Direction and Distance Test Practice Questions — State PSC Exam

Question 1Question →

Four delivery vans depart from a central depot. Each driver executes a specific sequence of movements and turns. Match each van's movement sequence with the final direction it is facing.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Van 1 matches with North-West, Van 2 matches with South-West, Van 3 matches with South-East, and Van 4 matches with North-East.

By mapping out each individual vehicle's rotational shifts on a standard compass rose, the exact final orientations correctly pair Van 1 to North-West, Van 2 to South-West, Van 3 to South-East, and Van 4 to North-East.

Step-by-Step Solution

1

Track Van 1's orientation sequentially.

North -> East -> North-West.

North is

0^\circ

. A

90^\circ

clockwise turn results in East (

90^\circ

). A

135^\circ

anti-clockwise turn from East subtracts

135^\circ

, ending at

315^\circ

, which is North-West.

2

Track Van 2's orientation sequentially.

South -> East -> South -> South-West.

South is

180^\circ

. A left turn from South faces East (

90^\circ

). A right turn from East returns to South (

180^\circ

). A

45^\circ

clockwise turn from South adds

45^\circ

, ending at

225^\circ

, which is South-West.

3

Track Van 3's orientation sequentially.

West -> North-East -> South-East.

West is

270^\circ

. A

135^\circ

clockwise turn adds

135^\circ

to reach

405^\circ

(equivalent to

45^\circ

, North-East). A

90^\circ

right turn from North-East adds

90^\circ

, ending at

135^\circ

, which is South-East.

4

Track Van 4's orientation sequentially.

East -> West -> North -> North-East.

East is

90^\circ

. A U-turn flips the direction to West (

270^\circ

). A right turn from West faces North (

0^\circ

). A

45^\circ

clockwise turn from North ends at

45^\circ

, which is North-East.

Key Concept

Angular Rotations and Cardinal Direction Mapping

Question 2Question →

A tourist is exploring a historical monument. He starts from the main gate and walks $30\text{ m}$ towards the West. He then turns $135^{\circ}$ to his left and walks $20\sqrt{2}\text{ m}$ . Next, he turns $45^{\circ}$ to his right and walks $10\text{ m}$ . Finally, he turns $90^{\circ}$ to his left and walks $50\text{ m}$ . What is his shortest distance and direction from the main gate?

50\text{ m}

, South-East

50\text{ m}

, North-East

50\text{ m}

, North-West

70\text{ m}

, South-East

Show answer & explanation

Answer: $50\text{ m}$ , South-East

Answer

The shortest distance is

50\text{ m}

and the direction is South-East.

By tracking the movements sequentially on a 2D coordinate plane from an origin of (0, 0), the tourist's positions are (-30, 0), (-10, -20), (-10, -30), and finally (40, -30). The straight-line distance from the origin to (40, -30) is the hypotenuse of a right triangle with legs of 40 and 30, which equals 50. The coordinate (40, -30) lies in the lower-right quadrant, indicating a South-East direction.

Step-by-Step Solution

1

Map the starting point as the origin (0, 0) and execute the first movement.

Walking

30\text{ m}

West places the tourist at (-30, 0) facing West.

Establishing a Cartesian coordinate system allows for precise tracking of movements.

2

Calculate the second movement: turn

135^{\circ}

left and walk

20\sqrt{2}\text{ m}

.

A

135^{\circ}

left turn from West changes the facing direction to South-East. Moving

20\sqrt{2}\text{ m}

South-East adds

+20

to the X-axis and

-20

to the Y-axis. New position: (-30 + 20, 0 - 20) = (-10, -20).

A left turn is counter-clockwise. West minus

135^{\circ}

equals South-East. Breaking this diagonal vector into components yields equivalent

+20\text{ m}

East and

-20\text{ m}

South.

3

Calculate the third movement: turn

45^{\circ}

right and walk

10\text{ m}

.

A

45^{\circ}

right (clockwise) turn from South-East points South. Walking

10\text{ m}

South subtracts 10 from the Y-axis. New position: (-10, -20 - 10) = (-10, -30).

Right turns add clockwise rotation, shifting the direction from diagonal to a cardinal line.

4

Calculate the final movement: turn

90^{\circ}

left and walk

50\text{ m}

.

A

90^{\circ}

left (counter-clockwise) turn from South points East. Walking

50\text{ m}

East adds 50 to the X-axis. Final position: (-10 + 50, -30) = (40, -30).

Standard orthogonal rotation maps the final displacement vector.

5

Calculate the direct distance and direction from the origin (0,0) to the final point (40, -30).

Distance =

\sqrt{40^2 + (-30)^2} = \sqrt{1600 + 900} = 50\text{ m}

. Since X is positive (East) and Y is negative (South), the direction is South-East.

The Pythagorean theorem determines the shortest straight-line Euclidean distance, and the quadrant dictates the cardinal direction.

Key Concept

Vector displacement combining angular turns, Cartesian mapping, and the Pythagorean theorem.

Question 3Question →

Match each autonomous robotic surveyor (P, Q, R, S) executing a multi-step navigation sequence with its exact shortest straight-line displacement from its initial starting point.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matching pairs are: Robotic Surveyor P matches with

13\text{ m}

, Robotic Surveyor Q matches with

15\text{ m}

, Robotic Surveyor R matches with

20\text{ m}

, and Robotic Surveyor S matches with

25\text{ m}

.

