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**Notes Paper 212.****Science and Technology****NIOS Class 10 Science and Technology Chapter 9 ****Motion and Its Description**

**NIOS Class 10 Science and Technology Chapter 9**

**Motion and Its Description**Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given **NIOS Class 10 Science and Technology Chapter 9 Motion and Its Description**, NIOS Secondary Course

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**Science and Technology****Motion and Its Description**

**Motion and Its Description****Chapter: 9**

**INTEXT QUESTIONS 9.1**

**Choose the correct answer in the followings:**

**1. For an object moving along a straight line without changing its direction.**

(a) distance travelled > displacement.

(b) distance travelled < displacement.

(c) distance travelled = displacement.

(d) distance is not zero but displacement is zero.

Ans: (c) distance travelled = displacement.

**2. In a circular motion the distance travelled is. **

(a) Always > displacement.

(b) Always < displacements.

(c) Always = displacement.

(d) Zero when displacement is Zero

Ans: (a) Always > displacement.

**INTEXT QUESTIONS 9.2**

**1. Some of the quantities are given in column I. Their corresponding values are written in column II but not in same order. You have to match these values corresponding to the values given in column:**

Column I | Column II |

(a) 1 kmh–1 | (i) 20 ms–1 |

(b) 18 km h–1 | (ii) 10 ms–1 |

c) 72 kmh–1 | (iii) 5/18 ms–1 |

(d) 36 km h–1 | (iv) 5 ms–1 |

Ans:

Column I | Column II |

(a) 1 kmh–1 | (iii) 5/18 ms–1 |

(b) 18 km h–1 | (iv) 5 ms–1 |

c) 72 kmh–1 | (i) 20 ms–1 |

(d) 36 km h–1 | (ii) 10 ms–1 |

**2. A cyclist moves along the path shown in the diagram and takes 20 minutes from point A to point B. Find the distance, displacement and speed of the cyclist.**

**3. Identify the situation for which speed and average speed of the objects are equal.**

(i) Freely falling ball.

(ii) Second or minute needle of a clock.

(iii) Motion of a ball on inclined plane .

(iv) Train going from Delhi to Mumbai.

(v) When object moves with uniform speed.

Ans: (v) When object moves with uniform speed.

**4. The distance-time graph of the motion of an object is given. Find the average speed and maximum speed of the object during the motion**

Ans: DO yourself.

**5. The distance travelled by an object at different times is given in the table below. Draw a distance-time graph and calculate the average speed of the object. State whether the motion of the object is uniform or non-uniform.**

Ans: Average speed = 0.2 ms–1, motion is uniform motion.

**6. A player completes his half of the race in 60 minutes and the next half of the race in 40 minutes. If he covers a total distance of 1200 m, find his average speed.**

Ans: **To find the average speed of the player, we can use the formula:**

Average speed = Total distance / Total time

Given that the total distance covered by the player is 1200 metres, and the total time taken is the sum of the time taken for the first half and the time taken for the second half:

Total time = 60 minutes + 40 minutes = 100 minutes

Converting the total time to hours:

Total time = 100 minutes / 60 minutes per hour = 5/3 hours

Now, we can calculate the average speed:

Average speed = 1200 metres / (5/3 hours)

≈ 720 metres per hour.

**7. A train has to cover a distance of 1200 km in 16 h. The first 800 km are covered by the train in 10 h. What should be the speed of the train to cover the rest of the distance? Also find the average speed of the train.**

Ans: **Given:**

Total distance to be covered = 1200 km.

Time taken to cover the first 800 km = 10 hours

So, the remaining distance to cover = Total distance – Distance covered in the first leg

Remaining distance = 1200 km – 800 km = 400 km

Time taken to cover the remaining distance = Total time – Time taken for the first leg

Time taken = 16 hours – 10 hours = 6 hours

Now, to find the speed for covering the remaining 400 km:

Speed = Distance / Time

Speed = 400 km / 6 hours = 66.67 km/h

So, the speed of the train to cover the remaining distance is 66.67 km/h.

To find the average speed of the train

Average speed = Total distance / Total time

Average speed = 1200 km / 16 hours

Average speed = 75 km/h

So, the average speed of the train is 75 km/h.

**8. A bird flies from a tree A to the tree B with the speed of 40 km h–1 and returns to tree A from tree B with the speed of 60 km h–1. What is the average speed of the bird during this journey?**

Ans: . 63 km h–1.

**9. Three players P, Q and R reach from point A to B in same time by following three paths shown in the Fig. 9.19. Which of the player has more speed, which has covered more distance?**

Ans: 48 km h–1.

