**Average Speed and Average Velocity**

**Average speed and average velocity-The The speed** is defined as the distance per unit time and **average speed** is defined as total distance covered in per unit of total time or the **average** rate of covering the total distance. The relationship between total **distance,** time and** average speed** is, the total distance is the product of total time taken and **average speed.** The** average speed** is the **total distance** covered per unit time and the **average velocity** is **total displacement** covered in per unit time is known as **average velocity.** The **average velocity** and **average speed** both are very important in Dynamics or Kinematics in the calculation of net energy. Before we study in detail what is **average** **speed** and **average velocity**, we are needed to study first what are **distance** and **displacement.**

**Distance.** The **total distance** covered by an object during its journey is known as **distance.** The **distance** is a scalar quantity since it is represented by a magnitude only. The S.I unit of **distance** is meter.

**NCERT Solutions Class 10 Science from chapter 1 to 16**

**Displacement.** The shortest **distance** between the initial and final position of an object during its journey is known as **displacement**. **Displacement** is the **vector** quantity since it represented by direction and magnitude. The S.I unit of **displacement** is meter.

**Laptops on very small EMI**

**Example. Mohan went 12 km west from his house and then turned 5 km towards the north, show the displacement and the distance covered by him**.

The **distance** covered covered by Mohan is = 12 + 5 =17 km

The** displacement** is = shortest distance = AC

Triangle ΔABC is the right triangle in which ∠B is the right angle

The **displacement** covered by Mohan = 13 km

When anything** changes its position** with respect to time then we say that the thing is in motion. According to the state of motions, it is of two kinds (i) Uniform Motion (ii) Non-Uniform Motion.

(i) Uniform Motion

- A moving car on a busy road.
- A Boy running in the field

(ii) Non-Uniform Motion.

- Moving fan
- Moon around the earth

**Distance.** The** distance** is the **total distance** covered by an object during its travel in a given time.

**Displacement.** The shortest **distance** covered by an object during the travel, in other words, the minimum **distance** from the initial point and final point during the journey of an object.

**Average Speed.** Per unit of the total **distance** covered in per unit of the total time is known as **average speed.**

**Average Velocity**. Per unit of the total displacement covered in per unit, time is known as **average velocity**.

**Example. Rohit traveled a distance of 24 km through a bicycle in 1 hour and then turned right and traveled 7 km in 10 minutes, find the average velocity and average speed of the bicycle he is cycling.**

Solution.

The total distance covered is = 24 + 7 = 31 km

Total time taken = 1 hour + 10 minutes

Therefore **average speed** is = 26.57 km/h

Total displacement = shortest distance in journey = BC

Total **displacement** = 25 km

**Total time** is taken = 7/6 Hr

Hence the **average velocity** of the bicycle is 21.43 km/h

**Science and Maths NCERT solution for Class 9 to 11 class**

**Example. Ramaswami Lingham went to his school with an average velocity of 6 km/h and returned to his home at an average speed of 8 km/h, find his average speed in his journey?**

Solution. Let the distance from his home to school is = x km

His speed from home to school = 6 km/h

Time is taken by him from home to school = x/6

**Distance** from school to home = x km

His speed from school to home = 8 km/h

Time is taken by him from school to home = x/8

Total time taken in total journey = x/6 + x/8 = 7x/24

∴ The total **distance** he covered = x + x = 2x km

Therefore the **average speed** of Ramswami Lingam is 6.86 km/h.

**Three equation of Motions-Class 9 ,Verification**

**Class 9 science NCERT solutions of chapter 2 -Is matter around us pure?**

**Example. Shyam makes a round of a circular ground in 5 minutes if the radius of the ground is 350 m, find how much distance, displacement, he will cover in 12.5 minutes?**

Solution. **Distance** covered by Shyam in one round of the ground = circumference of the ground = 2πr

Where r is the radius of the ground which is given to us = 350 m

**Distance** covered in one round = 2 × (22/7) × 350 = 2200 m

Time is taken to cover one round = 5 minutes

**Speed** = 440 meter/min

Time is given = 12.5 min.

**Distance** = **Speed × Time** = 440 × 12.5 = 5500 meter = 5.5 km

In 5 minutes Shyam covers → 1 round

In 1 minute he will cover → 1/5 round

In 12.5 minutes he will cover → 12.5 × 1/5 = 2.5 round

Suppose Shyam starts to move from A then after 2.5 round his **position** will be at B

The **displacement** = AB = Diameter of the ground = 400 meter

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