**Lesson 2**** Motion**

**9th Science Guide:** Greetings, students! On this page, you will discover the 9th Standard Science Solutions for the unit titled ‘Motion’ We have diligently addressed nearly all the book back questions from the second unit of Class 9 Science, focusing on the chapter Motion. Additionally, we have included some supplementary questions that we anticipate might appear in the final examination.

**Motion Book Back Answers**

Below, you can find the answers for the following sections: “Choose the Correct Answer,” “Fill in the Blanks,” “State Whether True or False,” and so on.

**I. Multiple choice Questions**

**1. The area under a velocity – the time graph represents the**

- Velocity of the moving object.
- Displacement covered by the moving object
- Speed of the moving object
- Acceleration of the moving object

Ans: Displacement covered by the moving object

**2. Which one of the following is most likely not a case of uniform circular motion?**

- Motion of the Earth around the sun
- Motion of a toy train on a circular track.
- Motion of a racing car on a circular track.
- Motion of hours hand on the dial of the clock.

Ans: Motion of a racing car on a circular track.

**3. Which of the following graph represents the uniform motion of a moving particle?**

Ans: b

**4. The centrifugal force is**

- Real force
- The force of reaction of centripetal force
- Virtual force
- Directed towards the center of the circular path.

Ans: Virtual force

**5. Slope of the velocity – time graph – gives**

- Speed
- Displacement
- Distance
- Acceleration

Ans: Acceleration

**6. A body moving with an initial velocity 5ms-1 and accelerates at 2ms-1. Its velocity after 10s is**

- 20ms
^{-1} - 25ms
^{-1} - 5ms
^{-1} - 22.55ms
^{-1}

Ans: 25ms^{-1}

**7. In a 100m race, the winner takes 10s to reach the finishing point. The average speed of the winner is**

- 5ms
^{-1} - 20ms
^{-1} - 40ms
^{-1} - 10ms
^{-1}

Ans: 10ms^{-1}

**8. A car is being driven at a speed of 20ms-1 when brakes are applied to bring it to rest in 5s. the deceleration produced in this case will be**

- +4ms
^{-2} - -4ms
^{-2} - -0.2ms
^{-2} - +0.25ms
^{-2}

Ans: +4ms^{-2}

**9. Unit of acceleration is**

- ms
^{-7} - ms
^{-2} - ms
- ms
^{2}

Ans: ms^{-2}

**10. The force responsible for drying clothes in a washing machine is**

- Centripetal force
- Centrifugal force
- Gravitational force
- Electrostatic force

Ans: Centrifugal force

**11. When a body starts from rest the acceleration of the body after 2 Seconds is ……………. Of its displacement.**

- Half
- Twice
- Four times
- One fourth

Ans: Half

**12. In a 100m race, the winner takes 10s to reach the finishing point. The average speed of the winner is ………………… ms ^{-1}**

- 5
- 10
- 20
- 40

Ans: 10

**II. Fill in the blanks**

**1. Speed is a …………….. quantity whereas velocity is a ……………quantity.**

Ans: Scalar, Vector

**2. The slope of the distance-time graph at any point gives …………**

Ans: Speed

**3. Negative acceleration is called ………….**

Ans: retardation (or) deceleration

**4. Area under a velocity-time graph shows ……………**

Ans: displacement

**5. Consider an object is a rest at position x = 20m, Then its displacement – time graph will be straight line to ……………**

Ans: Parallel

**III. State whether true or false. If false, correct the statement**

**1. The motion of a city bus on a heavy-traffic road is an example for uniform motion.**

Ans: False

Correct Ans: The motion of a city bus on a heavy-traffic road is an example for non-uniform motion.

**2. Acceleration can get a negative value also.**

Ans: True

**3. Distance covered by a particle never becomes zero but displacement becomes zero.**

Ans: True

**4. The velocity—time graph of a particle falling freely under gravity would be a straight**

**line parallel to the x-axis.**

Ans: False

Correct Ans: The velocity – time graph of a particle moving at uniform velocity, would be a straight line parallel to the x-axis.

**5. If the velocity-time graph of a particle is a straight line inclined to X-axis then its displacement – time graph will be a straight line.**

Ans : True

**IV. Assertion and Reason type questions**

**1. Assertion**: The accelerated motion of an object may be due to change in magnitude of velocity or direction or both of them.

