Class 11 NCERT PHYSICS • Chapter 3: Motion in a Plane • Question 10

QUESTION 10

On an open ground, a motorist follows a track that turns to his left by an angle of 60⁰ after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case

SOLUTION

We need to calculate the displacement and path length for a motorist on a hexagonal track at specific turns.

Concept: Regular Hexagon

The track forms a regular hexagon because the motorist turns left by $60^{\circ}$ every $500\,\text{m}$ (exterior angle $= 60^{\circ}$ , so number of sides $n = 360^{\circ}/60^{\circ} = 6$ ). Displacement is the vector distance from the start, while path length is the total distance traveled.

Given:

Side length $a = 500\,\text{m}$
Turn angle $\theta_{\text{ext}} = 60^{\circ}$
Turns to evaluate: $3^{\text{rd}}$ , $6^{\text{th}}$ , and $8^{\text{th}}$

Solving:

Geometry of the Path
The path is a regular hexagon $ABCDEF$ . Let the motorist start at vertex $A$ and move towards $B$ .
- Turn 1 occurs at $B$ .
- Turn 2 occurs at $C$ .
- Turn 3 occurs at $D$ .
- Turn 6 occurs at $A$ (completing one lap).
- Turn 8 occurs at $C$ (after starting the second lap).
At the Third Turn (Vertex $D$ )
- Path length $s_3$ :

s_3 = 3 \times 500\,\text{m} = 1500\,\text{m}

Displacement $d_3$ : This is the distance $AD$ , which is the diameter of the hexagon's circumcircle.

d_3 = 2a = 2 \times 500\,\text{m} = 1000\,\text{m}

The direction is $60^{\circ}$ to the left of the initial path $AB$ .

Comparison

\frac{d_3}{s_3} = \frac{1000}{1500} = \frac{2}{3} \approx 0.67

At the Sixth Turn (Vertex $A$ )
- Path length $s_6$ :

s_6 = 6 \times 500\,\text{m} = 3000\,\text{m}

Displacement $d_6$ : The motorist returns to the starting point $A$ .

d_6 = 0\,\text{m}

Comparison

\frac{d_6}{s_6} = 0

At the Eighth Turn (Vertex $C$ )
- Path length $s_8$ :

s_8 = 8 \times 500\,\text{m} = 4000\,\text{m}

Displacement $d_8$ : This is the distance $AC$ . In a regular hexagon, the distance between vertices separated by one vertex is $a\sqrt{3}$ .

\begin{aligned} d_8 &= \sqrt{a^2 + a^2 - 2a^2 \cos(120^{\circ})} \\ &= \sqrt{500^2 + 500^2 - 2(500^2)(-0.5)} \\ &= 500\sqrt{3} \approx 866.03\,\text{m} \end{aligned}

The direction is $30^{\circ}$ to the left of the initial path $AB$ .

Comparison

\frac{d_8}{s_8} = \frac{500\sqrt{3}}{4000} = \frac{\sqrt{3}}{8} \approx 0.22

Answer

At the $3^{\text{rd}}$ turn: Displacement is

1000\,\text{m}

( $60^{\circ}$ left of initial path); Path length is

1500\,\text{m}

At the $6^{\text{th}}$ turn: Displacement is

0\,\text{m}

; Path length is

3000\,\text{m}

At the $8^{\text{th}}$ turn: Displacement is

866.03\,\text{m}

( $30^{\circ}$ left of initial path); Path length is

4000\,\text{m}