A city planner is inspecting a newly constructed botanical garden. She starts at the Main … — State PSC Exam

A city planner is inspecting a newly constructed botanical garden. She starts at the Main Gate and walks $80\text{ m}$ straight North. She then turns $90^{\circ}$ to her right and walks $30\text{ m}$ . Following this, she turns $45^{\circ}$ to her left and walks $20\sqrt{2}\text{ m}$ . Finally, she turns $45^{\circ}$ to her left again and walks $20\text{ m}$ to reach the Central Greenhouse. What is the shortest distance between the Main Gate and the Central Greenhouse, and in which direction is the Greenhouse located with respect to the Main Gate?

$130\text{ m}$ , North-EastAnswer
B
$130\text{ m}$ , North-West
C
$130\text{ m}$ , South-West
D
$170\text{ m}$ , North-East

Answer

The shortest distance is

130\text{ m}

and the direction is North-East.

By plotting each movement sequentially on a Cartesian coordinate system, the final position resolves to

50\text{ m}

East and

120\text{ m}

North of the starting point. Using the Pythagorean theorem, the shortest straight-line distance is calculated as

\sqrt{50^2 + 120^2} = 130\text{ m}

. Since the final coordinates are located in the upper-right quadrant relative to the starting point, the final direction is North-East.

Step-by-Step Solution

Map the initial movement from the starting point.

The planner walks

80\text{ m}

North to the coordinate

(0, 80)

Establishing the Main Gate at the origin

(0,0)

and translating the first movement along the y-axis.

Calculate the second movement after a

90^{\circ}

right turn.

The planner now faces East and walks

30\text{ m}

. The new position is

(30, 80)

A right turn from North corresponds to the East direction (+x axis).

Calculate the third movement after a

45^{\circ}

left turn.

The planner now faces North-East. Walking

20\sqrt{2}\text{ m}

adds

20\text{ m}

to the East and

20\text{ m}

to the North. The new position is

(30 + 20, 80 + 20) = (50, 100)

Resolving the diagonal distance into horizontal (

\Delta x = 20\sqrt{2} \cos 45^{\circ}

) and vertical (

\Delta y = 20\sqrt{2} \sin 45^{\circ}

) components.

Calculate the final movement after another

45^{\circ}

left turn.

The planner now faces straight North and walks

20\text{ m}

. The final position is

(50, 100 + 20) = (50, 120)

45^{\circ}

left turn from North-East points exactly North again.

Calculate the shortest distance and final direction from the origin.

Distance =

\sqrt{50^2 + 120^2} = \sqrt{2500 + 14400} = \sqrt{16900} = 130\text{ m}

. Since the final coordinate

(50, 120)

is positive on both axes, the direction is North-East.

Applying the Pythagorean theorem to find the Euclidean distance from the starting point.

Key Concept

Vector displacement and directional rotation tracking on a 2D Cartesian plane.

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