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Motion

Class 9th Science CBSE Solution
In Text Questions-pg-100
  1. An object has moved through a distance. Can it have zero displacement? If yes, support…
  2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the…
  3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude…
In Text Questions-pg-102
  1. Distinguish between speed and velocity.
  2. Under what condition(s) is the magnitude of average velocity of an object equal to its…
  3. What does the odometer of an automobile measure?
  4. What does the path of an object look like when it is in uniform motion?…
  5. During an experiment, a signal from a spaceship from the ground station in 5 minutes.What…
In Text Questions-pg-103
  1. When will you say a body is in? (i) Uniform acceleration? (ii) Non - uniform acceleration?…
  2. A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s. Find the acceleration of the…
  3. A train starting from a railway station and moving with uniform acceleration attains a…
In Text Questions-pg-107
  1. What is the nature of the distance time graphs for uniform and non-uniform motion of an…
  2. What can you say about the motion of an object whose distance time graph is a straight…
  3. What can you say about the motion of an object if its speed-time graph is a straight line…
  4. What is the quantity which is measured by the area occupied below the velocity-time graph?…
In Text Questions-pg-109
  1. A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes.…
  2. A train is traveling at a speed of 90 km h-1. Brakes are applied so as to produce a…
  3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will…
  4. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s…
  5. A stone is thrown in vertically upward direction with a velocity of 5 m s-1. If the…
Exercise-pg-112
  1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be…
  2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30…
  3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h-1.…
  4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate…
  5. A driver of a car travelling at 52 km h-1 applies the brakes and accelerates uniformly in…
  6. The figure shows the distance-time graphs of three objects A, B and C. Study the graphs…
  7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the…
  8. The speed-time graph for a car is shown here. (a) Find how far does the car travel in the…
  9. State which of the following situations are possible and give an example for each of…
  10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its…
In Text Questions-pg-100
Question 1.

An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:

Yes, even if an object has moved through a distance, it can have zero displacement. This can happen if, after moving through a certain distance, the moving object comes back to the same position from where it started from. For example, in going from home to school and coming back to home, some distance is traveled but displacement is zero because the initial and final point are the same(The person is leaving from home and coming back to home)
Another Example:- Athlete running in a circular track after completing one round comes back to the starting position, he has covered a certain distance(equal to circumference of the circle ) but the displacement is zero(because the straight line distance between the athletes final and starting position will be zero)
Question 2.

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:
Side of the given square field = 10m

Perimeter of a square = 4Xside = 10 m X 4 = 40 m
Farmer takes 40 s to move along the boundary.
Displacement after 2 minutes 20 s = 2 X 60 s + 20 s = 140 seconds
since in 40 s farmer moves 40 m
Therefore, in 1s the distance covered by farmer = 40 / 40 m = 1m
Therefore, in 140s distance covered by farmer = 1 X140 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 the farmer will be equal to 14.14 m northeast from the initial position.

Question 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:

(a) The first statement is false as the displacement can be zero in the case where initial and final positions are common.

(b) The magnitude of displacement can never be greater than the distance travelled by the object it is either equal or smaller. So, the second statement is also false.
In Text Questions-pg-102
Question 1.

Distinguish between speed and velocity.
Answer:


Question 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 under following condition:

When an object moves along a straight line in the same direction, its total path length is the magnitude of displacement. Hence its average speed is equal to the magnitude of average velocity.
Question 3.

What does the odometer of an automobile measure?
Answer:

Odometer is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates. The odometer gives us only the number it does not show us the direction of velocity. The odometer measure the scalar quantity – the speed of the car.
Question 4.

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

An object possess a uniform motion if it travels equal distances in equal intervals of time then no matter how small these time intervals are. This refers that while having a uniform motion, speed is constant but the direction of motion may vary. Unless the speed remains same, the object may have any path: it may be straight line path, a curved path, a circular path or even a zig-zag path.Below is the example of an object moving in uniform motion:-
Question 5.

During an experiment, a signal from a spaceship from the ground station in 5 minutes.What was the distance of the spaceship from the ground station? The signal travels at the speed of light i.e., 3×108 m s-1.
Answer:

As we know,

Speed = (i)


Given, Speed = 3×108 ms-1


Distance travelled =?


