We know velocity very well but do not have information about average velocity. Moreover, average velocity helps us to determine the relationship between distance and time. Furthermore, in this topic, we will talk about average velocity formula, its derivation, and solved example.

**Velocity**

For understanding average velocity we first need to know velocity. Velocity refers to the rate of change of displacement with respect to time. Also, we calculate it using the velocity formula.

**Average Velocity**

It refers to the variation amidst the starting and ending position, which we divide by starting and ending time. Also, velocity has direction and magnitude. Furthermore, its unit for the measure is meters per second (m/s).

**Average Velocity Formula**

The formula of average velocity is as follows:

average velocity = \(\frac{(end position) – (start position)}{(end time) – (start

time)}\)

\(v_{avg}\) = \(\frac{x_{2} – x_{1}}{t_{2} – t_{1}}\)

**Derivation of the Formula**

\(v_{avg}\) = refers to the average velocity in meter per second

\(x_{1}\) = refers to the starting position of the object in meter/s

\(x_{2}\) = refers to the ending position of the object in meter/s

\(t_{1}\) = refers to the initial time of motion in second/s

\(t_{2}\) = refers to the final time of the motion in second/s

**Solved Example on Average Velocity Formula **

**Example 1**

While driving a man sees a road signboard that says Chennai- 220 km away. Moreover, an hour later he sees another signboard that says Chennai- 100 km away. Now, calculate the velocity of the vehicle which the man is driving?

**Solution:**

For solving any velocity problem choosing the direction is important for solving it. Also, in this question, the position is expressed as a distance away from the location. So, for the location of Chennai we choose x = 0, also we assume the distance values to be positive, so the vehicle is moving in a negative direction x (-x).

By using this definition for direction, the resulting value of velocity will be negative. Then, the starting position \(x_{1}\) = 220 km, and the end position will be \(x_{2}\) = 100 km. Besides, the travel time is given as difference, in this way we choose the starting time to be \(t_{1}\) = 0 hours, and the ending time be \(t_{2}\) = 1.0 hours. But the question has asked the velocity values to be in meter per second. For correcting this we will convert kilometer in meters and hours in seconds.

1 hour = (1 hour) \(\left ( \frac{60 min}{1 hour} \right )\) \(\left ( \frac{60 s}{1 min} \right )\)

1 hour = (1 hour) \(\left ( \frac{60 min}{1 hour} \right )\) \(\left ( \frac{60 s}{1 min} \right )\)

1 hour = 60 × 60 s = 3600 s

**Now let’s calculate the distance in meter**

1 km = (1 km) \(\left ( \frac{1000 m}{1 km} \right )\)

1 km = (1 km) \(\left ( \frac{1000 m}{1 km} \right )\)

1 km = 1000 m

**Now using this we can convert 220 km and 100 km to meter**

\(x_{1}\) = (220 km) \(\left ( \frac{1000 m}{1 km} \right )\)

therefore, \( x_{1}\) = 2,20,000 m

\(x_{2}\) = (100 km) \(\left ( \frac{1000 m}{1 km} \right )\)

therefore,\( x_{1}\) = 1,00,000 m

\(t_{1}\) = (0 hours) \(\left ( \frac{3600 s}{1 hour} \right )\)

therefore,\( t_{1}\) = 0 s

\(t_{2}\) = (1 hours) \(\left ( \frac{3600 s}{1 hour} \right )\)

therefore, \( t_{1}\) = 3600 s

**Now put all the values in the average velocity formula**

\(v_{avg}\) = \(\frac{x_{2} – x_{1}}{t_{2} – t_{1}}\)

\(v_{avg}\) = \(\frac{(100000 m) – (220000 m) } {(3600 s) – (0 s)}\)

\(v_{avg}\) = \(\frac{(-120000 m) } {(3600 s)}\)

\(v_{avg}\) \(\cong\) – 33.33 m/s

So, the average velocity of the car is -33.33 m/s according to the direction described above. Moreover, it can also be stated that the average velocity is 33.33 m/s towards Chennai.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…