Each robotic surveyor's final position is determined by establishing a Cartesian coordinate system with the starting point at

(0,0)

. By breaking each movement into orthogonal components

(x, y)

using trigonometry for diagonal movements (

45^\circ, 135^\circ

) and applying the Pythagorean theorem

d = \sqrt{x^2 + y^2}

, we obtain exact straight-line displacements of

13\text{ m}

for P,

15\text{ m}

for Q,

20\text{ m}

for R, and

25\text{ m}

for S.

Step-by-Step Solution

1

Calculate displacement coordinates for Robotic Surveyor P

Path vector components:
1.

15\text{ m}

North:

(0, 15)

2. Turn

135^\circ

CW (facing SE), walk

10\sqrt{2}\text{ m}

:

\Delta x = 10\sqrt{2}\sin(135^\circ) = 10

,

\Delta y = 10\sqrt{2}\cos(135^\circ) = -10 \implies (10, 5)

3. Turn

90^\circ

right (facing SW), walk

5\sqrt{2}\text{ m}

:

\Delta x = -5

,

\Delta y = -5 \implies (5, 0)

4. Walk

12\text{ m}

North:

\Delta x = 0

,

\Delta y = 12 \implies (5, 12)

Final distance:

d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13\text{ m}

.

Decomposing vectors into

x

(East) and

y

(North) components allows precise tracking of position.

2

Calculate displacement coordinates for Robotic Surveyor Q

Path vector components:
1.

20\text{ m}

East:

(20, 0)

2. Turn

135^\circ

CCW (facing NW), walk

8\sqrt{2}\text{ m}

:

\Delta x = -8

,

\Delta y = 8 \implies (12, 8)

3. Turn

45^\circ

CW (facing North), walk

7\text{ m}

:

\Delta x = 0

,

\Delta y = 7 \implies (12, 15)

4. Walk

6\text{ m}

South:

\Delta x = 0

,

\Delta y = -6 \implies (12, 9)

Final distance:

d = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = 15\text{ m}

.

Using trigonometry and coordinate tracking ensures correct evaluation of net displacement.

3

Calculate displacement coordinates for Robotic Surveyor R

Path vector components:
1.

10\text{ m}

West:

(-10, 0)

2. Turn

45^\circ

CCW (facing SW), walk

6\sqrt{2}\text{ m}

:

\Delta x = -6

,

\Delta y = -6 \implies (-16, -6)

3. Turn

135^\circ

CW from SW (facing North), walk

18\text{ m}

:

\Delta x = 0

,

\Delta y = 18 \implies (-16, 12)

Final distance:

d = \sqrt{(-16)^2 + 12^2} = \sqrt{256 + 144} = 20\text{ m}

.

Angular rotations must be measured relative to the current heading, not absolute cardinal directions.

4

Determine initial orientation and calculate displacement for Robotic Surveyor S

At sunrise, the Sun is in the East, so shadows fall toward the West. Facing its own shadow means S initially faces West.
1. Turn

90^\circ

right (facing North), walk

10\text{ m}

:

(0, 10)

2. Turn

45^\circ

right (facing NE), walk

7\sqrt{2}\text{ m}

:

\Delta x = 7

,

\Delta y = 7 \implies (7, 17)

3. Turn North, walk

7\text{ m}

:

\Delta x = 0

,

\Delta y = 7 \implies (7, 24)

Final distance:

d = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = 25\text{ m}

.

Shadow orientation establishes the initial cardinal direction from which relative turns are executed.

Key Concept

Vector Displacement, Angular Rotation, and Shadow Orientation

Question 4Question →

An archaeological surveyor starting from a central benchmark walks $20\text{ m}$ due East. He then turns $135^\circ$ to his left and walks $14\sqrt{2}\text{ m}$ . Next, he turns $45^\circ$ to his left and walks $18\text{ m}$ . Finally, he turns $90^\circ$ to his right and walks $2\text{ m}$ to reach an excavation site. What is the shortest direct distance (in meters) between the starting benchmark and the excavation site?

Show answer & explanation

Answer: 20

Answer

The shortest direct distance between the starting benchmark and the excavation site is 20 meters.

By resolving each sequential movement into Cartesian coordinate components

(x, y)

, the final position relative to origin

(0,0)

is found to be

(-12\text{ m}, 16\text{ m})

. The straight-line distance is computed using the distance formula

\sqrt{(-12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ m}

.

Step-by-Step Solution

1

Plot the initial displacement on a Cartesian coordinate plane with origin

(0,0)

at the starting benchmark.

Position after walking

20\text{ m}

East is

(20, 0)

facing East (

0^\circ

).

East represents the positive x-axis direction.

2

Calculate displacement after turning

135^\circ

left (facing North-West) and walking

14\sqrt{2}\text{ m}

.

Position becomes

(20 - 14, 0 + 14) = (6, 14)

.

In North-West direction, horizontal component is

-14\sqrt{2} \times \frac{1}{\sqrt{2}} = -14\text{ m}

and vertical component is

+14\sqrt{2} \times \frac{1}{\sqrt{2}} = +14\text{ m}

.

3

Calculate displacement after turning

45^\circ

left from North-West (facing West) and walking

18\text{ m}

.

Position becomes

(6 - 18, 14) = (-12, 14)

.

