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## Karnataka State Syllabus Class 8 Science Important Questions Chapter 8 Describing Motion

Question 1.

When is an object said to be in a state of motion?

Answer:

An object is said to be in motion if its position changes with time with reference to a fixed frame.

Question 2.

When do we say that an object is at rest?

Answer:

An object is said to be at rest if its position does not change with time in relation to another object.

Question 3.

The concepts of rest and motion are relative. Explain this statement with an example.

Answer:

Consider a situation in which a person is sitting inside a moving train. This person is in a state of motion with respect to people standing on the platform. However, the person is at rest with respect to a fellow passenger. This shows that the concepts of rest and motion are relative.

Question 4.

Give an illustration to show that motion is relative.

Answer:

The daily motion of the sun from east to west demonstrates that motion is relative. In fact, the earth rotates on its own axis from west to east. If we imagine that the earth is at rest, then, the sun appears to move from east to west.

Similarly, when we are travelling in a speeding bus, the trees and the road appear to move backwards. This again shows that motion is relative.

Question 5.

Is it possible for an object to be at rest with respect to one object and in motion with respect to another object? Explain.

Answer:

Yes, this is possible. A star appears stationary with respect to another star in its neighbourhood. The same star appears to be in a state of motion with respect to the earth.

Question 6.

What is scalar quantity?

Answer:

A scalar quantity is a physical quantity which requires only magnitude for its description.

Question 7.

What is vector quantity?

Answer:

A vector quantity is a physical quantity which requires both magnitude and direction for its description.

Question 8.

What is meant by ‘distance travelled’ by an object?

Answer:

When an object moves from one point to another, the length of the actual path traversed by it is called the distance travelled.

Question 9.

Define displacement.

Answer:

When an object moves from one position to another, the least distance between its initial position and final position is called displacement.

Question 10.

Under what condition does the displacement of an object become equal to the distance travelled?

Answer:

When an object moves along a straight line, the displacement is equal to the distance travelled by it.

Question 11.

What is the S.I. unit of ‘displacement’ and ‘distance travelled’?

Answer:

Both displacement and distance travelled have the same unit. The S.l. unit of displacement and distance travelled is ‘metre’ (m).

Question 12.

There is a square field ABCD. An object starts from A and reaches the opposite corner C through B. If the length of its side is 20 m, what is the displacement and the distance travelled?

Answer:

Displacement = AC

(AC)^{2} = (AB)^{2} + (BC)^{2}

= (20)^{2} + (20)^{2}

= 400 + 400

AC = \(\sqrt{800}\)

= 28.28 m

Distance travelled = AB + BC

= 20 m + 20 m

= 40 m

Question 13.

An object travels from AtoB and then from B to C as shown in the figure. What is the distance travelled? What is the displacement?

Answer:

Distance travelled = AB + BC

= 3 m + 4 m

= 7 m

Displacement = AC = 5 m from A to C.

Question 14.

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

Answer:

Distance travelled = \(\frac{1}{2}\) × Circumference

= \(\frac{1}{2}\) × 2 × π × r

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

= 22 m

Displacement = Diameter

= 2 × radius = 2 × 7

= 14 m.

Question 15.

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

Answer:

The object has started from a point and has come back to the same (original) point after moving in a circular path. Therefore, the displacement of the object is zero.

The distance travelled = Circumference of the path

= 2πr

= 2 × \(\frac{22}{7}\) × 3.5 m

= 22 m.

Question 16.

The least distance between P and Q is 10 m. An object takes path 1 and another object takes path 2 to reach from P to Q as shown in the figure. What is the displacement of the object in each case?

Answer:

Both the objects have changed position from P to Q. Therefore, their displacement is the least distance between P and Q. Thus, the displacement of both the objects is 10 m.This shows that the displacement is independent of the path length.

Question 17.

Distinguish between distance travelled and displacement.

