Class 9 Science Chapter 7 Important Questions of Motion. All the extra questions are important for school exams as well as class tests. Class 9 Science Extra Question answers covers the entire chapter 7 from NCERT Books. There are short answers as well as long answers questions given in the form of important questions. All questions are strictly confined to NCERT Books only.

## Class 9 Science Chapter 7 Important Questions

### Class 9 Science Chapter 7 Important Extra Questions Set – 1

### The phenomenon of motion was placed on a sound scientific footing by two scientists. Write their names.

Galileo Galilei and Isaac Newton.

### Are rest and motion absolute or relative terms?

They are relative terms.

### Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. What is the net displacement and distance travelled by the ball?

Displacement is zero. Distance is twice the distance between point P and Q.

### Which speed is greater: 30 m/s or 30 km/h?

30 m/s

### What do you mean by 2m/s?

The velocity of the body increases by 2m/s after every second.

Class: 9 | Science |

Chapter: 7 | Motion |

Contents: | Important Questions with Answers |

#### The Instantaneous Velocity

Instantaneous velocity is the velocity of the body at any particular instant during its motion. For example the instantaneous velocity of a motorcycle at a particular instant is 40kmh is it is moving 40kmh at that particular instant. It is measured by the speedometers on the vehicles.

### Class 9 Science Chapter 7 Important Extra Questions Set – 2

### Can uniform linear motion be accelerated?

No

### Define one radian.

It is the angle which is subtended at the centre by an arc having a length equal to the radius of the circle.

### What is relation between linear velocity an angular velocity?

Linear velocity = Angular velocity × Radius of circular path.

### Give an example when we infer the motion indirectly.

We infer the motion of air by observing the movement of dust particles or leaves and branches of trees, or simply by feeling the blowing air on our face.

### What is essential to describe the position of an object?

We need to specify a reference point called the origin.

### Class 9 Science Chapter 7 Important Extra Questions Set – 3

### What is the simplest type of motion?

Motion in a straight line.

### What indicates the motion of the earth?

The phenomenon like day and night indicates the motion of the earth.

### If the displacement of a body is zero, is it necessary that the distance covered by it is also zero?

No, when the body comes back to the same position after travelling a distance, its displacement is zero though it has travelled some distance.

### Can the displacement be greater than the distance travelled by an object?

No, it is always either be equal to or less than the distance travelled by the object.

### When do the distance and displacement of a moving object have the same magnitude?

The magnitude of distance and displacement of a moving object are same when the object moves along the same straight line in the same fixed direction.

### Class 9 Science Chapter 7 Important Extra Questions Set – 4

### Does the speedometer of a car measure its average speed?

No, it measures its instantaneous speed.

### A body is moving with a velocity of 10m/s. If the motion is uniform, what will be the velocity after 10s.

As the motion is uniform; the velocity remains 10m/s after 10s.

### Can a body have constant speed but variable velocity?

Yes, example a body in uniform circular motion has constant speed but due to the change in the direction of motion, its velocity changes at every point.

### When is the acceleration taken as negative?

Acceleration is taken as negative if it is in the direction of velocity.

### What is uniform acceleration?

Acceleration of an object is said to be uniform it is travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time. For example, motion of a freely falling body.

##### The uses of Distance-time Graph

The various uses of a distance-time graph are as follows:

- It tells us about the position of the body at any instant of time.
- From the graph we can see the distance covered by in the body during a particular intervals of time.
- It also gives us information about the velocity of the body at any instant of time.

#### Negative Acceleration or Retardation

If the velocity of a body decreases with time, then its final velocity is less than the initial velocity and thus its acceleration is negative. Negative acceleration is called retardation is called retardation or deceleration. For example when brakes are applied to a moving truck, its velocity gradually decreases. In other word it is under retardation.

### Class 9 Science Chapter 7 Important Extra Questions Set – 5

### Give an example of Non-uniform acceleration.

A car is travelling along a straight road increases its speed by unequal amounts in equal intervals of time.

In this case distance travelled by the object is directly proportional to the time taken.

### What would be acceleration of a body is its velocity-time graph is a line parallel to the time axis?

Zero as the body possesses uniform velocity.

### Is the motion of a body uniform or accelerated if it goes round the sun with constant speed in a circular orbit?

It is accelerated as its velocity changes due to change in direction.

### Give an example of a body which may appear to be moving for one person and stationary for the other.

The passenger in a moving bus observe that the trees, building as well as the people on the roadside appear to be moving backwards. Similarly, a person standing on the roadside observe that the bus is moving in forward direction. But at the same time each passenger in a moving bus or train observes his fellow passengers sitting and not moving. Thus we can tell that motion is relative.