Turning

45^\circ

left from North-West aligns facing direction with West (negative x-axis).

4

Calculate displacement after turning

90^\circ

right from West (facing North) and walking

2\text{ m}

.

Final position is

(-12, 14 + 2) = (-12, 16)

.

Turning

90^\circ

right from West aligns facing direction with North (positive y-axis).

5

Calculate the direct straight-line distance from starting point

(0,0)

to final position

(-12, 16)

.

Distance

D = \sqrt{(-12)^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ m}

.

Applying the Pythagorean theorem

D = \sqrt{x^2 + y^2}

yields the shortest distance.

Key Concept

Coordinate Geometry and Vector Resolution in Direction and Distance Problems

Estimated Time:2m 0s

Question 5Question →

At sunset, a surveyor facing the setting sun launches a drone from a fixed control pad. The drone executes the following sequence of movements:
1. It travels $40\text{ m}$ straight ahead in the direction the surveyor is facing.
2. It turns $135^\circ$ clockwise and flies $30\sqrt{2}\text{ m}$ .
3. It makes a $90^\circ$ right turn and flies $10\sqrt{2}\text{ m}$ .
4. Finally, it turns $135^\circ$ counter-clockwise and flies $15\text{ m}$ .

What is the shortest straight-line distance of the drone from the control pad, and in which direction is it located relative to the starting point?

35\text{ m}

, North

35\text{ m}

, South-West

25\text{ m}

, North-West

95\text{ m}

, North

Show answer & explanation

Answer: $35\text{ m}$ , North

Answer

The drone is

35\text{ m}

away from the control pad in the North direction.

At sunset, facing the setting sun means facing West. The drone travels

40\text{ m}

West (

x = -40, y = 0

). Turning

135^\circ

clockwise reorients it to North-East; flying

30\sqrt{2}\text{ m}

changes position to

x = -10, y = 30

. Turning

90^\circ

right reorients it to South-East; flying

10\sqrt{2}\text{ m}

shifts position to

x = 0, y = 20

. Turning

135^\circ

counter-clockwise reorients it North; flying

15\text{ m}

brings it to

x = 0, y = 35

. The drone is

35\text{ m}

North of the origin.

Step-by-Step Solution

1

Determine initial facing direction and first displacement vector

The surveyor faces West at sunset. Moving

40\text{ m}

straight ahead places the drone at coordinates

(-40, 0)

relative to origin

(0,0)

. Current heading: West.

Sun sets in the West, establishing the initial cardinal reference.

2

Apply

135^\circ

clockwise turn and calculate second displacement

Turning

135^\circ

clockwise from West reorients the drone to North-East. Flying

30\sqrt{2}\text{ m}

adds

+30\text{ m}

East and

+30\text{ m}

North. Position becomes

(-40 + 30, 0 + 30) = (-10, 30)

.

A

135^\circ

clockwise shift from West (

270^\circ

) leads to North-East (

45^\circ

). Vector components are

30\sqrt{2}\cos(45^\circ) = 30

and

30\sqrt{2}\sin(45^\circ) = 30

.

3

Apply

90^\circ

right turn and calculate third displacement

Turning

90^\circ

right from North-East reorients heading to South-East. Flying

10\sqrt{2}\text{ m}

adds

+10\text{ m}

East and

-10\text{ m}

South. Position becomes

(-10 + 10, 30 - 10) = (0, 20)

.

A

90^\circ

right turn rotates North-East to South-East. Vector components are

+10\text{ m}

along East and

-10\text{ m}

along North.

4

Apply

135^\circ

counter-clockwise turn and calculate final displacement

Turning

135^\circ

counter-clockwise from South-East reorients heading to North. Flying

15\text{ m}

North advances position to

(0, 20 + 15) = (0, 35)

.

Turning

135^\circ

counter-clockwise from South-East points directly North.

5

Calculate net displacement magnitude and cardinal direction

Net coordinates are

(0, 35)

. Distance =

\sqrt{0^2 + 35^2} = 35\text{ m}

due North.

X-component is

0\text{ m}

and Y-component is

+35\text{ m}

.

Key Concept

Direction and Distance Test - Multi-step Angular Rotation & Vector Displacement

Question 6Question →

An archaeological surveyor is mapping an excavation site using a miscalibrated compass. The compass needle that is supposed to point 'North' actually points to the true geographical East.

Starting from the base camp, she faces the direction the compass indicates as 'North' and walks $20\text{ m}$ straight. She then turns $135^\circ$ clockwise and walks $10\sqrt{2}\text{ m}$ in a straight line. Next, she turns $45^\circ$ anti-clockwise and proceeds $10\text{ m}$ forward. Finally, she turns $90^\circ$ clockwise and walks another $10\text{ m}$ to reach the main artifact chamber.

What is the true geographical direction and shortest direct distance of the main artifact chamber relative to the base camp?

20\text{ m}

, South

20\text{ m}

, East

20\text{ m}

, North

10\sqrt{5}\text{ m}

, South-East

Show answer & explanation

Answer: $20\text{ m}$ , South

Answer

The main artifact chamber is

20\text{ m}

directly to the South of the base camp.

By applying each movement as a vector on a coordinate plane, the starting path of

20\text{ m}

East is followed by

10\sqrt{2}\text{ m}

South-West (yielding

-10

East,

-10

North). The surveyor then goes

10\text{ m}

South, and finally

10\text{ m}

West. Summing the horizontal (East/West) components:

+20 - 10 + 0 - 10 = 0

. Summing the vertical (North/South) components:

0 - 10 - 10 + 0 = -20

. The net displacement is exactly

20\text{ m}

South.