Answer:

Displacement | Distance travelled |

1. Displacement is the shortest distance between the initial position and final position of a moving object. | Distance travelled is the length of the actual path covered by an object. |

2. Displacement has both quantity and direction. | Distance travelled has only quantity and no direction. |

Question 18.

How are motions of objects classified?

Answer:

Motions of objects are broadly classified as

- Uniform motion
- Non-uniform motion.

Question 19.

What is uniform motion? Explain with an example.

Answer:

An object is said to be in uniform motion if it covers equal distances in equal intervals of time, however small the intervals may be.

Consider an object moving from A to B along a straight line path. Let the object cover a distance of 16 m every second during its journey from A to B. Now, the object is said to be in uniform motion if it travels 8 m in every 0.5 s, 4 m in every 0.25 s, and so on. Here, the object is moving along a straight line covering equal distances in equal intervals of time. Hence, it is a case of uniform motion.

Question 20.

What is non-uniform motion? Explain with an example.

Answer:

An object is said to be in non-uniform motion if it travels unequal distances in equal intervals of time.

Consider an object travelling from A to B. Let the object cover 8 m during the first 1 second, 4 m in the next 1 second, 9 m during the next 1 second, 7 m during the next 1 second and so on.

Now, the object is traversing unequal distances in equal intervals of time. This is an instance of non-uniform motion. Non-uniform motion is sometimes referred to as variable motion.

Question 21.

Explain the term ‘speed’.

Answer:

We see a number of objects in a state of motion. All objects do not travel at the same rate. Different objects travel at different rates. Some are fast while others are slow. The fastness of motion of an object is expressed by the term ‘speed’. Speed of an object is defined as the distance travelled by the object in unit time. We can obtain the speed of an object by dividing the distance travelled by the time taken by the object to travel that distance.

Speed = \(\frac{\text { Distance travelled }}{\text { Time taken }}\)

When an object is in a state of uniform motion, we can use this formula to find its speed.

Question 22.

Why is it necessary to use the concept of ‘average speed’? How is the ‘average speed’ of an object calculated?

Answer:

Generally objects arc in a state of non-uniform motion. Therefore, it is necessary to use the concept of ‘average speed’. Average speed of an object is calculated by dividing the total distance travelled by the object by the total time taken.

Question 23.

What are uniform and non-uniform speeds?

Answer:

If an object covers equal distances in equal intervals of time, it is said to have uniform speed. If an object covers unequal distances in equal intervals of time, it is said to have non-uniform speed. It is also called average speed because the speed will not be uniform throughout.

Question 24.

What is the S.I. unit of speed?

Answer:

The S.I. unit of speed is ‘metre per second’. It is written as m s^{-1}.

Question 25.

An object travels a distance of 80 km in 5 hours. Calculate its average speed.

Answer:

Total distance travelled d = 80 km = 80 × 1000 m = 80000 m

Time taken to travel the distance t = 5h = 5 × 3600 s = 18000 s

Average speed of the object = ?

Question 26.

An object travels 80 m during the first 10 seconds and 100 m during the next 10 s. Calculate the average speed of the object.

Answer:

Total distance travelled by the object d = 80 m + 100 m = 180 m

Total time taken to travel 180 m t = 10 s + 10 s = 20 s

Average speed of the object = ?

Question 27.

A bus covers a distance of 30 km in one hour. In the next one hour, it 1covers a distance of 50 km. Calculate the average speed of the bus.

Answer:

Total distance travelled by the bus d = 30 km + 50 km = 80 km

Total time taken to travel t = 1 h + 1 h = 2 h

Average speed of the bus = ?

Question 28.

An object covers a distance oflm during the first 30 s. It further covers a distance of 4 m in the next 15 s. Calculate the total distance travelled and total time taken. Calculate the average speed of the object.

Answer:

Total distance travelled by the object d= 1 m + 4m = 5m

Total time taken to travel t = 30 s + 15 s = 45 s

Average speed of the object = ?

Question 29.