**INTEXT QUESTIONS 9.3**

**1. Describe the motion of an object shown in Fig. 9.28.**

Ans: For first five seconds object moves with constant speed i.e. 2ms–1. From 5 to 15 second it remains at rest and then from 15 to 20 seconds it moves with constant speed 2 ms–1. The motion of the object is not uniform.

**2. Compare the velocity of two objects where motion is shown in Fig. 9.29.**

Ans: Velocity of object A is 4 times the velocity of B.

**3. Draw the graph for the motion of object A and B on the basis of data given in Table 9.10.**

Ans:

**4. A car accelerates from rest uniformly and attains a maximum velocity of 2 ms–1 in 5 seconds. In next 10 seconds it slows down uniformly and comes to rest at the end of 10th second. Draw a velocity-time graph for the motion. Calculate from the graph.**

**(i) Acceleration.**

Ans: a = 0.4 ms–2.

**(ii) Retardation.**

Ans: – a = 0.4 ms–2.

**(iii) Distance travelled.**

Ans: 10 m.

**5. A body moving with a constant speed of 10 ms–1 suddenly reverses its direction of motion at the 5th second and comes to rest in next 5 second. Draw a position time graph of the motion to represent this situation.**

Ans:

**INTEXT QUESTIONS 9.4**

**1. A ball is thrown straight upwards with an initial velocity 19.6 ms–1. It was caught at the same distance above the ground from which it was thrown: **

**(i) How high does the ball rise?**

Ans: (i) 19.6 m.

**(ii) How long does the ball remain in air? (g = 9.8 ms–2).**

Ans: (ii) 4 s.

**2. A brick is thrown vertically upwards with the velocity of 192.08 ms–1 to the labourer at the height of 9.8 m. What are its velocity and acceleration when it reaches the labourer? **

Ans: Zero and 9.8 ms–2.

**3. A body starts its motion with a speed of 10 ms–1 and accelerates for 10 s with 10 ms–2. What will be the distance covered by the body in 10 s? **

Ans: 600 m.

**4. A car starts from rest and covers a distance of 50 m in 10 s and 100 m in next 10 s. What is the average speed of the car?**

Ans: 10 m.

**INTEXT QUESTIONS 9.5**

**1. In circular motion the point around which body moves. **

(a) always remain in rest.

(b) Always remain in motion.

(c) May or may not be in motion.

(d) Remain in oscillatory motion.

Ans: (a) Always remain in rest.

**2. In uniform circular motion.**

(a) Speed remain constant.

(b) Velocity remain constant.

(c) Speed and velocity both remain constant.

(d) Neither speed nor velocity remain constant.

Ans: (a) speed remain constant.

**3. A point on a blade of a ceiling fan has.**

(a) Always uniform circular motion.

(b) Always uniformly accelerated circular motion.

(c) May be uniform or non-uniform circular motion.

(d) Variable accelerated circular motion.

Ans: (d) Variable accelerated circular motion.

TERMINAL EXERCISE |

**1. An object initially at rest moves for t seconds with a constant acceleration a. The average speed of the object during this time interval is**

(a) a.t/2;

(b) 2a.t;

(c ) 1/2at^{2};

(d) ½ a^{2}t;

Ans: (a) a.t/2; .

**2. A car starts from rest with a uniform acceleration of 4 ms–2. The distance travelled in metres at the ends of 1s, 2s, 3s and 4s are respectively, **

(a) 4, 8, 16, 32.

(b) 2, 8, 18, 32.

(c) 2, 6, 10, 14.

(d) 4, 16, 32, 64.

Ans: (c) 2, 6, 10, 14.

**3. Does the direction of velocity decide the direction of acceleration?**

Ans: Yes, the velocity direction is decided the direction of the acceleration.

**4. Establish the relation between acceleration and distance travelled by the body.**

Ans: The relation between the acceleration and distance travelled of the body is moving at constant velocity, the distance travelled is simply the velocity multiplied by one second. This is the formula given in a lot of the answers: d = u+ 1/2 a(2n-1) where d is the displacement, u is the initial velocity, a is the acceleration.

**5. Explain whether or not the following particles have acceleration:**

**(i) a particle moving in a straight line with constant speed, and **

**(ii) a particle moving on a curve with constant speed.**

Ans: The particle moving in a straight line with constant speed does not accelerate, while the particle moving along a curve experiences acceleration. By Newton’s first law of motion, a body in motion does not change direction unless an external force (depicted as an acceleration) acts on it.

