# Motion in One Dimension Exe-2A Numericals Class-9 ICSE Physics Selina Publishers

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 sessionVisit 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

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### 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.

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.

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

Ans 54 km/h

(a) 1 km h-1

(b) 18 km min-1

#### 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]

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.

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.

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?

(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) ms2

= 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 s2

#### 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/s2

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

#### 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/s2

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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