A bus travels front A to B at a speed of 40 km per hour. It travels back from B to A at a speed of

60 km per hour. What is the average speed of the bus during its overall journey?

Answer:

Speed = \(\frac{\text { Distance travelled }}{\text { Time }}\) or Time = \(\frac{\text { Distance travelled }}{\text { Speed }}\)

Hence time taken to cover a distance from A to B at a speed of 40 km h^{-1} is

= \(\frac{\text { Distance travelled from A to B }}{\text { Speed }}\)

In short, T_{1} = \(\frac{d}{40}\), speed is 40 km hour^{-1}, ‘d’ is the distance.

Similarly, time taken to cover a distance from B to A at a speed of 60 km h^{-1} is

= \(\frac{\text { Distance travelled from } B \text { to } A}{\text { Speed }}\)

In short, T_{2} = \(\frac{d}{60}\), speed is 60 kmh^{-1}, ‘d’ is the distance.

Question 30.

An object moves from AtoB covering a distance of 100 km with an average speed of 50 km hr^{-1}. It further moves from B to C covering a distance of 180 km with a speed of 60 km hr^{-1}. What is the average speed of the object?

Answer:

Total distance travelled by the object d = 100 km + 180 km = 280 km

Time taken to travel from A to B = \(\frac{100 \mathrm{km}}{50 \mathrm{km} \mathrm{h}^{-1}}\) = 2h

Time taken to travel from B to C = \(\frac{180 \mathrm{km}}{60 \mathrm{km} \mathrm{h}^{-1}}\) = 3h

Total time taken to travel = 2h + 3h = 5h

Average speed of the object = ?

Question 31.

Define the term ‘velocity’. How is it different from speed?

Answer:

For an object moving along a straight line, the distance travelled by the object in unit time is called velocity. Velocity is speed in a definite direction.

Both speed and velocity ‘are a measure of rate of motion of an object. However, speed has only quantity but no direction. Velocity has both quantity and direction. The velocity of an object can be obtained by dividing the displacement by time.

Question 32.

A bus is travelling from Dharwad to Bijapur. To describe the rate of motion of the bus, is it appropriate to use the term ‘speed’ or ‘velocity’? Why?

Answer:

For a bus travelling from Dharwad to Bijapur, it is appropriate to use the term ‘average speed’. This is because the object is covering unequal distances in equal intervals of time. Further, the direction of the motion of the bus keeps on changing. This means the bus does not move in a fixed direction. Therefore, ‘average speed’ is appropriate to describe the rate of motion of the bus.

Question 33.

Define ‘average velocity’.

Answer:

For an object moving along a straight line in non-uniform motion, the average velocity is defined as the total distance travelled in a unit of time. It is calculated by dividing the total distance travelled by the total time taken.

Average velocity = \(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)

Question 34.

What is the S.I. unit of velocity?

Answer:

The S.I. unit of velocity is the same as that of speed. It is called ‘metre per second’. It is written as m s^{-1}.

Question 35.

An object is moving in a circular path of radius 7 m covering equal distances time. It starts from A and moves along the circular path and reaches B in 2 of the object?

Answer:

Radius of the path = 7 m

Distance travelled = πr

= \(\frac{22}{7}\) × 7 m

= 22 m

Time taken = 2s

Question 36.

Distinguish between speed and velocity.

Answer:

Speed | Velocity |

1. The distance travelled by an object in unit time is called speed. | The distance travelled by an object in unit time in a definite direction is called velocity. |

2. Speed has only quantity and no direction. | Velocity has both quantity and direction. |

Question 37.

Under which circumstance do we calculate average speed and average velocity?

Answer:

Average speed and average velocity are calculated when the object under consideration is in non-uniform motion.

Question 38.

To describe the motion of a swirling stone rotating at constant speed over our head, do we use uniform speed or uniform velocity? Why?