**6. Consider the following combination of signs for velocity and acceleration of an object with respect to a one dimensional motion along x-axis and give example from real life situation for each case.**

Velocity | Acceleration | Example |

(a) Positive | Positive | Ball rolling down on a slope like slide or ram. |

(b) Positive | Negative | |

(c) Positive | Zero | |

(d) Negative | Positive | |

(e) Negativ | Negative | |

(f) Negative | Zero | |

(g) zero | Positive | |

(h) Zero | Negative |

Ans:

Velocity | Acceleration | Example |

(a) Positive | Positive | Ball rolling down on a slope like slide or ram. |

(b) Positive | Negative | Bullet fired on water. |

(c) Positive | Zero | Car going on road with constant speed. |

(d) Negative | Positive | Ball falling down from upward compressing a spring. |

(e) Negativ | Negative | Compressing a spring. |

(f) Negative | Zero | For going backwards with constant speed. |

(g) zero | Positive | When a ball throw upward at peak velocity is zero but acceleration is 9.8m/s2 |

(h) Zero | Negative | When we hit ball towards the wall the time of striking. |

**7. A car travelling initially at 7 ms–1 accelerates at the rate of 8.0 ms–2 for an interval of 2.0 s. What is its velocity at the end of the 2 s?**

Ans: As we all know, V = u + at,

Where, v = final velocity, u = initial velocity = 7m /s.a = acceleration = 8.0 m/s^2, t = time=2s.

So, v = 7 + 8×2 = 2.5m/s.

**8. A car travelling in a straight line has a velocity of 5.0 ms–1 at some instant. After 4.0 s, its velocity is 8.0 ms–1. What is its average acceleration in this time interval?**

Ans: We know, v = u + at.

Here, v = 8 m/s, u = 5m/s, a = ? T = 4s.

So, average acceleration is = (8-5)/4

= ¾ = 0.75m/s^2.

**9. The velocity-time graph for an object moving along a straight line has shown in Fig. 3.32. Find the average acceleration of this object during the time interval 0 to 5.0 s, 5.0 s to 15.0 s and 0 to 20.0 s.**

Ans: we know acceleration = velocity / time

0 to 5.0s = -8+8/5-0

5.0 to 15.0 = (8-(8)/5-0 = 16/10 = 1.6 m/s^{2}

0 20.s = +1.6 + 0

= 1.6m/s^{2}.

**10. The velocity of an automobile changes over a period of 8 s as shown in the table given below:**

(i) Plot the velocity-time graph of motion.

(ii) Determine the distance the car travels during the first 2 s.

(iii) What distance does the car travel during the first 4 s?

(iv) What distance does the car travel during the entire 8 s?

(v) Find the slope of the line between t = 5.0 s and t = 7.0 s. What does the slope indicate?

(vi) Find the slope of the line between t = 0 s to t = 4 s. What does this slope represent?

Ans: Do yourself.

**11. The position-time data of a car is given in the table given below:**

(i) Plot the position-time graph of the car.

(ii) Calculate average velocity of the car during first 10 seconds.

(iii) Calculate the average velocity between t = 10 s to t = 20 s.

(iv) Calculate the average velocity between t = 20 s and t = 25.

**11. What can you say about the direction of the motion of car?**

Ans: DO yourself.

**12. An object is dropped from the height of 19.6 m. Draw the displacement-time graph for time when object reach the ground. Also find velocity of the object when it touches the ground.**

Ans: Do yourself.

**13. An object is dropped from the height of 19.6 m. Find the distance travelled by object in last second of its journey.**

Ans: we know, s = ut + ½ at^2;

Here, a = 9.8; s =19.6; u = 0;

So, 19.6 = 0 + 1/2× 9.8× t^2;

Or, t = 2s.

Distance travelled in n second be, S(n) = u + a/2(2n -1).

Here, n = 2;

So, S(2) = 0 + 9.8/2 × ( 4-1) = 14.7m.

**14. Show that for a uniformly accelerated motion starting from velocity u and acquiring velocity v has average velocity equal to arithmetic mean of the initial (u) and final velocity (v).**

Ans: Initial velocity = u, final velocity = v,

Let acceleration = a, time = t.

The arithmetic means of initial and final velocity = (v+u)/2.

We know, v = u + at.

Or, t = (v-u)/a.

Here, s = ut + 1/2at^2.

Or, s = (u-v)/a ×((u+v)/2).

Average velocity = s/t = u+ v/ 2.

**15. Find the distance, average speed, displacement, average velocity and acceleration of the object whose motion is shown in the graph (Fig. 9.33).**

Ans: DO yourself.

**16. A body accelerates from rest and attains a velocity of 10 ms-1 in 5 s. What is its acceleration?**

Ans: Do yourself.