Step-by-Step Solution

1

Determine the true starting direction based on the faulty compass.

The surveyor walks

20\text{ m}

towards the true East.

The compass's 'North' actually points East. Walking

20\text{ m}

in that direction results in a displacement of

+20\text{ m}

on the East-West axis (Coordinate: 20 East, 0 North).

2

Calculate the direction and displacement of the second movement.

She turns

135^\circ

clockwise to face South-West, walking

10\sqrt{2}\text{ m}

.

Facing East, a

135^\circ

clockwise turn lands exactly on South-West. A movement of

10\sqrt{2}\text{ m}

South-West equates to moving

10\text{ m}

West and

10\text{ m}

South (using a

45^\circ

-

45^\circ

-

90^\circ

triangle). New coordinate:

(20 - 10)

East,

(0 - 10)

North =

10\text{ m}

East,

10\text{ m}

South.

3

Calculate the direction and displacement of the third movement.

She turns

45^\circ

anti-clockwise to face South, walking

10\text{ m}

.

Facing South-West, turning

45^\circ

anti-clockwise (left) brings her to face South. Walking

10\text{ m}

South brings her coordinate to

10\text{ m}

East,

20\text{ m}

South.

4

Calculate the direction and displacement of the final movement.

She turns

90^\circ

clockwise to face West, walking

10\text{ m}

.

Facing South, a

90^\circ

clockwise turn points West. Moving

10\text{ m}

West shifts her East-West position by

-10\text{ m}

. Final coordinate:

(10 - 10)

East,

20\text{ m}

South =

0\text{ m}

East,

20\text{ m}

South.

5

Determine the final distance and direction from the starting point.

The final position is exactly

20\text{ m}

South of the base camp.

The East-West displacements cancel out completely, leaving a pure Southward displacement of

20\text{ m}

.

Key Concept

Vector displacement, angular rotation tracing, and relative reference framing.

Estimated Time:1m 30s

Question 7Question →

A botanist leaves her base camp to collect soil samples in a dense forest. She first walks $35\text{ m}$ straight East. She then turns $135^{\circ}$ to her left and walks $20\sqrt{2}\text{ m}$ . Finally, she turns $135^{\circ}$ to her left again and walks $12\text{ m}$ . What is the shortest straight-line distance, in meters, between her current position and the base camp?

Show answer & explanation

Answer: 17

Answer

The shortest straight-line distance between the botanist's current position and the base camp is 17 meters.

By breaking each movement into its horizontal and vertical components on a Cartesian plane, the final position is found to be 15 meters East and 8 meters North of the origin. The shortest direct distance is the hypotenuse of the triangle formed by these coordinates: √(15² + 8²) = 17.

Step-by-Step Solution

1

Plot the first movement on a coordinate plane.

Position 1 is at

(35, 0)

.

Setting the starting base camp at the origin

(0, 0)

, moving East means traveling along the positive x-axis.

2

Calculate the vector and position for the second movement.

Position 2 is at

(15, 20)

.

Facing East (0°), a left turn of 135° results in facing North-West. Moving

20\sqrt{2}\text{ m}

diagonally at a 45° angle to the axes translates to

\Delta x = -20\text{ m}

(West) and

\Delta y = +20\text{ m}

(North). New coordinate:

(35 - 20, 0 + 20)

.

3

Calculate the vector and position for the third movement.

Position 3 is at

(15, 8)

.

Facing North-West, another left turn of 135° places her facing directly South. Walking

12\text{ m}

South subtracts 12 from the y-coordinate. New coordinate:

(15, 20 - 12)

.

4

Calculate the shortest straight-line distance to the origin.

The final distance is

17\text{ m}

.

The straight-line distance from

(0, 0)

to

(15, 8)

forms a right-angled triangle. Using Pythagoras:

d = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17

.

Key Concept

Calculating continuous displacement using angular turns, vector breakdown, and the Pythagorean theorem.

Estimated Time:1m 30s

Question 8Question →

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-East

130\text{ m}

, North-West

130\text{ m}

, South-West

170\text{ m}

, North-East

Show answer & explanation

Answer: $130\text{ m}$ , 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

1

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.

2

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).

3

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.

4

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)

.

A

45^{\circ}

left turn from North-East points exactly North again.

5

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.

Question 9Question →

Rohan walks $12\text{ km}$ North from his house. He then turns right and walks $5\text{ km}$ , turns right again and walks $12\text{ km}$ , and finally turns left and walks $4\text{ km}$ . In which direction is Rohan located with respect to his starting house?

East

West

North

South

Show answer & explanation

Answer: East

Answer

Rohan is located directly East of his starting house.

Starting at

(0,0)

, moving North

12\text{ km}

reaches

(0,12)

. Turning right (East)

5\text{ km}

reaches

(5,12)

. Turning right (South)

12\text{ km}

returns to the baseline at

(5,0)

. Turning left (East)

4\text{ km}

ends at

(9,0)

. Since the y-coordinate is

0

and the x-coordinate is

+9

, Rohan is directly East of his starting house.

Step-by-Step Solution

1

Set the starting house as the origin

(0, 0)

and trace the first movement North.