Answer:

The motion of a stone swirling over the head with constant speed is best described in terms of uniform speed and not uniform velocity. Since the stone is changing its direction continually, its motion is an instance of non-uniform motion. Hence, we cannot use ‘uniform velocity’ in this situation.

Question 39.

What is distance-time graph?

Answer:

A graph obtained by plotting the ‘distance travelled’ along the Y-axis and the ‘time taken’ along the X-axis is called distance-time graph. It is simply called d-t

Question 40.

How do you interpret a distance-time graph?

Answer:

1. Uniform Motion:

2. Uniformly Accelerated Motion:

The distance-time graph of an object in uniform motion is a straight line. The distance-time graph of an object in non-uniform motion is a curved line.

Question 41.

What are the uses of distance-time graph?

Answer:

From a distance-time graph, we can know instantly whether the object is in uniform motion or non-uniform motion. We can also know the distance travelled by the object from its initial position. It is also possible to find out the distance travelled by the object between any two intervals of time.

We can know from the graph the time intervals during which the object was slowest or fastest. We can use the graph to find the speed of the object in case of uniform motion and average speed of the object in case of non-uniform motion.

Question 42.

Study the distance-time graph given aside and answer the following questions:

- What is the distance travelled by the object in 20 s?
- In how much time has the object traversed 5 m?
- What is the average speed of the object?

Answer:

- The object has travelled a distance of 4 m in 20 second
- The object has traversed 5 m in 30 s.
- Average speed = \(\frac{6 \mathrm{m}}{40 \mathrm{s}}\) = 0.15ms
^{-1}.

Question 43.

When do we say that an object is moving with uniform speed?

Answer:

An object is said to be moving with uniform speed if it covers equal distances in equal intervals of time, howsoever small the interval may be.

Question 44.

When do we say that an object is moving with uniform velocity?

Answer:

An object is said to be moving with uniform velocity if it is moving along a straight line covering equal distances in equal intervals of time.

Question 45.

Define ‘acceleration’. State its S.I. unit.

Answer:

The rate of change of velocity of an object is called acceleration. In other words, the change in the velocity of an object in unit time is called acceleration. The S.I. unit of acceleration is called ‘metre per second square’. It is written as m s^{-2}.

Question 46.

How do you calculate the acceleration of an object?

Answer:

The acceleration of an object is calculated by dividing the change in velocity by the time taken.

Acceleration = \(\frac{\text { Change in velocity }}{\text { Time taken }}\)

Question 47.

Give the formula to find the acceleration of an object when its initial velocity and final velocity are known.

Answer:

Consider an object whose velocity changes from u to v in time t. Let its acceleration be a.

Change in velocity = v – u

Question 48.

An object at rest gains a velocity of 40 m s^{-1} in 5 s. Find its acceleration.

Answer:

Initial velocity of the object u = 0

Final velocity of the object v = 40 m s^{-1}

Time taken t = 5 s

Acceleration a = ?

Question 49.

An object A is moving anti-clockwise in a circular path with centre O. B and C are two points on the circular path. What is the direction of its velocity at points B and C? Show this by a diagram.

Answer:

For an object in circular motion, the direction of motion of the object (the direction of its velocity) at any point is given by the tangent drawn to the path at that point.

The direction of motion of the object at points B and C are shown in the diagram.

Question 50.

Distinguish between ‘uniform acceleration’and ‘non-uniform acceleration’.

Answer:

An object is said to be in uniform acceleration if its velocity changes by equal amounts in equal intervals of time. An object is said to be in non-uniform acceleration if its velocity changes by unequal amounts in equal intervals of time.

Question 51.

If an object is moving with uniform speed in a given direction, then, the acceleration will be zero. Why?

Answer:

Change in velocity can happen when the speed of an object changes, direction of motion of the object changes, or both change. Since none of these are happening in the given situation, there is no change in velocity. Therefore, the acceleration is zero.

Question 52.

While expressing acceleration, the time is mentioned two times. Why?