Time taken = 5 minutes


= 5×60


= 300 seconds


Now we put the values in equation (i), we get


3 × 108 =


Distance travelled = 3×108 × 300


= 9 × 1010 m


Therefore, the distance of spaceship from the ground station is 9 × 1010 metres.
In Text Questions-pg-103
Question 1.

When will you say a body is in?

(i) Uniform acceleration?

(ii) Non - uniform acceleration?
Answer:

(i) If a body travels in a straight line and its velocity changes by equal amounts in equal intervals of time, however small these time intervals may be, then the body is said to be in uniform acceleration.

(ii) When a body moves with a unequal velocity in equal interval of time, the body is said to be moving with non-uniform acceleration.
Question 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:

When a moving vehicle decreases its speed by applying breaks, the phenomenon is called de-acceleration.
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 (Given)

(We know 1km=1000m and 1hour=3600 seconds)

=

= 22.22 m/s (i)


Final speed of bus, v = 60 km/h(Given)
(We know 1km=1000m and 1 hour=3600 seconds)

=

= 16.67 m/s (ii)


Time taken, t = 5 sec (iii)

As we know, from first equation of motion that v= u + a × t

Acceleration, a =


Putting the Values from (i) ,(ii) and (iii)
a =
a =

a = - 1.11 m/s2

Here, negative sign shows the retardation in the sped off bus due to application of brakes.
Question 3.

A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h-1 in 10 minutes. Find its acceleration.
Answer:

Given:

Initial speed, u = 0 (As it starts from rest)

Final speed, v = 40 km/h (Given in Question)
We know 1 km=1000 m and 1 hour=60 x 60 seconds)

we get final speed in terms of m/s as:

=

= 11.11 m/s

And,

Time taken, t = 10 minutes (Given in Question)

= 10 × 60 seconds (As 1 Minute= 60 Seconds)

= 600 s

Formula used:

by the first equation of motion,

v = u + a × t

where v, Final velocity
u, initial velocity
t, time taken to reach to final velocity
a, acceleration of the train

By rearranging the Equation , we get
a =

a =

a =

a = 0.0185 m/s2 or 0.067 km/ min
In Text Questions-pg-107
Question 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) As in uniform motion, the distance time graph would be a straight line along with some slope, because the equal distance is covered in equal units of time.



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

If the distance-time graph of an object is a straight line parallel to the time axis, it indicates that the distance of the object is same from its initial position at all intervals of time. Since the object makes no change in the distance from its initial position, hence this states that the object is not moving. The object is at rest.
Question 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:

If the speed-time graph of an object is a straight line parallel to the time axis, then the speed of the object remains constant at every instant of time. Hence, we can say that the object is moving with the constant speed (or uniform speed). There is no acceleration at all.
Question 4.

What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:

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 under the velocity-time graph.

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In Text Questions-pg-109
Question 1.

A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find:

(a) The speed acquired.

(b) The distance travelled.
Answer:

(a) The speed acquired:

Given,


Initial speed, u = 0 (As it starts from rest)


Final speed, v =?


Acceleration, a = 0.1 m/s2


Time, t = 2 minutes


= 2 × 60 seconds


= 120 s


Final velocity, v = 0 + 0.1 × 120


= 12 m/s


Therefore, the speed acquired by the bus is 12 m/s


(b) The distance travelled:


We know that,


Distance travelled, s = ut +at2


= 0 × 120 + × 0.1 × (120)2


= 0 + × 0.1 × 14400


= 720 m


Therefore, the distance travelled by the bus is 720 metres
Question 2.

A train is traveling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of, -0.5 ms-2. Find how far the train will go before it is brought to rest.
Answer:

Initial speed, u = 90 km/h2

(Converting it into m/s, We know 1km=1000m and 1 hour= 3600 seconds)

=

= 25 m/s

Final speed, v = 0 (As the train stops)

Acceleration, a = -0.5 m/s2(As the brakes are being applied the speed is reducing)

And,

Distance traveled, s =? (To be Calculated)

Now, v2 = u2 + 2as

(0)2 = (25)2 + 2 × (-0.5) × s

0 = 625 – 1 × s
0 = 625 - s

∴s = 625 m (Taking s on the other side will make it positive)

Therefore, 625 meters is the distance traveled by the train before it is brought to rest.
Question 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


Final velocity, v =?