Walking

12\text{ km}

North reaches coordinates

(0, 12)

.

North represents movement along the positive y-axis.

2

Turn right and walk

5\text{ km}

.

A

90^\circ

clockwise turn from North is East, reaching coordinates

(5, 12)

.

Right turn when facing North changes orientation to East.

3

Turn right again and walk

12\text{ km}

.

A

90^\circ

clockwise turn from East is South, reaching coordinates

(5, 0)

.

Right turn when facing East changes orientation to South. Moving

12\text{ km}

South cancels out the initial

12\text{ km}

North.

4

Turn left and walk

4\text{ km}

.

A

90^\circ

counter-clockwise turn from South is East, reaching final coordinates

(9, 0)

.

Left turn when facing South reorients towards East.

Key Concept

Direction and Distance Vector Tracking

Estimated Time:45s

Question 10Question →

A delivery cyclist starts from a warehouse and rides $12\text{ km}$ towards the South. He then turns left and rides $9\text{ km}$ . What is the shortest direct distance, in kilometers, between his current location and the warehouse?

Show answer & explanation

Answer: 15

Answer

The shortest direct distance between the current location and the warehouse is

15\text{ km}

.

The straight-line displacement forms the hypotenuse of a right triangle with perpendicular sides of length

12\text{ km}

and

9\text{ km}

. Applying the Pythagorean theorem yields

\sqrt{12^2 + 9^2} = \sqrt{225} = 15\text{ km}

.

Step-by-Step Solution

1

Determine the cardinal directions of each movement segment

First displacement is

12\text{ km}

South. Turning left while facing South directs the cyclist towards East, so the second displacement is

9\text{ km}

East.

A 90-degree left turn from South points towards East.

2

Set up the right-angled displacement triangle

The two perpendicular legs of the right triangle are

12\text{ km}

(South) and

9\text{ km}

(East).

South and East form a perpendicular angle (

90^\circ

).

3

Calculate the straight-line distance using the Pythagorean theorem

Distance

= \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ km}

.

The direct distance corresponds to the hypotenuse:

c = \sqrt{a^2 + b^2}

.

Key Concept

Shortest direct distance calculation using the Pythagorean theorem

Estimated Time:45s

Question 11Question →

A field researcher starting at a base station walks $15\text{ km}$ towards the North. He then turns $135^\circ$ clockwise and walks $8\sqrt{2}\text{ km}$ . Next, he turns $90^\circ$ counter-clockwise and walks $7\sqrt{2}\text{ km}$ . From this location, he turns West and walks $3\text{ km}$ , and finally turns South and walks $9\text{ km}$ to reach an observation post. What is the shortest straight-line distance (in km) between the observation post and the base station?

Show answer & explanation

Answer: 13

Answer

The shortest straight-line distance between the observation post and the base station is

13\text{ km}

.

Resolving each directional segment into horizontal and vertical components gives a net final coordinate of

(12, 5)

relative to the starting origin

(0, 0)

. Applying the Pythagorean theorem

\sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ km}

yields the exact straight-line distance.

Step-by-Step Solution

1

Represent the trajectory using 2D coordinate vectors starting from origin (0, 0).

Initial position = (0, 0). After moving 15 km North, position = (0, 15).

North movement aligns with the positive y-axis.

2

Decompose the 135-degree clockwise turn vector (South-East displacement).

Displacement is (+8, -8), placing the researcher at (8, 7).

A 135-degree clockwise turn from North faces South-East (45 degrees below positive x-axis).

3

Decompose the 90-degree counter-clockwise turn vector (North-East displacement).

Displacement is (+7, +7), placing the researcher at (15, 14).

A 90-degree counter-clockwise turn from South-East faces North-East (45 degrees above positive x-axis).

4

Apply cardinal direction adjustments for West and South movements.

Moving 3 km West and 9 km South gives final coordinates (12, 5).

West subtracts from x-coordinate, and South subtracts from y-coordinate.

5

Calculate net displacement using Pythagorean theorem.

Distance = sqrt(12^2 + 5^2) = 13 km.

The direct line between (0, 0) and (12, 5) forms a right-angled triangle with sides 12 and 5.

Key Concept

Vector resolution of cardinal and angular displacements using Cartesian coordinates and the Pythagorean theorem

Estimated Time:3m 0s

Question 12Question →

At sunrise, an environmental officer stands facing her own shadow. She then executes the following sequence of movements:
1. She turns $135^\circ$ anti-clockwise and walks $10\sqrt{2}\text{ m}$ .
2. She turns $90^\circ$ to her right and walks $6\sqrt{2}\text{ m}$ .
3. She turns $135^\circ$ to her left and walks $15\text{ m}$ .
4. She turns $135^\circ$ to her left again and walks $7\sqrt{2}\text{ m}$ .

What is her shortest direct distance from the starting point, and in which direction is she positioned relative to her starting point?

15\text{ m}

, North-East

15\text{ m}

, South-East

25\text{ m}

, North-East

21\text{ m}

, North-West

Show answer & explanation

Answer: $15\text{ m}$ , North-East

Answer

15\text{ m}

in the North-East direction

At sunrise, a person facing her shadow faces West. Executing the four angular turns results in successive vectors:

(10, 10)

North-East,

(6, -6)

South-East,

(0, 15)

North, and

(-7, -7)

South-West. Summing these vector components gives a net East-West displacement of

X = 9\text{ m}

East and a net North-South displacement of

Y = 12\text{ m}

North. Using the Pythagorean theorem, the direct distance is

\sqrt{9^2 + 12^2} = 15\text{ m}

in the North-East quadrant.

