# Motion Chapter 8 Class 9 Science NCERT Textbook PDF

NCERT Solutions for Class 9 Science Chapter 8‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 9 Science Chapter 8 Exercise Solution’ using the download button.

### Chapter 8: Motion

In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins and arteries and cars move.

Atoms, molecules, planets, stars, and galaxies are all in motion. We often perceive an object to be in motion when its position changes with time.

However, there are situations where the motion is inferred through indirect evidence.

For example, we infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees.

What causes the phenomena of sunrise, sunset, and changing of seasons? Is it due to the motion of the earth? If it is true, why don’t we directly perceive the motion of the earth?

An object may appear to be moving for one person and stationary for some other. For
the passengers in a moving bus, the roadside trees appear to be moving backward.

A person standing on the roadside perceives the bus along with the passengers as moving. However, a passenger inside the bus sees his fellow passengers at rest. What do these observations indicate?

Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate.

There may be situations involving a combination of these. In this chapter, we shall first learn to describe the motion of objects along a straight line.

We shall also learn to express such motions through simple equations and graphs. Later, we shall discuss ways of describing circular motion.

#### 8.1 Describing Motion

We describe the location of an object by specifying a reference point. Let us understand this by an example.

Let us assume that a school in a village is 2 km north of the railway station. We have specified the position of the school with respect to the railway station.

In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience.

Therefore, to describe the position of an object we need to specify a reference point called the origin.

### 8.1.2 Uniform Motion And Nonuniform Motion

Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second, and 5 m in the fourth second.

In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion.

The time interval in this motion should be small. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example, when a car is moving on a crowded street or a person is jogging in a park. These are some instances of non-uniform motion.

#### 8.2.1 Speed With Direction

The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed.

The quantity that specifies both these aspects is called velocity. Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable.

It can be changed by changing the object’s speed, the direction of motion, or both.

When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as we calculate average speed.

### NCERT Solutions Class 9 Science Chapter 8 Motion

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

Solution

Yes, an object which has moved through a distance can have zero displacements if it comes back to its initial position.

Example: If a person jogs in a circular park which is circular and completes one round. His initial and final position is the same.

Hence, his displacement is 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?

Solution

Given,

Side of the given square field = 10m

Hence, the perimeter of a square = 40 m

Time taken by the farmer  to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of 1 m

Now,

Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140 m

So,

The total number of rotations taken by the farmer to cover a distance of 140 m = total distance/perimeter

= 3.5

At this point, let us say the farmer is at point B from the origin O

Therefore,  from Pythagoras theorem, the displacement s = √(102+102)

s = 102

s = 14.14 m

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.

Solution

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 traveled.