** Reason** : Acceleration can be produced only by change in magnitude of the velocity. It does not depend the direction.

- If both assertion and reason are true and reason is the correct explanation of assertion.
- If both assertion and reason are true but reason is not the correct explanation of assertion.
- If assertion is true but reason is false.
- If assertion is false but reason is true

Ans: If assertion is true but reason is false.

**2. Assertion** : The Speedometer of acar ora motor-cycle measures its average speed.

** Reason** ; Average velocity is equal to total displacement divided by total time taken.

- If both assertion and reason are true and reason is the correct explanation of assertion.
- If both assertion and reason are true but reason is not the correct explanation of assertion.
- If assertion is true but reason is false.
- If assertion is false but reason is true

Ans: Assertion is false but reason is true

**3. Assertion** : Displacement of a body may be zero when distance travelled by it is not zero.

** Reason** : The displacement is the shortest distance between initial and final position.

- If both assertion and reason are true and reason is the correct explanation of assertion.
- If both assertion and reason are true but reason is not the correct explanation of assertion.
- If assertion is true but reason is false.
- If assertion is false but reason is true

Ans: Both assertion and reason are true and reason is the correct explanation of assertion

**V. Match the following**

1. Motion of a body covering equal distances in equal interval | |

2. Motion with non-uniform Acceleration | |

3. Constant retardation | |

4. Uniform acceleration. | |

Ans: 1 – D, 2 – C, 3 – A, 4 – B |

**VI. Short Questions & Answers**

**1. Define velocity?**

Velocity is the rate of change of displacement. It is the displacement in unit time.

**2. Distinguish distance and displacement?**

Distance | Displacement |

The actual length of the path traveled by a moving body irrespective of the direction. | The change in position of a moving body in a particular direction. |

It is a scalar quantity. | It is a vector quantity. |

**3. What do you mean by uniform motion?**

An object is said to be in uniform motion if it covers an equal distance in equal intervals of time how so ever bit or small these time intervals may be. A particle is in uniform motion when it moves with constant velocity.

**4. Compare speed and velocity?**

Speed | Velocity |

It is the rate of change of distance with respect to time | It is the rate of change of displacement with respect to time |

It is a scalar quantity having magnitude only | It is a vector quantity having both magnitude and direction |

Speed is velocity without a particular direction | Velocity is the speed in a particular direction |

It is measured in ms^{-1} in the SI system | It is also measured in ms^{–1} in a particular the direction in SI system |

**5. What do you understand about negative acceleration?**

- If the final velocity is less than the initial velocity, the velocity decreases with time and the value of acceleration is negative. It is called negative acceleration.
- Negative acceleration is called retardation (or) deceleration.

**6. Is the uniform circular motion accelerated? Give reasons for your Answer**

Yes, the uniform circular motion is accelerated.

A body is said to be accelerated if the velocity of the body changes either in magnitude (or) in direction. But in this case, the change in velocity is only due to the change in direction.

**7. What is meant by uniform circular motion? Give two examples of uniform circular motion.**

When an object moves with constant speed along a circular path, the motion is called uniform circular motion.

Examples:

- Revolution of the earth around the sun.
- Revolution of the moon around the earth.
- In an atom, an electron moves around the nucleus in a circular path.

**8. What remains constant in uniform circular motion? And what changes are continuously in uniform circular motion?**

- Objects are moving at a constant speed along a circular path.
- The direction changes continuously in a uniform circular motion.

**VIII. Paragraph Questions**

1. **Derive equations of motion by graphical method.**

Equations of motion from velocity-time graph:

The graph shows the change in velocity with time for a uniformly accelerated object. The object starts from point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph.

The initial velocity of the object = u =OD = EA

The final velocity of the object = v = OC = EB

Time = t = OE = DA

Also from the graph, we know that, AB = DC

#### The first equation of motion

By definition, acceleration = change in velocity/time

= (final velocity – initial velocity)/time | |

= (OC – OD) / OE | |

= DC / OE | |

a | = DC / t |

DC | = AB = at |

From the graph EB | = EA + AB |

v = u + at ………………. (1) |

This is the first equation of motion.