Acceleration, a = 2 cm/s2


And,


Time, t = 3 s


We know that,


v = u + at


= 0 + 2 × 3


= 6 cm/s


Therefore, after 3 seconds the velocity of trolley will be 6 cm/s.
Question 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,

Initial velocity, u = 0


Acceleration, a = 4 m/s2


And,


Time, t = 10 s


Distance covered, s =?


Now, s = ut + at2


= 0 × 10 +× 4 × (10)2


= 0 + 2 × 100


= 200 m


Therefore, in 10 seconds racing car will covered a distance of 200 m.
Question 5.

A stone is thrown in 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: u, = 5
a, = -10
v, = 0
By using
v= u +at
0 = 5+(-10)T
-5 = -10 T
t = 1/2=0.5
s=ut+1/2at ^2
s = 5(1/2) + 1/2(-10) ×1/4
s = (5/2-5/4) m
s = (10-5 / 4) m
s = 5/4 m
s = 1.25m
Exercise-pg-112
Question 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, Total time = 2 minutes 20 seconds

= 2 × 60 + 20 (As 1 Minute= 60 Seconds)

= 120 + 20

= 140 seconds

According to Question,

In 40 Seconds⇒ 1 Round is completed
∴In 140 Seconds⇒



(a) Calculation of Distance Covered in 3.5 Rounds
Diameter of circular track = 200m (Given)

∴Radius of circular track, (r) = 100m( Diameter/2)

Now, Distance covered in 1 round = Circumference of the circular track

= 2πr

=

= 628.57 m
∴ Distance covered in 3.5 Rounds = 628.57 X 3.5= 2200 m

(b) As the athlete makes 3.5 rounds of the circular track, Suppose he starts from point A then after completing the 3 rounds he will be at the same point A. On starting again from the same point(Point A) and makes the remaining half round then he will reach at the point B. Therefore, the displacement of the athlete will be equal to the diameter of the track i.e., 200m.
Question 2.

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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

(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 from A to B = 2 minutes 30 seconds

= 2 × 60 + 30

= 120 + 30

= 150 s

Average speed (from A to B) === 2.0 m/s (i)

The displacement and the time were taken of Joseph is the same in going from A to B i.e., 300m and 170 s respectively.

Therefore, Average velocity (from A to B) == = 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 are the same.


(b) Given,

Total distance from A to C = 300 + 100= 400 m
Total time = 2 minutes 30 seconds + 1 minutes

= 150 s + 60 s (As we Know 1 Minute= 60 seconds)

= 210 s

Average Speed (from A to C) = = = 1.90 m/s (iii)


Now the average velocity of Joseph from A to C:

Displacement = 300 – 100= 200 m

Total time taken is the same as that of the time taken from A to C i.e., 210 s (Calculated Above)

Average velocity (from A to C) =

From (iii) and (iv) it is clear that the average speed of Joseph is different from his average velocity.
Question 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 us assume the school to be at distance x km from where Abdul starts

Suppose the time taken while driving to school is t1

Given,

Average speed = 20 km/h



Time taken,

On returning from the trip the average speed is 30 km/h

Let, the time taken for the return trip is t2



Time taken,


Now, total distance of the whole trip from going to trip to returning back to school is:

Total distance = x + x

= 2x Km (3)

Adding (1) and (2) to get the total time

Total Time taken = Time taken to go to college +Time taken to return back to school



Solving this equation

=

=(4)


Now, we will calculate the average speed of the whole trip:

Average speed =

=


= 24 km/h


Therefore, the average speed for Abdul’s whole trip from going from to school to return back is 24 Km/h.
Question 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 speed, u = 0 (As the boat is starting from rest)

Time, t = 8.0 s(Given)

Acceleration, a = 3.0 m/s2 (Given)

We know that,

s =


Putting the values in above equation


s=


Therefore, total distance travelled by boat is 96m.
Question 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:

Given:

Firstly, for the first car:

Initial speed, u = 52 km/h

(We know that 1 km=1000 m and 1 hour=3600 seconds)

=


= 14.4 m s-1 (i)


Final speed, v = 0 (As the car stops) (ii)


Time taken, t = 5 s (iii)


Now, for the second car:

Initial speed, u = 3 km/h

(We know that 1 km=1000 m and 1 hour=3600 seconds)

= 0.833 m/s (iv)

Final speed, v = 0 (As the car stops) (v)

Time taken, t = 10 s (vi)

In the graph, we will plot time taken in X-axis and speed in the Y-axis. In the graph the sloping point AB is the speed-time graph for the first car and the sloping point CD is the speed-time graph for the second car.