Answer:

Acceleration is obtained by dividing the change in velocity by the time taken. The unit of velocity is m s^{-1}. The unit of time is ‘s’ (second). Therefore, the unit of time is mentioned twice while expressing acceleration.

Question 53.

An object changes its velocity from 30 m s^{-1} to 40 m s^{-1} in 2 second. What is the acceleration of the object?

Answer:

Initial velocity u = 30 m s^{-1}; Final velocity v = 40 m s^{-1}; Time taken t = 2 s; Acceleration a = ?

v = u + at

Question 54.

The average speed of a bus is 40 km hr^{1}. What do you understand by this statement?

Answer:

The statement means that the bus is in non-uniform motion and it is on an average covering a distance of 40 km every hour.

Question 55.

An object at rest starts moving. It covers a distance of 2 m in 1 second. It covers a further distance of 5 m in two second in the same direction. What is its average velocity and acceleration?

Answer:

Total distance travelled d = 2m + 5m = 7m

Total time taken r=ls + 2s = 3s

Average velocity v = ?

Velocity v = \(\frac{d}{t}=\frac{7 m}{3 s}\) = 2.33 m s^{-1}

Initial velocity of the object u = 0

Final velocity of the object v = 2.33 m s^{-1}

Time taken t = 3 s

Question 56.

What is ‘velocity-time’ graph? How is it different from ‘speed-time’ graph?

Answer:

A graph obtained by plotting ‘velocity’ along the Y-axis and ‘time taken’ along the X-axis is called velocity-time graph. It is also called v-t graph. This type of graph is generally used when the object is moving along a straight line.

A graph obtained by plotting ‘speed’ along the Y-axis and ‘time taken’ along the X-axis is called speed-time graph. This type of graph is used when the object is not moving in any one particular direction.

Since the numerical value of velocity is speed itself, there is no difference between speed-time graph and velocity-time graph.

Question 57.

An object moves at a uniform velocity of 6 m s^{-1} for 50 s. Draw a v-t graph to represent this. What is the acceleration of the object?

Answer:

The acceleration of the object is zero. This is because the object is in uniform motion and hence there is no change in the velocity of the object.

Question 58.

Write the three equations of motion. Explain the meaning of the symbols used in them.

Answer:

The three equations of motion are:

- v = u + at
- s = ut + \(\frac{1}{2}\) at
^{2} - v
^{2}= u^{2}+ 2as

where u = initial velocity, v = velocity after time t, t = time taken, s = distance travelled, a = acceleration.

Question 59.

An object at rest starts moving and attains a velocity of10 m s^{-1} after 5 s. What is the acceleration?

Answer:

Initial velocity u = 0 (rest)

Final velocity v = 10 m s^{-1}

Time interval t = 5 s

Question 60.

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

Answer:

Initial velocity u = 10 ms^{-1}

Final velocity v = 0

Time interval t = 5s

a = \(\frac{0-10}{5}\) = – 2ms^{-2}

The negative sign indicates that acceleration is against the direction of motion.

Question 61.

An object at rest starts moving with uniform acceleration of 1 m s^{-2}. Calculate the distance travelled by it in 4 second.

Answer:

Initial velocity u = 0

Acceleration a = 1 m s^{-2}

Time taken t = 4 s

Distance travelled s =?

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

= 0 × 4 + \(\frac{1}{2}\) × 1 × 4^{2}

= 0 + \(\frac{1}{2}\) × 1 × 16 = 8m.

Question 62.

An object starts from rest and moves with uniform acceleration of 4 m s^{-2}. What is the velocity of the object after it has traversed 0.5 m?

Answer:

Initial velocity u = 0

Acceleration a = 4 m s^{-2}

Distance travelled 5 = 0.5 m

Final velocity v = ?

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

= 0^{2} + 2 × 4 m s^{-2} × 0.5 m = 4

v = \(\sqrt{4}\) = 2 m s^{-1}.

Question 63.

