the second car is given by the area of the triangle COD
= × (OD) × (OC)
OC = 10 seconds
OC = 3km.h^-1 = 0.83 m/s
Therefore, area of triangle COD = = × (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) travelled 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?
Answer –
(a) Since the slope of line B is the greatest, B is travelling 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 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 –  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 =20 m
Downward acceleration,  a =10 m s^-2
As we know, 2as =v²-u²
v² = 2as + u²
= 2 × 10 × 20 + 0
= 400
∴ Final velocity of ball, v = 20 ms^-1
t = (v – u)/a
∴ Time taken by the ball to strike = (20 – 0)/10
= 20/10
= 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 shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:
× 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 6^th to the 10^th 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.
Answer –
(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 possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time Circular motion is an example of an object moving with acceleration but with uniform speed.
An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(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.
Answer –
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.

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