GSEB Solutions Class 9 Science Chapter 8 Motion

Gujarat Board GSEB Textbook Solutions Class 9 Science Chapter 8 Motion Textbook Questions and Answers, Notes Pdf.

Gujarat Board Textbook Solutions Class 9 Science Chapter 8 Motion

Gujarat Board Class 9 Science Motion InText Questions and Answers

Page 100

Question 1.
An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Answer:
Yes, if the final and initial position of the object is the same, i.e., the object has returned to its initial position.

Question 2.
A farmer motos along the boundary of a square field of side 10 m in 40 sec. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Answer:
The total distance traveled in one round is 40 min 40 seconds.
Total time = 2 minutes 20 seconds = 140 seconds

The object will complete 3 rounds in 120 (40 x 3) seconds and in the next 20 seconds.
it will be at diametrically opposite point (point C if it is started from point A).
Net displacement will be AC using Pythagoras theorem
(AC)² = (AB)² + (BC)²
(AC)² = (10)² + (10)²
(AC)² = 100 + 100
(AC) = 10\(\sqrt{2}\)
Displacement = 14.14 m

Question 3.
Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance traveled by the object.
Answer:
(a) The statement is not true. Displacement can be zero.
(b) The statement is not true. The magnitude of displacement is never greater than the distance traveled by the object.

Page 102

Question 2.
Distinguish between speed and velocity.
Answer:
Speed:

• Speed is the ratio of distance and time.
• Speed is always positive
• Speed is a scalar quantity.

Velocity:

• Velocity is the ratio of displacement and time.
• Velocity may be negative or positive.
• Velocity is a vector quantity.

Question 3.
Under what condition(s) is the magnitude of average velocity an object equal to its average speed?
Answer:
In a straight line motion in one direction, the magnitude of the average velocity of an object equal to its average speed.

Question 4.
What does the odometer of an automobile measure?
Answer:
The odometer of an automobile measures distance traveled by a vehicle.

Question 5.
What does the path of an object look like when it is in uniform motion?
Answer:
In uniform motion, the speed of a body is constant and the direction of motion may be changed. Therefore the path of an object can be a straight line, curved line, zig-zag, or even a circle.

Question 6.
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, which is 3 x 10⁸ ms¹.
Answer:
Given,
Speed u = 3 x 10⁸ m/s^-1
Time t = 5min = 5 x 60 = 300s
Distance =?
Distance = speed x time
Distance, s = 3 x 10⁸ x 300
s = 9 x 10¹⁰ ml

Page 103

Question 1.
When will you say a body is in

1. Uniform acceleration?
2. Non-uniform Acceleration?

Answer:

• A body is said to be in uniform acceleration when the velocity of the body changes by equal amounts in equal intervals of time. For example, the motion of a body falling freely under gravity.
•  A body is said to be in non-uniform acceleration when the velocity of the body changes by unequal amounts in equal intervals of time. For example, the motion of a cycle on a normal road.

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:
Given,
Initial speed, u = 80 kmh^-1
u = 80 x \(\frac{5}{18}\) = 22.22 m/s
Final speed, υ = 60 kmh^-1
υ = 60 x \(\frac{}{t}\) 16.66 m/s²
Time taken, t = 5s
Acceleration, a = \(\frac{υ – u}{t}\) = \(\frac{16.66-22.22}{5}\)
a = \(\frac{-5.56}{5}\) = – 1.11m/s
Negative sign is for retardation.

Question 3.
A train starting from a railway station and moving with uniform acceleration attains speed 40 kmh^-1 in 10 minutes. Find its acceleration.
Answer:
Initial speed, u = 0
Final speed, υ = 40 km h^-1 = 40 x \(\frac{5}{18}\) = 11.11 m/s
Timetaken t = 10 min = 10 x 60 s = 600s
accelerabon, a = \(\frac{υ – u}{t}\) = \(\frac{υ – u}{t}\) = \(\frac{υ – u}{t}\) = 0.0185
a = 1.85 x 10^-2 m/s²

Page 107

Question 1.
What is the nature of the distance-time graph for uniform and non-uniform motion of an object?
Answer:

• In uniform motion, the distance-time graph is a straight line with some slope.
• In non-uniform motion, the distance-time graph is a curved line.

Question 2.
What can you say about the motion of an object if its distance-time graph is a straight line parallel to the time axis?
Answer:
The object is stationary as the position of the object is changing with time.

