Q9CQ - College Physics (Urone)

Question

There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.

Step-by-Step Solution

Answer

Average speed is the total distance traveled by the body to the time taken. Average velocity is the total displacement of the body to that of the time taken to travel.

Step 1: Average speed and average velocity

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. 

 $\mathbf{\text{Average\hspace{0.17em}speed=}}\frac{\mathbf{\text{Total\hspace{0.17em}distance}}}{\mathbf{\text{Total\hspace{0.17em}Time}}}$

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.

 $\mathbf{\text{Average\hspace{0.17em}velocity=}}\frac{\mathbf{\text{Total\hspace{0.17em}displacement}}}{\mathbf{\text{Tota\hspace{0.17em}Time}}}$

The unit of average velocity is also m/s. 

Step 2: Example that illustrates the difference between speed and velocity

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 $\mathbf{300}\mathbf{sec}$ . 

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

 $\mathbf{\text{Average\hspace{0.17em}speed = }}\frac{\mathbf{\text{Total\hspace{0.17em}distance}}}{\mathbf{\text{Total\hspace{0.17em}Time}}}$

 Here distance = perimeter of the circle

                        $\begin{array}{l}=2\mathrm{\pi R}\\ =2\times \mathrm{\pi }\times 5\\ =31.4\mathrm{m}\end{array}$

Time is given $\mathbf{\text{300\hspace{0.17em}sec}}$ .

 

     $\begin{array}{rcl}\text{Average\hspace{0.17em}speed}& =& \frac{31.4}{300}\\ & =& 0.104\text{\hspace{0.17em}m/s}\end{array}$   

                 

Hence the average speed is  $=\mathbf{0}.\mathbf{104}\text{\hspace{0.17em}m/s}$

Now let’s find out the average velocity:

 $\text{Average\hspace{0.17em}velocity = }\frac{\text{Total\hspace{0.17em}displacement}}{\text{Total\hspace{0.17em}Time}}$

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

Hence the average velocity will be zero.

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.