#### The second equation of motion

From the graph, the distance covered by the object during time t is given by the area of quadrangle DOEB

s | = area of the quadrangle DOEB |

= area of the rectangle DOEA + area of the triangle DAB | |

= (AE × OE) + (1/2 × AB × DA) | |

s = ut + ½ at^{2} …………………. (2) |

This is the second equation of motion.

#### Third equation of motion

From the graph, the distance covered by the object during time t is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. Then

S | = area of trapezium DOEB |

= ½ × sum of length of parallel side × distance between parallel sides | |

= ½ × (OD + BE) × OE | |

S | = ½ × (u + v) × t |

since a | = (v – u) / t or t = (v – u)/a |

Therefore s | = ½ × (v + u) × (v – u)/a |

2as | = v ^{2} – u^{2} |

v^{2} = u2 + 2 as ………………. (3) |

This is the third equation of motion.

**2. Explain different types of motion.**

(i) **Linear Motion **

The motion of an object along a straight line is known as linear motion. Ex: Car moving on a straight road.

(ii)** Circular Motion **

The motion of an object is a circular path is known as circular motion. Ex: Earth revolving around the sun.

(iii) **Oscillatory Motion **

Repetitive and for motion of an object at regular intervals of time is called oscillatory motion. Ex: Motion of the pendulum of a clock.

(iv) **Random Motion **

The disordered or irregular motion of a body is called random motion. Ex: Movement of fish underwater.

**IX. Exercise Problems**

**1. A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10 ms ^{-2} with what velocity will it strike the ground? After what time will it strike the ground?**

Ans: Here we have

Initial velocity | = 0 |

Distance, s | = 20m |

Acceleration, a | = 10m/s^{2} |

Final velocity, v | = ? |

Time, t | = ? |

a) Calculation of final velocity,

We know that,v^{2} | = u^{2}+2as |

v^{2} | = 0+2~x 10 m/s^{2} x 20m |

v^{2} | = 400 m^{2}/s^{2} |

= √400m^{2} /s^{2} | |

v | = 20 m/s |

b) Calculation of time, t

We know that,v | = ut+at |

20m/s | = 0+10m/s^{2} x t |

t | 20m/s^{2 }/ 20m/s = 2s |

‘. Ball will strike the ground at a velocity of 20 ms^{-1}Time is taken to reach the ground = 2s. |

**2. 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 m and 20 s?**

Ans: Here we have

Diameter | = 200m |

Radius | = 200 m/2 = 100 m |

Time of one rotation | = 40s |

Time after 2m 20 s | 2 x 60 s +20 s= 140 s |

Distance after 140 s | = ? |

Displacement after 140 s | = ? |

Circular track with a diameter of 200m |

We know that, velocity

along a circular path | = circumference/time |

v | = 2mr / 40 s |

v | = 2 x 3.14 x 100 m / 40 s |

v | = 628 m / 40 s |

v | = 15.7 m/s |

a) Distance after 140 s

We know that, distance | = velocity x time |

=> Distance | = 15.7 m/s x 140s |

= 2198m |

b) Displacement after 2 min 20s i.e, in 140s

We know that, distance | = velocity x time |

Since, rotation in 4o s | = 1 |

ஃ Rotation in l s | = 1 / 40 |

ஃ Rotationin 140 s | = 1 / 40 x 140 = 3.5 |

ஃ In 3.5 rotation athlete will be just at the opposite side of the circular track.

i.e. at a distance equal to the diameter of the circular track which is equal to 200 m

ஃ Distance covered in 2 min 20 *s *= 2198 m

Displacement after 2 min 20 *s* = 200 m.

**3. Aracing car has a uniform acceleration of 4ms? What distance does it cover in 10s after the start?**

Here we have

Acceleration, a | = 4 m/s^{2} |

Initial velocity u | = 0 |

Time t | = 10 s |

Distance (s) covered | = ? |

We know that, s | = uf +½at^{2} |

s | = (0 x10s) + [½ x 4 m/s^{2} x (10s)^{2}] |

= ½ x 4 m/s^{2} x 100s^{2} | |

= 2x 100 m=200m | |

Thus, the racing car will cover a distance of 200 m after starting in 10s with a given acceleration. |