A forest ranger leaves her monitoring station facing North. She walks 100\text{ m} due Nor… — State PSC Exam

A forest ranger leaves her monitoring station facing North. She walks $100\text{ m}$ due North, turns $90^\circ$ clockwise, and walks $150\text{ m}$ . She then turns $135^\circ$ anti-clockwise and walks $30\sqrt{2}\text{ m}$ . Finally, she turns $45^\circ$ clockwise and walks $30\text{ m}$ straight ahead to reach a observation tower. What is the shortest distance between the observation tower and her starting monitoring station, and in which direction is the tower with respect to the starting point?

$200\text{ m}$ , North-EastAnswer
B
$200\text{ m}$ , North-West
C
$260\text{ m}$ , North-East
D
$160\text{ m}$ , North-East

Answer

200\text{ m}

, North-East

The net horizontal displacement is

150\text{ m} - 30\text{ m} = 120\text{ m}

(East), and net vertical displacement is

100\text{ m} + 30\text{ m} + 30\text{ m} = 160\text{ m}

(North). Applying the Pythagorean theorem yields

\sqrt{120^2 + 160^2} = 200\text{ m}

. Because both coordinates are positive, the position relative to the origin is North-East.

Step-by-Step Solution

Calculate displacement for the first two legs of movement.

After walking

100\text{ m}

North and turning

90^\circ

clockwise to walk

150\text{ m}

East, the ranger is at position

(150, 100)

relative to origin

(0,0)

North corresponds to

+y

direction and East corresponds to

+x

direction.

Resolve the

135^\circ

anti-clockwise turn and

30\sqrt{2}\text{ m}

movement.

Facing East (

0^\circ

), an anti-clockwise turn of

135^\circ

points towards North-West. Displacement is

\Delta x = -30\sqrt{2} \cos(45^\circ) = -30\text{ m}

and

\Delta y = +30\sqrt{2} \sin(45^\circ) = +30\text{ m}

. New position is

(150 - 30, 100 + 30) = (120, 130)

North-West direction has negative horizontal component and positive vertical component.

Resolve the

45^\circ

clockwise turn and final

30\text{ m}

movement.

Facing North-West (

135^\circ

), a

45^\circ

clockwise turn reorients facing direction to due North (

90^\circ

). Walking

30\text{ m}

North adds

+30\text{ m}

to the

y

-coordinate, resulting in final position

(120, 160)

Turning

45^\circ

clockwise from North-West aligns movement directly along the positive y-axis.

Calculate total straight-line distance and final direction relative to origin.

Distance

= \sqrt{120^2 + 160^2} = \sqrt{14400 + 25600} = \sqrt{40000} = 200\text{ m}

. Since both

x = +120

and

y = +160

are positive, the direction relative to starting point is North-East.

Pythagorean theorem applies to perpendicular orthogonal displacements

(x, y)

Key Concept

Vector addition of multi-leg paths combining cardinal directions and angular rotations

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic