91Ó°ÊÓ

Average Speed: The rate of total distance traveled by the body to that of the time taken to travel that distance is known as average speed.

$\mathbf{\text{´¡±¹±ð°ù²¹²µ±ð s±è±ð±ð»å=}}\frac{\mathbf{\text{°Õ´Ç³Ù²¹±ô d¾±²õ³Ù²¹²Ô³¦±ð}}}{\mathbf{\text{°Õ´Ç³Ù²¹±ôâ€Í¿¾±³¾±ð}}}$

The unit of average speed is also m/s.

Average velocity: The rate of the total displacement of the body to that of the time taken to travel that distance is known as average velocity.

$\mathbf{\text{´¡±¹±ð°ù²¹²µ±ð v±ð±ô´Ç³¦¾±³Ù²â=}}\frac{\mathbf{\text{°Õ´Ç³Ù²¹±ô d¾±²õ±è±ô²¹³¦±ð³¾±ð²Ô³Ù}}}{\mathbf{\text{°Õ´Ç³Ù²¹â€Í¿¾±³¾±ð}}}$

The unit of average velocity is also m/s.

Suppose a man is walking on a circular ground of radius$\mathbf{5}\mathbf{meter}$. He does one round of the circular ground and returns to the position from where he started in$\mathbf{300}\text{ }\mathbf{sec}$.

So, let’s find out the average speed of the man and the average velocity.

$\mathbf{\text{´¡±¹±ð°ù²¹²µ±ð s±è±ð±ð»å=}}\frac{\mathbf{\text{°Õ´Ç³Ù²¹±ô d¾±²õ³Ù²¹²Ô³¦±ð}}}{\mathbf{\text{°Õ´Ç³Ù²¹±ôâ€Í¿¾±³¾±ð}}}$

Here distance = perimeter of the circle

role="math" localid="1655552164845" $\begin{array}{l}=2\mathrm{Ï€¸é}\\ =2\text{ }×\mathrm{Ï€}×5\\ =31.4\mathrm{m}\end{array}$

Time is given$\mathbf{\text{300 s±ð³¦}}$.

$\begin{array}{rcl}\text{´¡±¹±ð°ù²¹²µ±ð s±è±ð±ð»å}& =& \frac{31.4}{300}\\ & =& 0.104\text{″¾/²õ}\end{array}$

Hence the average speed is $=\mathbf{0}.\mathbf{104}\text{″¾/²õ}$

Now let’s find out the average velocity:

$\text{´¡±¹±ð°ù²¹²µ±ð v±ð±ô´Ç³¦¾±³Ù²â=}\frac{\text{°Õ´Ç³Ù²¹±ô d¾±²õ±è±ô²¹³¦±ð³¾±ð²Ô³Ù}}{\text{°Õ´Ç³Ù²¹±ôâ€Í¿¾±³¾±ð}}$

Here displacement is the shortest distance between the initial and final position. As the man returns to his initial position, the displacement will bezero.

Hence the average velocity will bezero.

Therefore, it is not necessary that the average speed will always be equal to the average velocity. Average speed is a scalar quantity, and average velocity is a vector.