Step-by-Step Solution

1

Determine initial facing direction

Facing West (

180^\circ

)

At sunrise, the Sun is in the East, so shadows fall toward the West. Facing her own shadow means she initially faces West.

2

Calculate vector displacement for Movement 1

Direction: North-East (

45^\circ

), Vector:

(+10, +10)

From West (

180^\circ

), turning

135^\circ

anti-clockwise gives

180^\circ - 135^\circ = 45^\circ

(North-East). Displacement

= (10\sqrt{2}\cos 45^\circ, 10\sqrt{2}\sin 45^\circ) = (10, 10)

.

3

Calculate vector displacement for Movement 2

Direction: South-East (

315^\circ

), Vector:

(+6, -6)

From North-East (

45^\circ

), turning

90^\circ

right (clockwise) gives

45^\circ - 90^\circ = -45^\circ = 315^\circ

(South-East). Displacement

= (6\sqrt{2}\cos 315^\circ, 6\sqrt{2}\sin 315^\circ) = (6, -6)

.

4

Calculate vector displacement for Movement 3

Direction: North (

90^\circ

), Vector:

(0, +15)

From South-East (

-45^\circ

), turning

135^\circ

left (anti-clockwise) gives

-45^\circ + 135^\circ = 90^\circ

(North). Displacement

= (0, 15)

.

5

Calculate vector displacement for Movement 4

Direction: South-West (

225^\circ

), Vector:

(-7, -7)

From North (

90^\circ

), turning

135^\circ

left (anti-clockwise) gives

90^\circ + 135^\circ = 225^\circ

(South-West). Displacement

= (7\sqrt{2}\cos 225^\circ, 7\sqrt{2}\sin 225^\circ) = (-7, -7)

.

6

Sum total net displacement

(X, Y)

and find direct distance and final cardinal direction

X = 9\text{ m}

East,

Y = 12\text{ m}

North; Distance

= 15\text{ m}

, Direction = North-East

Total East displacement

X = 10 + 6 + 0 - 7 = 9\text{ m}

. Total North displacement

Y = 10 - 6 + 15 - 7 = 12\text{ m}

. Direct distance

= \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ m}

. Since both

X > 0

and

Y > 0

, the final direction relative to the start is North-East.

Key Concept

Direction and Distance Test - Vector Decomposition & Shadow Orientation

Estimated Time:3m 0s

Question 13Question →

An inspector starts walking from a police station and walks $30\text{ m}$ towards the East. He then turns right and walks $40\text{ m}$ to reach a checkpoint. What is the shortest direct distance between the police station and the checkpoint?

50\text{ m}

70\text{ m}

10\text{ m}

10\sqrt{7}\text{ m}

Show answer & explanation

Answer: $50\text{ m}$

Answer

The shortest direct distance between the police station and the checkpoint is

50\text{ m}

.

The inspector's path consists of

30\text{ m}

East followed by a perpendicular

40\text{ m}

South. The shortest direct path forms the hypotenuse of a right-angled triangle with legs of

30\text{ m}

and

40\text{ m}

. Applying

\sqrt{30^2 + 40^2} = \sqrt{2500}

gives

50\text{ m}

.

Step-by-Step Solution

1

Determine the path and orientations.

The inspector moves

30\text{ m}

East and then turns right to walk

40\text{ m}

South.

A right turn when traveling East directs movement towards the South.

2

Calculate the shortest straight-line distance using the Pythagorean theorem.

Distance

d = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\text{ m}

.

The East and South movements are perpendicular to each other, forming a right-angled triangle where the direct distance is the hypotenuse.

Key Concept

Shortest Distance using Pythagorean Theorem

Question 14Question →

A forest conservation officer starting from a watchtower walks $5\text{ km}$ due North. She then turns $135^\circ$ clockwise and walks $4\sqrt{2}\text{ km}$ . Next, she turns $45^\circ$ counter-clockwise and walks $8\text{ km}$ straight. Finally, she turns $90^\circ$ counter-clockwise and walks $4\text{ km}$ to reach a research station. What is the shortest straight-line distance between the watchtower and the research station, and in which direction is the research station relative to the watchtower?

13\text{ km}

, North-East

13\text{ km}

, South-East

21\text{ km}

, North-East

13\text{ km}

, South-West

Show answer & explanation

Answer: $13\text{ km}$ , North-East

Answer

The research station is

13\text{ km}

away from the watchtower in the North-East direction.

The net horizontal displacement along the East-West axis is

4 + 8 = 12\text{ km}

East, and the net vertical displacement along the North-South axis is

5 - 4 + 4 = 5\text{ km}

North. Applying the Pythagorean theorem yields a direct direct distance of

\sqrt{12^2 + 5^2} = 13\text{ km}

. Because both net displacement coordinates are positive, the research station lies to the North-East of the starting watchtower.

Step-by-Step Solution

1

Set up a Cartesian coordinate system with the starting watchtower at

(0, 0)

.

Initial position:

(0, 0)

, facing North.

Establishing standard coordinate axes (

x

-axis for East/West,

y

-axis for North/South) simplifies vector calculation.