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:
The object is moving with a uniform speed as the speed is not changing with time.

Question 4.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
The displacement of an object is measured by the area under the velocity-time graph.

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 ms^-1. If the acceleration of the stone during its motion is 10 ms^-2 in the downward direction, what will be height attained by the stone and how much time it will take to reach there?
Answer:
Given, Initial velocity, u = 5 ms^-1
Final velocity, υ = 0 ms^-1
Acceleration, a = -10 ms^-2

1. Height attained by the stone
Using equation, υ² – u² = 2as
0 ² – (-5)² = 2 (-10)h
h = 1.25 m

2. Time taken
Using equation, υ = u + at
0 = 5 – 10 x t
t = 0.5 s

Pages 109 – 110

Question 1.
A bus starting from rest moves with a uniform acceleration of 0.1 ms² for 2 minutes. Find (a) the speed acquired (b) the distance traveled.
Answer:
Given, ïnitial speed, u = 0 m/s^-2
final speed, υ = ?
acceleration, a = 0.1 ms^-2
time, t = 2 minutes = 2 x 60 = 120 s
Using equation,
υ = u + at
υ = 0 + 0.1 x 120
υ = 12 m/s

(b) Using equation
s = ut + \(\frac {1}{2}\) at²
s = 0 x 120 + \(\frac {1}{2}\) x 0.1 x (120)²
∴ s = 7.2 x 10²m

Question 2.
A train is traveling at a speed of 90 kmh^-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:
Given,
Initial speed, u = 90 km h^-1
u = 90 x \(\frac {5}{8}\) = 25 ms^-1
Final velocity u = 0 m/s
acceleration, a = – 0.5 m/s²
Using equation,
u² – u² = 2 as
0² – (25)² = 2(-0.5)(s)
∴ s = 625 m

Question 3.
A trolley, while going down an inclined plane, has an acceleration of 2 cm². What will be its velocity 3s after the start?
Answer:
Given,
acceleration, a = 2 cm r²
time, t = 3s
initial velocity, u = 0 ms^-1
Using equation, υ = u + at
υ = 0 + 2 x 3
υ = 6 cms^-1

In-Text Activities Solved

(Textbook Page 98)

Activity 8.1
Discussion and conclusion
Answer:
The walls of our classroom are at rest with respect to the Earth. But Earth is in motion therefore, when classroom walls are seen from outside the Earth, the walls are in motion. States of rest and motion are relative
phenomenon.

(Textbook Page 98)

Activity 8.2
Discussion and conclusion
Answer:
Yes, when we are sitting in a train at rest, and the train on the adjoining track moves, we feel as if our train is moving in the opposite direction. It is due to relative motion, we consider adjoining train at rest and hence our train appears to move in the opposite direction to that of the other train.

(Textbook Page 99)

Activity 8.3
Discussion and conclusion
Answer:
Let us take the dimensions of a basketball court = x x y
Distance travelled from A to C = x + y
Displacement from A to C = \(\sqrt{x^{2}+y^{2}}\)

(Textbook Page 99)

Activity 8.4
Discussion and conclusion
Answer:
We can locate the position of New Delhi and Bhubaneshwar from the map of India. After measuring the length of the line joining New Delhi and Bhubaneshwar, we can convert it using a suitable map scale (given map scale). It will be less than 1850 km.

(Textbook Page 100)

Activity 8.5
Discussion and conclusion
Answer:
Object A travels the equal distance in equal intervals of time and hence the motion of object A is uniform motion.
Object B travels the unequal distance in equal intervals of time and hence the motion of Object B is non-uniform motion.

(Textbook Page 101)

Activity 8.6
Discussion and conclusion
Answer:
Suppose, we take 10 minutes to walk from our house to the bus stop or school.
Distance speed x time
Speed = 4 kmh^-1 (given)
Distance =4 \(\frac {km}{h}\) x \(\frac {10}{60}\)h = \(\frac {1}{2}\)km = 0.67km

(Textbook Page 102)

Activity 8.7
Answer:
The speed of light is greater than the speed of sound. Therefore thunder takes more time to reach.

(Textbook Page 103)

Activity 8.8
Discussion and conclusion
Answer:

• When the speed of the car is increasing.
• When a bus is retarding. (applying brakes)
• When a body is falling freely.
• The motion of a car on a normal road.

