Before learning about average speed and average velocity, we must know the difference between distance and displacement.

Distance is a scalar quantity which generally implies how much ground has been covered by the object.

Displacement is a Vector quantity, and it is the shortest possible distance between the start and end point.

**Ex: **If a particle is moving in a circle, after one revolution, the distance will be the perimeter of the circle while the displacement would be zero.

**Speed Vs Velocity**

Speed is the distance travelled per second.Velocity is the distance travelled per second in a Specific Direction.

**Speed: **Speed is a scalar quantity which means it has no direction. It denotes how fast an object is moving. If the speed of the particle is high it means the particle is moving fast and if it is now, it means the particle is moving slow.

**Velocity: **Velocity is a vector quantity which means it has both magnitude and direction. It denotes the rate at which the object is moving or changing position. The direction of the velocity vector is easy to find. Its direction is same as the direction of the moving object. Even if the object is slowing down, and the magnitude of velocity is decreasing, its direction would still be same as the direction in which the object is moving.

**When an object is in motion, its position changes with time. But how fast is the position changing with time and in what direction? To describe this, we define the quantity “average velocity”?**

**Average Velocity: **Average velocity is the measure of the average change of position of an object in any given time interval.

Average Velocity (V_{avg}) = \(\frac{\Delta x}{\Delta t}\,=\,\frac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}\)

Where,

Δx = Total change in position

Δt = Total time Interval.

Average velocity does not depend on the actual path followed by the object, It only depends on the Initial and final position of the object in a time period.

If we draw a graph “X” versus “t” for the motion of an object.

Where,

X = Position of an object

t = Time.

Then the average velocity of the object in a given time interval is the slope of the line joining its initial and final position in the time interval.

**How to find Average Velocity:**

**Example: **Find the Average Velocity at a specific time interval of a particle if it is moves 5 m at 2 sec and 15 m at 4 sec along x – axis?

**Solution: **Given,

Initial distance travelled, x_{i} = 5m

Final distance travelled, x_{f} = 15m

Initial time interval, t_{i} = 2sec

Final time interval, t_{f} = 4sec

Now,

\(Average\,Velocity\,({{V}_{avg}})=\frac{{{x}_{f}}-{{x}_{i}}}{{{t}_{f}}-{{t}_{i}}}\)

\(Average\,Velocity\,({{V}_{avg}})=\frac{15-5}{4-2}=\frac{10}{2}=5m/\sec \)

∴ Average Velocity = 5 m/sec

**Average Velocity as defined above only the displacement of the object. We have seen earlier that the magnitude of displacement may be different from the actual path length. To describe the rate of motion over the actual path. We introduce another quantity called “Average Speed”?**

**Average Speed: **It is defined as the total path length travelled divided by the total time interval during which the motion has taken place.

\(Average\,Speed=\frac{Total\,path\,length}{Total\,time\,Interval}\)

It has obviously the same unit (m/sec) as that of velocity. But it does not tell us in what direction an object is moving. Thus, it always positive (In contrast to the average velocity which can be positive of negative). If the motion of an object is along a straight line and in the **same direction**. The magnitude of displacement is equal to the total path length. In that case, the magnitude of average velocity is equal to the average speed. This is not always the case.

**How to find Average Speed:**

**Example: **A car is travelling with the speed of 30 mph from city A to B and back from city B to A with the speed of 40 mph. find its average speed?

**Solution: **For finding the average speed of the car, we need to first identify total distance which is equal to two times the distance between cities A and B.

Time taken from A to B is = \(\frac{D}{{{S}_{1}}}\)

Time taken from B to A is = \(\frac{D}{{{S}_{2}}}\)

Total distance travelled is = 2D

So,

\({{S}_{(avg)}}=\frac{(2D)}{\left( \frac{D}{{{S}_{1}}}+\frac{D}{{{S}_{2}}} \right)}\)

Putting the value and Solving we have

\({{S}_{(avg)}}\,=\,2\times \frac{\left( {{S}_{1}}\times {{S}_{2}} \right)}{\left( {{S}_{1}}+{{S}_{2}} \right)}\)

\({{S}_{(avg)}}\,=\,2\times \frac{\left( 30\times 40 \right)}{\left( 30+40 \right)}=2\times \frac{\left( 30\times 40 \right)}{70}\)

S_{(avg)} = 34.29 mph.