A vehicle starts from rest and moves with uniform acceleration of 5 in s^{-2}. What is the distance travelled by it in 6 minutes?

Answer:

Initial velocity u = 0

Acceleration a = 5 m s^{-2}

Time taken t = 6 minute = 6 × 60 s = 360 s

Distance travelled s = ?

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

s = 0 + \(\frac{1}{2}\) × 5 ms^{-2} × 360 s

= 324000 m = 324 km.

Multiple Choice Questions

Question 1.

Uniform circular motion is called accelerated motion because

(a) speed changes

(b) direction of motion changes continuously

(c) velocity remains the same

(d) direction of motion does not change.

Answer:

(b) direction of motion changes continuously

Question 2.

A cricketer hits a six. The cricket ball moves up with a velocity of 2 m s^{-1} and then falls down Its initial velocity during its downward journey is

(a) 1ms^{-1}

(b) 1 m s^{-2}

(c) zero

(d) 2ms^{-1}

Answer:

(c) zero

Question 3.

The S. I. unit of displacement is

(a) metre

(b) m s^{-2}

(c) km h^{-1}

(d) m s^{-1}

Answer:

(a) metre

Question 4.

When an object is travelling with constant speed,

(a) its position does not change with time

(b) it covers equal distances in equal intervals of time

(c) its acceleration is zero

(d) it is moving along a straight line.

Answer:

(b) it covers equal distances in equal intervals of time

Question 5.

The distance-time graph of an object in non-uniform motion is

(a) a straight line parallel to the X-axis

(b) a straight line parallel to the Y-axis

(c) a straight line inclined to the X-axis

(d) a curved line

Answer:

(d) a curved line

Question 6.

A car travels from A to B at a speed of 20 km hr1 and returns from B to A at an average speed of 30 km h^{-1}. The average speed for the entire journey is

(a) 5 km h^{-1}

(b) 24 km h^{-1}

(c) 25 km h^{-1}

(d) 50 km h^{-1}

Answer:

(c) 25 km h^{-1}

Question 7.

The numerical ratio of displacement to distance is

(a) always less than 1.

(b) always equal to 1.

(c) always more than 1.

(d) equal to or less than 1.

Answer:

(d) equal to or less than 1.

Question 8.

The acceleration of an object moving with uniform velocity is always

(a) equal to 1

(b) zero

(c) -1

(d) more than 1

Answer:

(b) zero

Question 9.

In the v – t graph given aside, the distance travelled by the object during the first 10 s is

(a) 30 m

(b) 200 m

(c) 100 m

(d) 300 m

Answer:

(c) 100 m

Question 10.

Which one of the following has both quantity and direction?

(a) Distance travelled

(b) Displacement

(c) Speed

(d) Time

Answer:

(b) Displacement

Question 11.

Speed and velocity are different because,

(a) both magnitude and direction are the same

(b) magnitude may be same but direction may be different

(c) both have same magnitude

(d) both have same direction.

Answer:

(b) magnitude may be same but direction may be different

Question 12.

If an object moving in a circular path starts from a point ‘A’ and returns to ‘A’, its distance travelled will be,

(a) πr^{2}

(b) 2πr

(c) πr

(d) πr^{3}

Answer:

(b) 2πr

Question 13.

In the diagram, if an object starts from ‘A’, reaches ‘B’ and then reaches ‘C’, its displacement will be

(a) 17 m

(b) 30 m

(c) 13 m

(d) 12 m

Answer:

(c) 13 m

Fill In The Blanks

1. The distance travelled by an object in unit time is called speed

2. The S.l. unit of acceleration is m s^{-2}

3. Velocity has both speed and direction

4. Displacement of an object which moves from A and comes back to A is zero

5. The rate of change of velocity is called acceleration

6. Speed in a definite direction is called velocity

7. In velocity-time graph, velocity is taken along X-axis

8. An object that travels equal distances in equal intervals of time is said to be in a state of uniform motion