## Karnataka Board Class 8 Science Chapter 8 Describing Motion

### KSEEB Class 8 Science Chapter 8 Textual Questions & Answers

I. Four alternatives are given to each of the following incomplete statement/ question, choose the right answer.

Question 1.

Uniform circular motion is called continuously accelerated motion mainly because

(a) direction of motion changes

(b) Speed remains the same

(c) Velocity remains the same

(d) direction of motion does not change

Answer:

a) direction of motion changes

Question 2.

A cricketer hits a sixer, the cricket ball moves up with a velocity of 2 ms^{-1} and falls down. Its initial velocity while falling down will be

a) 1 ms^{-1
}b) 1 ms^{-2}

c) 0 ms^{-1}

d) 2 ms^{-1}

Answer:

c) 0 ms^{-1}

II. Fill in the blanks with suitable words:

Question 1.

S.I. unit of acceleration is ………….

Answer:

meter per second^{2}/ms^{-2}

Question 2.

Velocity has both speed and ……….

Answer:

direction

Question 3.

If an object starts from A and comes back to A. It displacement will be ………….

Answer:

(0) zero

III. Solve

Questions 1.

An object is moving in a circular path of radius 3.5 m. If completes one full cycle, what will be the displacement and what is the distance traveled?

Answer:

radius = 3.5 m

starting point is A and ending point also A.

the displace = 0

Distance traveled = π × 3.5 × 2

Question 2.

An object changes its velocity from 30 ms^{-1} to 40 ms^{-1 }in a time interval of 2 seconds what is its acceleration?

Answer:

u = 30 ms^{-1}

v = 40 ms^{-1}

t = 2 sec

a = ?

Question 3.

An object at rest starts moving. It covers a distance of 2 m in one second. It covers a, further distance of 5 m in two seconds in the same direction what is the average velocity and acceleration?

Answer:

distance traveled on 1 sec = 2 mts

distance traveled on 2 sec = 5 mts

total distance traveled = 2 + 5 = 7 mts

total time taken to cover 7 mts = 1+2 = 3 sec

= \(\frac { 7 }{ 3 }\) = 2.3 ms^{-1}

Average velocity = 2.3 ms^{-1
}s = ut + \(\frac { 1 }{ 2 }\) at^{2
}7 = 0 3 +\(\frac { 1 }{ 2 }\) a × 3^{2
}7 =0 + \(\frac { 1 }{ 2 }\) a × 9

= \(\frac { 9 }{ 2 }\) a = 7

9a =7 × 2 = 14

a = \(\frac { 14 }{ 9 }\) = 1.5 ms^{-2}

IV. Answer the following.

Question 1.

If a body is moving with uniform velocity in a given direction its acceleration will be zero, why?

Answer:

The acceleration will be zero in the above situation since there is no change in velocity in unit time.

Question 2.

Distinguish between speed and velocity.

Answer:

- Speed
- Speed is the distance traveled by an object in unit time
- Speed = \(\frac { distance }{ time }\)
- Speed is a scalar quantity
- Speed has the only magnitude

- Velocity
- Velocity is the rate of displacement of the body in unit time.
- Velocity =\(\frac { displacement }{ time }\)
- Velocity is a vector quantity
- It has both magnitude and direction.

Question 3.

Distinguish between distance traveled and displacement.

Answer:

- Distance traveled
- Distance traveled is the length of actual path traveled by a body from one position to another.
- It is a scalar quantity
- It has an only numerical value

- Displacement
- Displacement is the shortest path between initial and final position of the moving body.
- It is a vector quantity
- If has both numerical value and direction.

Question 4.

What are uniform and non-uniform speed?

Answer:

- If an object covers equal distance in equal intervals of time, it is said to be uniform motion.
- non-uniform motion: If an object covers unequal distance in equal intervals of time, it is said to be non uniform motion.

Question 5.

While mentioning acceleration the time is mentioned two times, why?

Answer:

Velocity is the rate of displacement of a body in unit time.

V =\(\frac { displacement }{ time }\)

Acceleration is the change in velocity of a body in unit time.

While mentioning acceleration time is mentioned two times. (Acceleration occurs twice)

V. Extended activity

Represent the following motion by a graph.

Question 1.

Find the acceleration

Ans. Acceleration from the graph

= \(\frac { 40-10 }{ 4-1 }\) = \(\frac { 30 }{ 3 }\) = 10 ms^{-1}

Question 2.

Find the time taken when the velocity in 35 ms^{-1}

Answer:

Time taken to travel a distance of 35 ms^{-1 } is 1.5 seconds.

### KSEEB Class 8 Science Chapter 8 Additional Questions & Answers

Question 1.

Define motion?

Answer:

The change in position of a body with time when compared with that of another body is called motion.

Question 2.

Motion is reflective-substantiate this statement with a suitable example.

Answer:

Imagine that you are sitting inside the train when the train starts moving you a fee that the persons on the platform are moving backward you feel that you and the other persons inside the train are not moving.

