Question

A cyclist moves in such a way that he takes $72^\circ$ turn towards left after travelling $200\text{ m}$ in a straight line. His displacement when he just takes fourth turn like that is:

• A. Zero
• B. $200\text{ m}$
• C. $400\text{ m}$
• D. $600\text{ m}$

Hint:

An $N$-sided regular polygon path has $N-1$ turns.
At the $(N-1)^{\text{th}}$ turn, the particle is exactly one side length away from the starting point.
Here, $N=5$, so the $4^{\text{th}}$ turn leaves the cyclist exactly one side length ($200\text{ m}$) from the start.

Answer 1

The Correct Option is B

Step 1: Set up direction angles.
Since the cyclist turns by $72^\circ$ each time, his direction of travel for each of the five equal legs points along $0^\circ, 72^\circ, 144^\circ, 216^\circ, 288^\circ$ (taking the first leg as reference). Five such equal vectors placed head to tail close up into a regular pentagon, so all five must sum to zero.
Step 2: Use vector cancellation.
After the fourth turn, only four of the five $200\text{ m}$ legs have been covered. Since the sum of all five vectors is zero, the sum of any four of them must be equal and opposite to the missing fifth one.
Step 3: Read off the magnitude.
The missing (fifth) leg also has length $200\text{ m}$, so the resultant of the four already travelled legs has the same magnitude, just pointing the opposite way.
\[ \boxed{\text{Displacement} = 200\text{ m}} \]