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 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:
Show 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.
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.
Updated On: Jul 22, 2026
- Zero
- $200\text{ m}$
- $400\text{ m}$
- $600\text{ m}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
A cyclist travels in equal straight-line segments of $200\text{ m}$, turning left by $72^\circ$ at the end of each segment.
We need to calculate his net displacement from the starting point when he is at the position of the fourth turn.
Step 2: Key Concept and Approach:
The path of the cyclist can be represented as a regular polygon.
The number of sides $N$ of a regular polygon with exterior angle $\theta$ is:
\[ N = \frac{360^\circ}{\theta} \] Let us trace the vertices reached at each turn to find the final position relative to the start.
Step 3: Detailed Explanation:
• Determine the geometry of the path:
Using the exterior angle of $72^\circ$:
\[ N = \frac{360^\circ}{72^\circ} = 5 \] The cyclist's path forms a regular pentagon of side length $s = 200\text{ m}$.
• Trace the movement:
Let the vertices of the pentagon be $A$, $B$, $C$, $D$, and $E$.
- Starts at $A$, moves to $B$ (travels $200\text{ m}$). Takes the 1st turn at $B$.
- Moves from $B$ to $C$ (travels $200\text{ m}$). Takes the 2nd turn at $C$.
- Moves from $C$ to $D$ (travels $200\text{ m}$). Takes the 3rd turn at $D$.
- Moves from $D$ to $E$ (travels $200\text{ m}$). He is now at $E$, where he "just takes the 4th turn".
• Find the displacement:
The starting position is $A$, and the position after the fourth segment is $E$.
Since $A$ and $E$ are adjacent vertices of the regular pentagon, the straight-line distance between them is equal to the side length of the pentagon.
\[ \text{Displacement} = |\vec{AE}| = 200\text{ m} \]
Step 4: Final Answer:
The cyclist's displacement is $200\text{ m}$, which corresponds to Option (B).
A cyclist travels in equal straight-line segments of $200\text{ m}$, turning left by $72^\circ$ at the end of each segment.
We need to calculate his net displacement from the starting point when he is at the position of the fourth turn.
Step 2: Key Concept and Approach:
The path of the cyclist can be represented as a regular polygon.
The number of sides $N$ of a regular polygon with exterior angle $\theta$ is:
\[ N = \frac{360^\circ}{\theta} \] Let us trace the vertices reached at each turn to find the final position relative to the start.
Step 3: Detailed Explanation:
• Determine the geometry of the path:
Using the exterior angle of $72^\circ$:
\[ N = \frac{360^\circ}{72^\circ} = 5 \] The cyclist's path forms a regular pentagon of side length $s = 200\text{ m}$.
• Trace the movement:
Let the vertices of the pentagon be $A$, $B$, $C$, $D$, and $E$.
- Starts at $A$, moves to $B$ (travels $200\text{ m}$). Takes the 1st turn at $B$.
- Moves from $B$ to $C$ (travels $200\text{ m}$). Takes the 2nd turn at $C$.
- Moves from $C$ to $D$ (travels $200\text{ m}$). Takes the 3rd turn at $D$.
- Moves from $D$ to $E$ (travels $200\text{ m}$). He is now at $E$, where he "just takes the 4th turn".
• Find the displacement:
The starting position is $A$, and the position after the fourth segment is $E$.
Since $A$ and $E$ are adjacent vertices of the regular pentagon, the straight-line distance between them is equal to the side length of the pentagon.
\[ \text{Displacement} = |\vec{AE}| = 200\text{ m} \]
Step 4: Final Answer:
The cyclist's displacement is $200\text{ m}$, which corresponds to Option (B).
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