# NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science (physics) Chapter 8 Motion are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 8 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

## Class 9 Science Chapter 8 Textbook Questions and Answers

**INTEXT QUESTIONS**

**PAGE NO. 100**

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

**Answer:** Yes, zero displacement is possible if an object has moved through a distance. Suppose a body is moving in a circular path and starts moving from point A and it returns back at same point A after completing one revolution, then the distance will be equal to its circumference while displacement will be zero.

**Question 2:** A farmer moves along the boundary of a square field of side 10 m 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?

**Answer:**

Given, side of the square field = 10 m

Therefore, perimeter = 10 m × 4 = 40 m

Farmer moves along the boundary in 40 second

Time = 2 minutes 20 second = 2 × 60 + 20 = 140 s

Since, in 40 second farmer moves 40 m

Therefore, in 1 second distance covered by farmer = 40 ÷ 40 = 1m.

Therefore, in 140 second distance covered by farmer = 1 × 140 m = 140 m

Thus, after 2 minute 20 second the displacement of farmer will be equal to 14.1 m north east from initial position.

**Question 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.

**Answer:** (a) Not true

Displacement can become zero when the initial and final position of the object is the same.

(b) Not true

Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

__PAGE NO. 102__

**Question 1:** Distinguish between speed and velocity

**Answer:** Speed has only magnitude while velocity has both magnitude and direction. So, speed is a scalar quantity but velocity is a vector quantity.

**or**

Velocity | Speed |

It refers to the displacement of a given object over a time interval. | It refers to the distance moved by an object over a time interval. |

It has a specific direction | It does not have any direction. |

Velocity = displacement/time | Speed = distance / time |

Velocity can hold a negative value | Speed cannot hold a negative value. |

**Question 2:** Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

**Answer:** The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity in a straight line motion.

**Question 3:** What does the odometer of an automobile measure?

**Answer:** In automobiles, odometer is used to measure the distance.

**Question 4:** What does the path of an object look like when it is in uniform motion?

**Answer:** In the case of uniform motion, the path of an object will look like a straight line.

**Question 5:** During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3×10^{8} ms^{-1}.

**Answer:** Here we have, speed = 3 × 10^{8} m/s

Time = 5 minute = 5 × 60 s = 300 s

Using, Distance = Speed × Time

⇒ Distance = 3 × 10^{8 }× 300 m

= 900 × 10^{8 }m

= 9.0 × 10^{10} m

__PAGE NO 103__

**Question 1:** When will you say a body is in **(i) **uniform acceleration? **(ii) **non-uniform acceleration?

**Answer:** **(i) **A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.

**(ii)** A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

**Question 2:** A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

**Question 3: **A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

**PAGE NO. 107**

**Question 1: **What is the nature of the distance – time graphs for uniform and non-uniform motion of an object?

**Answer:**(a) The slope of the distance-time graph for an object in uniform motion is a straight line.

(b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

**Question 2:** What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

**Answer:**** **When the slope of distance-time graph is a straight line parallel to time axis, the object is stationary.

**Question 3:** What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

**Answer:** When the graph of a speed time graph is a straight line parallel to the time axis, the object is moving with constant speed.

**Question 4:** What is the quantity which is measured by the area occupied below the velocity- time graph?

**Answer:** The quantity of distance is measured by the area occupied below the velocity time graph.

**PAGE NO 109-110**

**Question 1: **A bus starting from rest moves with a uniform acceleration of 0.1ms^{–2} for 2 minutes. Find **(a)** the speed acquired, **(b)** the distance travelled.

**Answer: ** Here we have,

Initial velocity (u) = 0 m/s

Acceleration (a) = 0.1ms^{–2}

Time (t) = 2 minute = 120 seconds

**(a)** The speed acquired:

We know that, v = u + at

⇒ v = 0 + 0.1 × 120 m/s

⇒ v = 12 m/s

Thus, the bus will acquire a speed of 12 m/s after 2 minute with the given acceleration.

**(b)** The distance travelled:

Thus, bus will travel a distance of 720 m in the given time of 2 minute.

**Question 2: **A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s^{2}. Find how far the train will go before it is brought to rest.

**Question 3: **A trolley, while going down an inclined plane, has an acceleration of 2 cm/s^{2}. What will be its velocity 3 s after the start?

**Answer:** Here we have,

Initial velocity, u = 0 m/s

Acceleration (a) = 2 cm/s^{2} = 0.02 m/s^{2}

Time (t) = 3 s

Final velocity, v = ?

We know that, v = u + at

Therefore,

v = 0 + 0.02 × 3 m/s

⇒ v = 0.06 m/s

Therefore, the final velocity will be 0.06 m/s after start.

**Question 4: ** A racing car has a uniform acceleration of 4 m/s^{2}. What distance will it cover in 10 s after start?

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

**Question 5: **A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s^{2} in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Thus, stone will attain a height of 1.25 m and time taken to attain the height is 0.5 s.

__EXERCISES__

**Question 1: **An athlete completes one round of circular track of diameter 200 m in 40 sec. What will be the distance covered and the displacement at the end of 2 minutes 20 sec?