### Class 9 Science Chapter 7 Important Extra Questions Set – 6

### What is the difference between uniform velocity and non-uniform velocity?

Uniform velocity: An object with uniform velocity covers equal distances in equal distances in equal intervals of time in a specified direction example an object moving with speed of 40 kmh toward towards west has uniform velocity.

Non-uniform velocity: When an object covers unequal distances in equal intervals of time in a specified direction or if the direction of motion changes it is said to be moving with a non-uniform or variable velocity for example revolving fan at a constant speed has variable velocity.

### How will the equations of motion for an object moving with a uniform velocity change?

Acceleration a= 0, v= u

So, the equations of motion will become

s = ut

v² – u² = 0

### Express average velocity when the velocity of a body changes at a non-uniform rate and a uniform rate.

When the velocity of a body changes at a non-uniform rate its average velocity is found by dividing the net displacement covered by the total time taken.

Average velocity = (Net dispalcement )/(Total time taken )

In case the velocity of a body changes at a uniform rate than the average velocity is given by arithmetic mean of initial velocity and final velocity for a given period of time.

Average velocity = (Intitaial velocity+Final vellocity )/(2 )

### How will you show that slope of displacement-time graph gives velocity of the body?

The adjoining figure shows the displacement-time graph for a body moving with uniform velocity. Clearly it covers distance S₁ and S₂ at time t₁ and t₂ respectively.

Slope of line PQ = tanφ = QR/PR

= (S₂-S₁)/(t₂-t₁) = (Displacement )/(Time )

As (Displacement )/(Time ) is velocity so the slope of the distance-time graph gives velocity of the body.

### What are the characteristic of distance-time graph for an object an object moving with a non-uniform speed?

The characteristic of distance of distance-time graph for a non-uniform speed are:

(i) It is always a curve (parabola)

(ii) The speed of the moving object at any point is given by the slope of the tangent to the curve at that point.

##### Average Speed

Average speed is defined as the average distance travelled per unit time and the is obtained by dividing the total distance travelled by the total time taken.

The unit average speed is the same as that of the speed, that is m/s.

### Class 9 Science Chapter 7 Important Extra Questions Set – 7

### Given below is the velocity-time graph for the motion of the car. What does the nature of the graph show? Also find the acceleration of the car.

The nature of the graph shows that changes by equal amounts in equal intervals of time. For a uniformly accelerated motion, velocity-time graph is always a straight line.

As we know accelerated is equal to the slope of the graph.

a = BC/AC or a = (v₂ – v₁)/(t₂ – t₁)

a = (10.0 – 7.5)ms/(20 – 15)s

a = 2.5 ms/5s

a = 0.5 ms

### With the help of a graph derive the relation v = u + at.

Consider the velocity-time graph of an object that move under uniform acceleration.

From this graph we can see that initial velocity of an object (at point A) is u and then it increases to v (at point B) in time t, The velocity changes at a uniform rate a. The line BC and BE are drawn from point B on the time and the velocity axes respectively so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC.BD = BC – CD, represents the changes in velocity in the time intervals t.

If we draw AD parallel to OC we observe that

BC = BD + DC = BD + OA

Substituting BC with v and OA with u

We get v = BD + u

Or BD = v –u …(1)

Thus from the given velocity time graph the acceleration of the object is given by

a = (Chnage in velocity )/(Time taken )

= BD/AD = BD/OC

Substituting, OC with t we get

a = BD/t or BD = at …(2)

From equation (1) and (2) we have

v – u = at or v = u + at

### Deduce the following equations of motion: (i) s = ut + 1/2 at² (ii) v² = u² + 2as

(i) Consider a body which starts with initial velocity u and due to uniform acceleration a, its final velocity becomes v after time t. Then its average velocity is given by

Average velocity = (Initial velocity+FInal velocity)/2 = (u + v)/2

The distance covered by the body in time t is given by

Distances s = Average velocity × Time

Or s = (u + v)/2 × t or s = (u + (u + at))/2 × t

s = (2ut + at²)/2 or s = ut + 1/2at2

(ii) We know that

s = ut + 1/2 at²

a = (v – u)/(t )

t = (v – u)/(a )

Putting the value of t in (1) we have

s = u((v – u)/a) + 1/2a((v – u)/a)²

s= (uv – u²)/a + (v² + u² – 2uv)/2a

or 2as = 2uv -2u² + v² -2uv

or v² – u² = 2as

### Obtain a relationship for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.

Using the equation of motion

s = ut + 1/2 at²

Distance travelled in 5 seconds s = u× 5 + 1/2 a×5²

Or s = 5u + 25/2 a

Similarly, distance travelled in 4 second, s’ = 4u 16/2 a

Distance travelled in the internal between 4th and 5th seconds

= ( s – s’) (u + 9/2 a) m

### The driver of train A travelling at a speed of 54 km/h applies brakes and retards the train uniformly. The train stops in 5 seconds. Another train B is travelling on the parallel with a speed of 3km/h. Its driver applies the break and the train retards uniformly: train B stops in 10 seconds. Plot speed-time graphs for both the trains on the same axis. Which of the trains travelled farther after the brakes were applied?