2

Calculate position after the first leg (

5\text{ km}

North).

Position:

(0, 5)

, facing North (

90^\circ

relative to positive

x

-axis).

Moving

5\text{ km}

North adds

5

units along the positive

y

-axis.

3

Calculate position after turning

135^\circ

clockwise and walking

4\sqrt{2}\text{ km}

.

New orientation is South-East (

-45^\circ

). Displacement components are

\Delta x = 4\sqrt{2}\cos(-45^\circ) = 4\text{ km}

East,

\Delta y = 4\sqrt{2}\sin(-45^\circ) = -4\text{ km}

North (South). Position becomes

(0 + 4, 5 - 4) = (4, 1)

.

Turning

135^\circ

clockwise from North points facing to South-East.

4

Calculate position after turning

45^\circ

counter-clockwise and walking

8\text{ km}

.

New orientation is due East (

0^\circ

). Displacement

\Delta x = 8\text{ km}

,

\Delta y = 0

. Position becomes

(4 + 8, 1 + 0) = (12, 1)

.

Turning

45^\circ

counter-clockwise from South-East realigns facing direction to East.

5

Calculate position after turning

90^\circ

counter-clockwise and walking

4\text{ km}

.

New orientation is due North (

90^\circ

). Displacement

\Delta x = 0

,

\Delta y = 4\text{ km}

. Final position becomes

(12, 1 + 4) = (12, 5)

.

Turning

90^\circ

counter-clockwise from East points facing to North.

6

Compute direct shortest distance and final cardinal orientation relative to origin

(0, 0)

.

Shortest distance

d = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ km}

. Since both

x = 12 > 0

and

y = 5 > 0

, the final direction relative to the watchtower is North-East.

The Pythagorean theorem gives straight-line displacement in a 2D Euclidean plane.

Key Concept

Direction and Distance Test: 2D Vector Displacement and Angular Rotation

Question 15Question →

Four individuals ( $P$ , $Q$ , $R$ , and $S$ ) start from the same central point and walk according to different movement paths. Match each person in List-I with their final direction relative to the starting point given in List-II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Person P matches North-East; Person Q matches South-East; Person R matches North-West; Person S matches South-West.

Each person's movement path consists of an initial cardinal direction displacement followed by a single

90^\circ

turn. Analyzing each turn relative to their facing direction yields coordinates in four distinct ordinal quadrants: Person P at

(+10, +10)

[North-East], Person Q at

(+10, -10)

[South-East], Person R at

(-10, +10)

[North-West], and Person S at

(-10, -10)

[South-West].

Step-by-Step Solution

1

Determine the final position coordinates for Person P.

Walking North gives

+10\text{ m}

along the Y-axis. Turning right while facing North directs facing East; walking

10\text{ m}

adds

+10\text{ m}

along the X-axis. Final position is

(+10, +10)

, which is North-East.

Determine spatial position relative to origin

(0,0)

.

2

Determine the final position coordinates for Person Q.

Walking South gives

-10\text{ m}

along the Y-axis. Turning left while facing South directs facing East; walking

10\text{ m}

adds

+10\text{ m}

along the X-axis. Final position is

(+10, -10)

, which is South-East.

Turning left when facing South points toward East.

3

Determine the final position coordinates for Person R.

Walking West gives

-10\text{ m}

along the X-axis. Turning right while facing West directs facing North; walking

10\text{ m}

adds

+10\text{ m}

along the Y-axis. Final position is

(-10, +10)

, which is North-West.

Turning right when facing West points toward North.

4

Determine the final position coordinates for Person S.

Walking West gives

-10\text{ m}

along the X-axis. Turning left while facing West directs facing South; walking

10\text{ m}

adds

-10\text{ m}

along the Y-axis. Final position is

(-10, -10)

, which is South-West.

Turning left when facing West points toward South.

Key Concept

Direction and Distance Test - Cardinal and Ordinal Directions

Question 16Question →

A municipal engineer begins inspecting a road pipeline from point $A$ by walking $12\text{ m}$ towards the East. She then turns to her right and walks $5\text{ m}$ to reach point $B$ . What is the shortest direct distance between point $A$ and point $B$ , and in which direction is point $B$ relative to point $A$ ?

13\text{ m}

, South-East

17\text{ m}

, South-East

13\text{ m}

, North-East

17\text{ m}

, North-West

Show answer & explanation

Answer: $13\text{ m}$ , South-East

Answer

The shortest direct distance between point

A

and point

B

is

13\text{ m}

, and point

B

lies in the South-East direction relative to point

A

.

The engineer moves

12\text{ m}

East and then turns right (South) to walk

5\text{ m}

. The straight-line distance forms the hypotenuse of a right triangle with perpendicular sides of

12\text{ m}

and

5\text{ m}

, calculated as

\sqrt{12^2 + 5^2} = 13\text{ m}

. Since point

B

lies to the East and South of point

A

, its position relative to point

A

is South-East.

Step-by-Step Solution

1

Determine the direction of movement.

Initial movement is

12\text{ m}

East. Turning right while facing East leads to facing South, followed by walking

5\text{ m}

South.

A right turn relative to East points directly South.

2

Identify the relative cardinal direction of point

B

from point

A

.

Point

B

is displaced Eastward and Southward from point

A

, placing it in the South-East quadrant relative to point

A

.