(Textbook Page 106)

Activity 8.9
Discussion and conclusion
Answer:
Distance-time graph:

FromAtoBspeedoftrain= \(\frac {120 km}{3h}\) = 40km/h
From B to C speed of train = \(\frac {60 km}{1.5h}\) = 40 km/h

(Textbook Page 107)

Activity 8.10
Discussion and conclusion
Answer:
Feroz travels faster than Sania

(Textbook Page 111)

Activity 8.11
Discussion and conclusion
Answer:
1. The moment the stone is released, it moves tangent to the circular path at that moment.
2. By releasing the stone at different positions of the circular path, we will find that direction in which the stone moves is always different, but always tangent to the circular path.

Gujarat Board Class 9 Science Motion Textbook Questions and Answers

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:
Time taken = 2 min 20 sec = 140 sec.
Radius, r = 100 m
In 40 seconds the athlete completes one round.
In 140 seconds the athlete completes = \(\frac {140}{40}\) = 3.5 round
So, distance travelled in 140 seconds = 3.5 x 2πr
=2 x \(\frac {22}{7}\) x 100 x 3.5 = 2200m
At the end of this motion, the athlete will on diametrically opposite position.
So, Displacement = diameter = 200 m.

Question 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 speed and velocities in jogging (a) from A to B and (b) from A to C?
Answer:

(a) For motion from A to B.
Distance covered = 300 m
Displacement = 300 m
Time taken = 150 s
Total distance 300 m

(b) For motion from A to C.
Distance covered = 400 m
Displacement = 200 m
Time taken = 3 min 30 s = 210 s.

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 the distance while driving to school be x km.
Time taken for the forward trip at a speed of 20 km/h

Time taken in return trip at a speed of 30 km/h
t₂ = \(\frac {x}{30}\) h
Time taken forthe whole trip = \(\frac {x}{20}\) + \(\frac {x}{30}\) = \(\frac {3x + 20}{60}\) = \(\frac {5x}{60}\)h
Total distance travelled = 2x km

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^-2 for 8.0 s. How far does the boat travel during this time?
Answer:
Here, u = 0 m/s
a = 3 ms^-2
t = 8s
∴ s = ut + \(\frac {1}{2}\) at²
∴ s = 0 + \(\frac {1}{2}\) x 3 x 64
∴ s = 96 m

Question 5.
A driver of a car traveling at 52 km/h applies brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 km/h 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 traveled farther after brakes were applied?
Answer:
Given, initial speed of first
car, u₁ = 52 km/h = 52 x \(\frac {5}{18}\) = 14.4m/s
time, t = 5 s; final speed, υ₁ = 0
a₁ = \(\frac{v_{1}-u_{1}}{t_{1}}=\frac{0-14.4}{5}\) = – 2.88 m/s²
Initial speed of second car,
u₁ = 34 km/h = 34 x \(\frac {5}{18}\) =9.4m/s
time, t = 10 s ; final speed, υ₂ = 0
a₂ = \(\frac{v_{2}-u_{2}}{t_{2}}=\frac{0-9.4}{10}\) = -0.94 m/s²

Question 6.
Given figure 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 traveling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C traveled when B passes A?
(d) How far has B traveled by the time it passes C?
Answer:
(a) B is traveling fastest.
(b) All three are never at the same point on the road.
(c) 6.86km
(d) 5.71km

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:
u = 0 m/s
s = 20 m
a = 10 m/s² υ = ? t = ?
υ² – u² = 2as
υ² = 2 x 10 x 20
υ = 20 m/s
As υ = u + at
20 = 0 + 10 x t
t = 2s.

Question 8.
The speed-time graph for a car is shown in the figure.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance traveled by car during the period.
(b) Which part of the graph represents the uniform motion of the car?
Answer:
(a) Distance covered = area under speed – time graph
Distance = Area of 1 square x total number of square under the distance – time graph
= 2/5 x 2/3 x 62 = 16.53m
(b) After 6 seconds the car moves with uniform motion.

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 with zero velocity
(b) An object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
(a) Yes, a body can have constant acceleration even when its velocity is zero. When a body is thrown up, at the highest point its velocity is zero but it has acceleration equal to the acceleration due to gravity.
(b) Yes, when a body is moving horizontally like an airplane a perpendicular acceleration, acceleration due to gravity acts on it.

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:
r = 42250 km
t = 24 h = 24 x 60 x 60 s

υ = 3070.9 m/s
∴ υ = 3.07 km/s.