Question

A particle moves along the sides $AB,BC,$ and $CD$ of a square of side $25\text{m}$ shown in the figure with a speed of $15\text{m/s}$. The magnitude of average velocity of the particle is

• 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

Speed of the particle $=15\text{m/s}$

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

So total time taken $=\frac{75}{15}=5\text{sec}$

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

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

So, magnitude of average velocity of the particle is
$=\frac{25}{5}=5\text{m/s}$