Now we have to find the distance travelled under the graph line AB = Area under triangle AOB


= × base × height


= × 5 × 14.4


= 36 m (vii)


Now we will find the distance travelled under the line CD = Area under triangle COD


= × base × height


= × 0.833 × 10


= 4.1 m (viii)


Therefore, from (vii) and (viii) it is clear that the first car travels farther as compared to the second car after the application of brakes.
Question 6.

The figure shows the distance-time graphs of three objects A, B and C. Study the graphs 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) The slope of distance time graph of the moving objects represents the speed of those objects.Greater the slope of an object, higher is its speed.

Now, in the given figure we can see that the slope of distance time graph of object B is the maximum, hence object B has the maximum speed.

In other words the object B is traveling with the fastest speed.


(b) In order to be at the same point on the road, the respective distance and time values for all the three moving objects need to be the same. Since the distance time graph lines of the three objects A, B, and C do not cross at any point, therefore, the three objects are never at the same point on the road.

(c) We can see from the given figure that when B passes A at point D, then the C is at point E.

So, if we locate the distance corresponding to point E on the Y-axis, we would find that it is 6.5 km.

Thus, C has traveled 6.5 km when B passes A.


(d) The distance time graphs of B and C meet at point F.

If we locate the distance corresponding to point F on the Y-axis, we will find that it is 5 km.

Thus, B has traveled 5 km by the time it passes C.
Question 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, u = 0 (As the ball is dropped from rest)


Final velocity, v =?


Acceleration, a = 10 m/s2


Distance, s = 20 m


We know that,


v2 = u2 + 2as


v2 = (0)2 + 2 × 10 × 20


v2 = 0 + 400


v2 = 400


v =


v = 20 m/s


Now we have to calculate the time taken:


We know that:


v = u + at


20 = 0 + 10 × t


10t = 20


t =


t = 2 s


Therefore, after 2 seconds the ball will strike the ground.
Question 8.

The speed-time graph for a car is shown here.

(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 this period.

(b) Which part of the graph represents uniform motion of the car?
Answer:

(a) The distance travelled by the car in the first 4 seconds is given by the area between the speed time curve and the time axis from t = 0 to t = 4 s

This area of the distance time graph which represents the distance travelled by the car



In order to find the distance travelled by the car in the first 4 seconds, we have to count the number of squares in the shaded part of the graph and also calculate the distance represented by one square of the graph paper.


We will now calculate the distance represented by 1 square of the graph.


If we look at the X-axis, we find that 5 squares on X-axis represent a time of 2 seconds.


5 squares on X-axis = 2 s


1 square on X-axis = s (i)


Now, if we look at the Y-axis,


We find that 3 squares on Y-axis represent a speed of 2 m/s


Now, 3 squares on Y-axis = 2 m/s


1 square on Y-axis = m/s (ii)


Since, 1 square on X-axis represents s and 1 square on Y-axis represent ms-1


Therefore, Area of 1 square on graph = ×


= m


63 square represents distance = × 63


= 16.8 m


Therefore, the car travels a distance of 16.8 m in the first 4 seconds.


(b) In uniform motion, the speed of car becomes constant. The constant speed is represented by a speed time graph line which is parallel to the time axis. In the given figure, the straight line graph from t = 6 s to t= 10 s represents the uniform motion of the car. The part of graph representing uniform motion has been labelled AB.
Question 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 zero velocity.

(b) An object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:

(a) The above given situation is possible. For example, when a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s 2. This situation can be explained by the given graph below

(b) The above given situation can be seen in the following example:

When a car is moving in a circular track, its acceleration is perpendicular to its direction.This situation can be explained by the given graph below.
Question 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:

We know that,
(i)
As the Satellite is moving in a circular orbit ∴ the Distance = Circumference of the Orbit
We know that Circumference= 2πr (ii)
where r= radius o

π= (It is a constant)
According to Question

Radius, r = 42250 km

Putting the value of r in (ii)
We get Circumference= = 265571.42 km

Putting these values in (i), we get


Speed, v=


speed = 11065.4 km/h


= km s-1


= 3.07 km s-1