Motion in One Dimension **Exe-2A** Some Terms Related to Motion Numericals Type Questions for Class-9 ICSE Concise Selina Physics Solutions. There is the solutions of Numericals Type Questions of Motion for your latest textbook which is applicable in 2023-24 academic session**. **Visit official Website CISCE for detail information about ICSE Board Class-9.

**Ch-2 Motion in One Dimension Exe-2A Some Terms Related to Motion Numericals **

**(ICSE Class – 9 Physics Concise Selina Publishers)**

Board | ICSE |

Class | 9 |

Subject | Physics |

Writer / Publication | Concise selina Publishers |

Chapter-2 | Motion in One Dimension |

Exe – 2 A | Some Terms Related to Motion |

Topics | Solution of Exe-2(A) Numericals Type |

Academic Session | 2023-2024 |

Note : You can get here answer of these type questions such as what are Some basic term related to motion, Numericals formula of motion, how do you solve motion numericals what are 3 term associated with motion. Solved numerical problems of class 9 physics motion pdf, equation of motion numericals , extra questions for practice on motion numericals problems.

### Exe-2A Some Terms Related to Motion Numericals

**Ch-2 Motion in One Dimension Physics Class-9 ICSE Concise**

**Page 39**

**Question 1. **The speed of a car is 72 km h^{-1}. Express it in m s^{-1}.

**Answer :**

Speed of car = 72 km h^{-1}

Speed of car in ms^{-1}

^{=72×100/3600}

^{ = 20m/s}

**Question 2. **Express 15 m s^{-1} in km h^{-1}.

**Answer :**

15m/s =from 15m/s multiply by 18/5 to make km/h

Ans 54 km/h

**Question 3. **Express each of the following in m s^{-1}.

(a) 1 km h^{-1}

(b) 18 km min^{-1}

**Answer :**

**Question 4. **Arrange the following speeds in increasing order.

10 m s^{-1}, 1 km min^{-1} and 18 km h^{-1}.

[Hint: 1 km min^{-1} = 16.65 m s^{-1}, 18 km h^{-1}= 5 m s^{-1}]

**Answer :**

18 km h^{-1} < 10 m s^{-1} < 1 km min^{-1}

**Question 5. **A train takes 3 hours to travel from Agra to Delhi with a uniform speed of 65 km h^{-1}. Find the distance between the two cities.

**Answer :**

Total time taken = 3 hours

Speed of the train = 65 km/hr

Distance travelled = speed x time

= 65 x 3 = 195 km

**Question 6. **A car travels the first 30 km with a uniform speed of 60 km h^{-1} and the next 30 km with a uniform speed of 40 km h^{-1}. Calculate: (i) The total time of journey, (ii)The average speed of the car.

**Answer :**

For the first 30 km travelled, speed = 60 km/h.

Thus time taken (t1) = Distance / speed

= (30/60) h^{-1}

= 0.5 h^{-1} or 30 min.

For the next 30 km travelled, speed = 40 km/h

Thus time taken (t2) = Distance/speed

= (30/40) h^{-1}

= 0.75 h^{-1} or 45 min.

**(i) Total time = (30 + 45) min**

= 75 min or 1.25 h.

**(ii) Average speed of the car = Total distance travelled/total time taken**

=60km/1.25h

=48km/h

**Question 7. **A train takes 2 h to reach station B from station A, and then 3 h to return from station B to station A. The distance between the two stations is 200 km. Find: (i) The average speed, (ii) The average velocity of the train.

**Answer : **Here, total distance = (200 + 200) km = 400 km

Total time taken = (2 + 3) h = 5 h

**(i) Average speed = Total distance travelled/total time taken**

=400km/5h

=80km/h

**(ii) Average velocity of the train is zero because the train stops at the same point from where it starts, i.e. the displacement is zero.**

**Question 8. **A car moving on a straight path covers a distance of 1 km due east in 100 s. What is (i) the speed and (ii) velocity of the car?

