Question

A particle moves along the sides $AB,BC,$ and $CD$ of a square of side $25\text{m}$ shown in the figure. Time taken to reach point $D$ is $5\text{s}$. What is the average speed and magnitude of average velocity of the particle?

• A. $5\text{m/s}$
• B. $15\text{m/s}$
• C. $10\text{m/s}$
• D. $20\text{m/s}$

Answer 1

The correct option is A $5\text{m/s}$

Given, path followed by the particle is

Total distance travelled $=75\text{m}(25\text{m}+25\text{m}+25\text{m})$

Total time taken is $5\text{sec}.$

So, average speed$=\frac{totaldistance}{totaltime}=\frac{75}{5}=15\text{m/s}$

Now, total displacement is $A$ to $D$ only

i.e. $25\text{m}$ in $5\text{sec}$

So, magnitude of average velocity$=\frac{totaldisplacement}{totaltime}=\frac{25}{5}=5\text{m/s}$