For train A, the initial velocity.

u = 54 km/h = 54 = 5/18 = 15 ms⁻¹

Final velocity, v = 0 and time, t = 5s

For train B, u = 36 kmh = 36 × 5/18 = 10 ms⁻¹

v = 0; t = 10s

Distance travelled by train A

= Area under straight line graph RS

= Area of ORS

= 1/2 × OR × OS = 1/2 × 15 ms⁻¹ × 5 s = 37.5 m

Distance travelled by train B = Area under PQ = Area of OPQ

= 1/2 × OP × OQ = 1/2 × 10 ms⁻¹ × 10 s = 50 m

Thus, train B travelled farther after the brakes were applied

### Class 9 Science Chapter 7 Important Extra Questions Set – 8

### A body starts to slide over a horizontal surface with an initial velocity of 0.5m/s. Due to friction, its velocity decreases at the rate 0.05 m/s². How much time will it take for the body to stop?

Initial velocity, u = 0.5 m/s

Find velocity, v = 0

Acceleration, a = 0.05 m/s²

Now from the first equation of the motion,

v = u + at

or, 0 = 0.5 + (-0.05t) 0.5 = 0.05t,

t = 0.5/0.05 = 10s

Thus the body will take 10s to stop.

### A particle moves in a circle with O as center and AO = OB = 5cm, as radius. It starts from A. Calculate: (a) the distance covered and (b) the displacement, when it reaches B.

(a) Distance covered = π × OA = π × 5 = 5π cm

(b) Displacement = 2 × OB

= 2 × 5 = 10 cm along AB

### A cheetah is the fastest land animal and can achieve a peak velocity of 100km/h up to distances less than 500m. If a cheetah spots its prey at a distance of 100m what is the minimum time it will take to get its prey if the average velocity attained by it is 90km/h?

Average velocity = 90 km/h = (90 km)/1h

= (90×1000 m)/(60×60 s) =25 m/s

Also Average velocity = (Displacement )/(Time taken )

Cheetah moves in a straight line displacement is equal to 100m

Therefore time taken = 100/25 = 4s

### The brakes applied to a car produce an acceleration of 6 ms-2 in the opposite direction to the motion. If the car takes 2s to stop after the application to brakes, calculates the distance it travels during this time.

We have been given

a = -6ms⁻²; t = 2s and v= 0ms⁻¹

We know that

v = u + at

0 = u + (-6ms⁻²) × 2s or u= 12 ms⁻¹

We get s = ut + 1/2 at²

= (12 ms⁻¹) × (2s) + 1/2 (-6ms⁻² ) (2s)²

= 24 m – 12m = 12m

Thus the car will move 12 m before it stops after the application of brakes.

### A car starts from rest and moves along the x axis with constant acceleration 5ms-2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it. Started from the rest?

Initial velocity, u = 0

Using s = ut + 1/2 at²

The distance travelled in first 8s,

s₁ = 0 + 1/2 × 5 ×8² = 160m

At this point the velocity v = u + at

= 0 + 5 × 40 ms⁻¹

So, the distance covered in last four seconds

s₂ = 40 × 4 = 160 m

Hence total distance, s = s₁ + s₂

= 160 m + 160 m = 320 m

###### How can we describe the location of an object?

**Answer**: To describe the position of an object we need to specify a reference point called origin. For example suppose that a library in a city is 2km north of the railway station. We have specified the position of the library with respect to the railway station i.e. in this case the railway station acts as the reference point.

##### Difference between Distance and Displacement

Distance | Displacement |
---|---|

1. It is the length of the actual path covered by an object irrespective of its direction of motion. | 1. Displacement is the shortest distance between the initial and final positions of an object in a given direction. |

2. Distance is a scaler quantity. | 2. Displacement is a vector quantity. |

3. Distance covered can never be negative. It is always positive or zero. | 3. Displacement may be positive, negative or zero. |

4. Distance between two given points may be same or different path chosen. | 4. Displacement between two given points is always the same. |