**Answer :**

**(i) Speed of the car = Distance/time taken**

=1km/100s

=1000m/100s

hence =10m/s

**(ii) Velocity of car = Speed with direction**

= 10 m/s due east

**Question 9. **A body starts from rest and acquires a velocity 10m s^{-1} in 2 s. Find the acceleration.

**Answer : **Here, final velocity = 10 m/s

Initial velocity = 0 m/s

Time taken = 2s

Acceleration = (Final Velocity – Initial Velocity)/time

= (10/2) ms^{–}^{2}

= 5 ms^{-2}

**Question 10. **A car starting from rest acquires a velocity 180m s^{-1} in 0.05 h. Find the acceleration.

**Answer : **Here, final velocity = 180 m/s

Initial velocity = 0 m/s

Time taken = 0.05 h or 180 s

Acceleration = (Final Velocity – Initial Velocity)/time

= (180-0)/180 m s^{-2}

= 1 m s^{–}^{2}

**Question 11. **A body is moving vertically upwards. Its velocity changes at a constant rate from 50 m s^{-1} to 20 m s^{-1} in 3 s. What is its acceleration?

**Answer : **Here, final velocity = 20 m/s

Initial velocity = 50 m/s

Time taken = 3 s

Acceleration = (Final Velocity – Initial Velocity)/time

= (20 – 50)/3 m/s^{-2}

= -10 m/s

Negative sign here indicates that the velocity decreases with time, so retardation is 10 m/s.

**Question 12. **A toy car initially moving with uniform velocity of 18 km h^{-1} comes to a stop in 2 s. Find the retardation of the car in S.I. units.

**Answer : **Here, final velocity = 18 km/h or 5 m/s

Initial velocity = 0 km/h

Time taken = 2 s

Acceleration = (Final Velocity – Initial Velocity)/time

= (5 – 0) / 2 m s^{-2}

= 2.5 m s^{-2}

**Question 13. **A car accelerates at a rate of 5 m s^{-2}. Find the increase in its velocity in 2 s.

**Answer : **Acceleration = Increase in velocity/time taken

Therefore, increase in velocity = Acceleration × time taken

= (5 × 2) m/s

= 10 m/s

**Question 14. **A car is moving with a velocity 20 m s^{-1}. The brakes are applied to retard it at a rate of 2 m s^{-2}. What will be the velocity after 5 s of applying the brakes?

**Answer: **Initial velocity of the car, u = 20 m/s

Retardation = 2 m/s^{2}

Given time, t = 5 s

Let ‘v’ be the final velocity.

We know that, Acceleration = Rate of change of velocity /time

= (Final velocity – Initial velocity)/time

Or, -2 = (v – 20) / 5

So , -10 = v – 20

Or, v = -20 + 10 m/s

Therefore , v = -10 m/s

Negative sign indicates that the velocity is decreasing.

**Question 15. **A bicycle initially moving with a velocity 5.0 m s^{-1} accelerates for 5 s at a rate of 2 m s^{-2}. What will be its final velocity?

**Answer : **Initial velocity of the bicycle, u = 5 m/s

Acceleration = 2 m/s^{2}

Given time, t = 5 s

Let ‘v’ be the final velocity.

We know that, acceleration = Rate of change of velocity/time

= (Final velocity – Initial velocity)/time

Or 2 = (v – 5)/5

and , 10 = (v – 5)

so, v = 5 + 10

therefore, v = 15 m/s

**Question 16. **A car is moving in a straight line with speed 18 km h^{-1}. It is stopped in 5 s by applying the brakes. Find (i) the speed of car in m s^{-1}, (ii) the retardation and (iii) the speed of car after 2 s of applying the brakes.

**Answer : **Initial velocity of the bicycle, u = 18 km/hr

Time taken, t = 5 s^{-1}

Final velocity, v = 0 m/s (As the car comes to rest)

**(i) Speed in m/s**

=18×1000/1×3600

=5m/s

(ii) Retardation = (Final velocity – Initial velocity)/time taken

Or, Retardation

= (0-5)/5

=1m/s²

**(iii) Let ‘V’ be the speed of the car after 2 s of applying the brakes.**

Then, Acceleration = (V – 5)/ 2

Or, -1 = (V – 5)/2

Hence V = -2 + 5

Or, V = 3 m/s

— : End of Motion in One Dimension Exe-2A Some Terms Related to Motion Numericals :–

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