**Answer: **Time taken = 2 min 20 sec = 140 sec.

Radius, r = 100 m.

In 40 sec the athlete complete one round.

So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round.

At the end of his motion, the athlete will be in the diametrically opposite position.

Displacement = diameter = 200 m.

**Question 2: **Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging.

**(a)** from A to B**(b)** from A to C?

**Answer: **For motion from A to B: Distance covered = 300 m = Displacement = 300 m.

Time taken = 150 sec.

**(a)** We know that,

Average speed = Total distance covered ÷ Total time taken

= 300 m ÷ 150 sec

= 2 ms^{-1}

Average velocity = Net displacement ÷ time taken

= 300 m ÷ 150 sec

= 2 ms^{-1}

**(b)** For motion from A to C:

Distance covered = 300 + 100 = 400 m.

Displacement = AB – CB

= 300 – 100

= 200 m.

Time taken = 2.5 min + 1 min

= 3.5 min

= 3.5 × 60 min

= 210 sec.

Therefore,

Average speed = Total distance covered ÷ Total time taken

= 400 ÷ 210

= 1.90 ms^{-1}.

Average velocity = Net displacement ÷ time taken

= 200 m ÷ 210 sec

= 0.952ms^{-1}.

**Question 3: **Abdul, while driving to school, computes the average speed for his trip to be 20 kmh^{-1}. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed of Abdul’s trip?

**Question 4: **A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^{-2} for 8.0 s. How far does the boat travel during this time?

**Answer: ** Here,

u = 0 m/s

a = 3 ms^{-2}

t = 8 s

Using, s = ut + ½ at^{2}

⇒ s = 0 × 8 + ½ × 3×8^{2}

⇒ s = 96 m.

**Question 5:** A driver of a car travelling at 52 kmh^{-1} applies the brakes and accelerates uniformly in the opposite direction. The car stops after 5 s. Another driver going at 34 kmh^{-1} in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for two cars. Which of the two cars travelled farther after the brakes were applied?

**Answer: **In in the following graph, AB and CD are the time graphs for the two cars whose initial speeds are 52 km/h(14.4 m/s) and 34 km/h(8.9 m/s), respectively.

Distance covered by the first car before coming to rest

= Area of triangle AOB

= ½ × AO × BO

= ½ × 52 kmh^{-1} × 5 s

= ½ (52 × 1000 × 1/3600) ms^{-1} × 5 s

= 36.1 m

Distance covered by the second car before coming to rest = Area of triangle COD

= ½ × CO × DO

= ½ × 34 km h^{-1} × 10 s

= ½ × (34 × 1000 × 1/3600) ms^{-1} ×10 s

= 47.2 m

Thus, the second car travels farther than the first car after they applied the brakes.

**Question 6: **Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

**Answer:** **(a)** B is travelling fastest as he is taking less time to cover more distance.**(b)** All three are never at the same point on the road.**(c)** Approximately 6 kms. [as 8 – 2 = 6]**(d)** Approximately 7 kms. [as 7 – 0 = 7]

**Question 7: **A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms^{-2}, with what velocity will it strike the ground? After what time will it strike the ground?

**Answer: **Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms^{-2}

As per the third motion equation,

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

⇒ v^{2} = 2 × (10ms^{‒2}) × (20m) + 0

⇒ v^{2 }= 400m^{2}s^{‒2}

⇒ v = 20ms^{-1}

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation,

t = (v-u)/a

= (20-0)ms^{‒1} / 10ms^{‒2}

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

**Question 8: **The speed – time graph for a car is shown in Figure:

(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

**Answer: **Shaded area representing the distance travelled is as follows:

**(a) **The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

Distance = 1/2 × 4 × 6 = 12 m

Therefore, the car travels a total of 12 meters in the first four seconds.

**(b) **Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6^{th} to the 10^{th} second.

**Question 9: **State which of the following situations are possible and give an example of each of the following:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving with an acceleration but with uniform speed.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

**Answer:** (a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

**Question 10: **An artificial is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hrs to revolve around the earth.

**Answer:**** **Here,

Given, radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2 × π × 42250 km = 265571.42 km

Time taken for the orbit = 24 hours

Therefore, speed of the satellite = 265571.42 ÷ 24 = 11065.4 km.h^{-1}

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

### Class 9 Science NCERT Solutions Chapter 8 Motion

CBSE Class 9 Science NCERT Solutions Chapter 8 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

### NCERT Solutions for Class 9 Science Chapter 8 PDF

Below we have listed the topics discussed in NCERT Solutions for Class 9 Science Chapter 8. The list gives you a quick look at the different topics and subtopics of this chapter.

Section in NCERT Book | Topics Discussed |
---|---|

8.1 | Describing Motion |

8.2 | Measuring the Rate of Motion |

8.3 | Rate of Change of Velocity |

8.4 | Graphical Representation of Motion |

8.5 | Equations of Motion by Graphical Method |

8.6 | Uniform Circular Motion |