At sunrise, a researcher standing at initial point P walks 12\text{ m} directly towards th… — State PSC Exam

At sunrise, a researcher standing at initial point $P$ walks $12\text{ m}$ directly towards the shadow cast by a vertical pole. He then turns $135^\circ$ clockwise and walks $10\sqrt{2}\text{ m}$ . From that point, he makes a $90^\circ$ right turn and walks $5\sqrt{2}\text{ m}$ . Next, he rotates $225^\circ$ anti-clockwise and walks $15\text{ m}$ . Finally, he turns and walks in the South-East direction until he reaches a point that is due West of his starting point $P$ . What is his shortest straight-line distance from point $P$ , and in which direction is he located relative to point $P$ ?

$7\text{ m}$ , WestAnswer
B
$17\text{ m}$ , West
C
$7\text{ m}$ , East
D
$12\text{ m}$ , South-West

Answer

7\text{ m}

, West

By placing point

P

at origin

(0,0)

, walking towards a sunrise shadow means walking West to

(-12,0)

. A

135^\circ

clockwise turn faces North-East, moving

10\sqrt{2}\text{ m}

(-2,10)

. A

90^\circ

right turn faces South-East, moving

5\sqrt{2}\text{ m}

(3,5)

. A

225^\circ

anti-clockwise rotation turns to West, moving

15\text{ m}

(-12,5)

. Finally, walking South-East until reaching

y=0

(due West of

P

) brings the position to

(-7,0)

. This is

7\text{ m}

directly West of point

P

Step-by-Step Solution

Determine initial movement direction using sunrise shadow context.

At sunrise, the Sun is in the East, so shadows fall towards the West. Walking

12\text{ m}

towards the shadow places the researcher at position

(-12, 0)

facing West (

180^\circ

Implicit cardinal orientation based on solar position at sunrise.

Apply

135^\circ

clockwise turn and displacement of

10\sqrt{2}\text{ m}

Facing West (

180^\circ

)

+ 135^\circ\text{ CW} = 315^\circ

(or

45^\circ

North-East).

\Delta x = 10\sqrt{2}\cos(45^\circ) = 10\text{ m}

\Delta y = 10\sqrt{2}\sin(45^\circ) = 10\text{ m}

. New position is

(-12+10, 0+10) = (-2, 10)

facing North-East.

Vector decomposition of sub-cardinal movement.

Apply

90^\circ

right turn and displacement of

5\sqrt{2}\text{ m}

Right turn from North-East (

45^\circ

) leads to South-East (

-45^\circ

\Delta x = 5\sqrt{2}\cos(-45^\circ) = 5\text{ m}

\Delta y = 5\sqrt{2}\sin(-45^\circ) = -5\text{ m}

. New position is

(-2+5, 10-5) = (3, 5)

facing South-East.

Re-orienting facing direction by

90^\circ

clockwise.

Apply

225^\circ

anti-clockwise turn and

15\text{ m}

displacement.

Facing South-East (

315^\circ

)

+ 225^\circ\text{ CCW} = 540^\circ \equiv 180^\circ

(West). Walking

15\text{ m}

West shifts position by

\Delta x = -15\text{ m}, \Delta y = 0

. New position is

(3-15, 5) = (-12, 5)

facing West.

Angular addition for multi-turn rotation.

Calculate final South-East displacement until due West of starting point

P(0,0)

Due West of origin means

y = 0

. Moving South-East from

(-12, 5)

follows path

(x, y) = (-12 + t, 5 - t)

. Setting

y = 0 \implies 5 - t = 0 \implies t = 5

. Final position is

(-12+5, 5-5) = (-7, 0)

Finding line-axis intersection for South-East vector direction.

Compute final distance and relative direction from

P(0,0)

The final point

(-7, 0)

is at a straight-line distance of

\sqrt{(-7)^2 + 0^2} = 7\text{ m}

in the West direction relative to

P(0,0)

Euclidean distance formula on Cartesian plane.

Key Concept

Multi-step vector displacement with implicit shadow direction and angular rotations

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