But for a person standing outside the train, the feeling will be that the train, you and the other members inside the compartment are all moving.

Question 3.

An object travels from A to B and then from B to C as shown in the figure.

1. Find the distance traveled and

2. Find the displacement

3. Write the conclusion

Answer :

Let ABC is a right-angled triangle with B as shown in the actual figure,

- If a person travels from A to B and then to C the total distance traveled is = (3 m + 4 m) = 7 mts.
- The displacement is the change in position from A to C

- displacement value is lesser than the distance traveled.

Question 4.

An object moves from A to A in a circular path of radius 7 m. What is the distance traveled? What is the displacement?

Answer:

Displacement = Diameter = 2 × radius = 2 × 7= 14 m

Distance traveled = \(\frac { 1 }{ 2 }\) × circumference

= \(\frac { 1 }{ 2 }\) × 2 × π × r

= \(\frac { 1 }{ 2 }\) × 2 × \(\frac { 22 }{ 7 }\) × 7

= 22 m

Question 5.

Define the following

a) Speed

b) Velocity

Answer:

- Speed: is the distance traveled by an object in unit time
- Velocity: is the rate of displacement of a body in unit time.

Question 6.

An object is moving in a circular path of radius 7 m. To travel from ‘A’ to ‘B’ along the circumference it takes 2 seconds?

Answer:

Data : radius = 7 m

time = 2 s

average speed = ?

speed = \(\frac { distance traveled }{ time }\)

Question 7.

When can we say that speed and velocity are synonyms.

Answer:

Speed and velocity will be equal for uniform motions along a straight line. Then it will be one and the same.

Question 8.

What is meant by acceleration?

Answer:

A change in velocity of a body in unit time or the rate of change of velocity is called acceleration.

Question 9.

For an object moving at a uniform speed in a circular path, though the speed is uniform it is continuously accelerated, why? Give reason.

Answer:.

For an object moving in a circular path, the direction is changing at every point even though the speed is the same. Therefore velocity changes and hence it is continuously accelerated.

Question 10.

An object at rest gains an average velocity of 40 ms-1 in 5 seconds. What will be its acceleration?

Answer:

Its initial velocity, u = 0

Its final velocity, v = 40 ms^{-1
}Time taken t = 5 second the

rate of change of velocity

\(\frac { 40 }{ 8 }\) = 8 ms^{-2}

acceleration, a = 8 ms^{-2}

Question 11.

An object at rest starts moving and attains a velocity of 10 ms^{-1 }after 5 seconds. What is the acceleration?

Answer:

Initial velocity, u = 0

Final velocity, v = 10 ms^{-1}

Time interval, t = 5 second

a = \(\frac { u-v }{ t }\) = \(\frac { 10 – 0 }{ 5 }\) = 2 ms^{-2}

Question 12.

An object moving with a uniform velocity of 10 ms^{-1 }comes to rest after 5 seconds. What is the acceleration?

Answer:

Initial velocity, v = 10 ms^{-1}

Final velocity, u = 0

Time interval, t = 5 second

a = \(\frac { u-v }{ t }\) = \(\frac { 0 – 10 }{ 5 }\) = – 2 ms^{-2}

Question 13.

An object at rest starts moving with a uniform acceleration of 1 ms^{-2} Calculate the distance traveled by it in 4 seconds.

Answer:

u = 0

a = 1 ms^{-2}

t = 4 second ,

Distance traveled s = ut + \(\frac { 1 }{ 2 }\) at^{2}

S = 0 × 4 + \(\frac { 1 }{ 2 }\) × 1 × 4^{2}

=0 + \(\frac { 1 }{ 2 }\) × 1 × 16 = 18 m

Question 14.

An object starts from rest and attains a uniform acceleration of 4 ms^{-2} what will be its velocity at the end of half a meter?

Answer:

u = 0

a = 4 ms^{-2}

s = \(\frac { 1 }{ 2 }\)

v^{2} = ?

v^{2}= u^{2} + 2as

v^{2 }= 0^{2 }+ 2 × 4^{2 } × \(\frac { 1 }{ 2 }\)

v^{2 } = 4

v^{2} = \(\sqrt { 4 } \) = 2 ms^{-1}

Question 15.

What is the distance-time graph? What are its uses?

Answer:

Motion can be represented by a time graph. If it represents time on the x-axis and distance traveled on the y-axis, it is called a distance-time graph.

Uses of the distance-time graph are,

- The distance at which the body is at present from its starting point can be found out.
- Time taken to cover a definite distance can be determined
- Since it is uniform motion speed of the body can also be calculated.

Question 16.

What do you infer about the velocity when the velocity-time graph is parallel to the x-axis.

Answer:

The body is at rest no acceleration.

Question 17.

Define negative acceleration?

Answer:

It is a Negative acceleration as the moving body reduces its velocity due to applied force, denoted by “-a”.