Combining an Eastward vector with a Southward vector results in a South-East displacement.

3

Calculate the direct straight-line distance using the Pythagorean theorem.

\text{Distance} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ m}

.

The perpendicular legs of

12\text{ m}

and

5\text{ m}

form a right-angled triangle where the hypotenuse represents the direct distance.

Key Concept

Direction reorientation and displacement calculation using the Pythagorean theorem

Question 17Question →

A geological survey team is navigating an uncharted terrain using a miscalibrated digital compass. The compass incorrectly displays 'North' when it is actually pointing towards true North-West.

According to the readings on this faulty compass, the team leader executes the following sequence of movements:
- First, he travels $12\sqrt{2}$ km towards the 'East' indicated by the compass.
- Second, he rotates $135^\circ$ to his right and travels $12$ km straight.
- Finally, he rotates $90^\circ$ to his left and travels another $12$ km.

What is the team's shortest straight-line distance and true geographic direction from their starting point?

24

km, True South-East

24

km, True South

24

km, True East

24

km, True North

Show answer & explanation

Answer: $24$ km, True East

Answer

The team's shortest straight-line distance is

24

km, and their true geographic direction from the starting point is True East.

The miscalibrated compass means all readings are shifted. Compass North is True North-West, so Compass East is True North-East. A

12\sqrt{2}

km movement True North-East equals

(12, 12)

on a coordinate plane. A

135^\circ

right turn from Compass East faces Compass South-West, which is True South. Moving

12

km True South changes the position to

(12, 0)

. A

90^\circ

left turn from Compass South-West faces Compass South-East, which is True East. Moving

12

km True East changes the position to

(24, 0)

. This represents a distance of

24

km in the True East direction.

Step-by-Step Solution

1

Determine the true geographic bearing for the initial movement.

The compass reads 'North' when pointing True North-West. This means true directions are

45^\circ

counter-clockwise from compass readings. The team moves towards Compass 'East', which corresponds to True North-East. The displacement of

12\sqrt{2}

km True North-East translates to

12

km True East and

12

km True North.

Establishing the relationship between the faulty compass and true geographic directions is required to map the exact path correctly.

2

Calculate the true bearing and displacement for the second movement.

Turning

135^\circ

right (clockwise) from Compass East results in facing Compass South-West. Compass South-West corresponds to True South. Moving

12

km True South cancels out the previous

12

km True North, placing the team at exactly

12

km True East from the starting point.

Angular turns apply to the current facing direction. A

135^\circ

clockwise rotation from East lands on South-West.

3

Calculate the true bearing and displacement for the final movement.

Turning

90^\circ

left (counter-clockwise) from Compass South-West results in facing Compass South-East. Compass South-East corresponds to True East. Moving

12

km True East adds to the previous position, resulting in a total displacement of

24

km True East.

A

90^\circ

counter-clockwise rotation from South-West lands on South-East. Summing the true vectors provides the final destination.

Key Concept

Vector Addition and Relative Bearings in Miscalibrated Systems

Question 18Question →

Shortly after sunrise, a wildlife photographer begins tracking an animal. She starts walking in a straight line, noticing that her shadow falls exactly to her left. After walking $40\text{ m}$ , she turns $135^\circ$ to her right and walks $30\sqrt{2}\text{ m}$ . Finally, she turns $90^\circ$ to her right and walks another $10\sqrt{2}\text{ m}$ to reach the animal. What is her shortest distance and direction from her starting point?

20\text{ m}

towards East

20\text{ m}

towards West

20\sqrt{5}\text{ m}

towards North-East

20\text{ m}

towards South

Show answer & explanation

Answer: $20\text{ m}$ towards East

Answer

The photographer is

20\text{ m}

towards East from her starting point.

Based on sunrise shadow rules, the initial direction is North. Walking

40\text{ m}

North to

(0, 40)

, turning

135^\circ

right to travel

30\sqrt{2}\text{ m}

South-East to

(30, 10)

, and finally turning

90^\circ

right to travel

10\sqrt{2}\text{ m}

South-West results in a final coordinate of

(20, 0)

. This point is exactly

20\text{ m}

East of the starting point.

Step-by-Step Solution

1

Determine the initial facing direction using the position of the sun and the shadow.

The photographer is initially facing North.

At sunrise, the sun is in the East, so light travels West and shadows are cast to the West. Since her shadow falls to her left, her left side points West. Therefore, she must be facing North.

2

Plot the first movement on a coordinate plane.

Her position becomes

(0, 40)

.

Starting at the origin

(0,0)

, walking

40\text{ m}

North translates to an upward movement of

+40

on the y-axis.

3

Calculate the second movement after a

135^\circ

right turn.

Her new position is

(30, 10)

.

Turning

135^\circ

to the right (clockwise) from North puts her facing South-East. Moving

30\sqrt{2}\text{ m}

South-East yields a displacement of

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

and

\Delta y = 30\sqrt{2} \times \sin(-45^\circ) = -30

. Adding this to

(0, 40)

gives

(30, 10)

.

4

Calculate the final movement after a

90^\circ

right turn.

Her final position is

(20, 0)

.

Turning

90^\circ

to the right from South-East puts her facing South-West. Moving

10\sqrt{2}\text{ m}

South-West yields a displacement of

\Delta x = -10

and

\Delta y = -10

. Adding this to

(30, 10)

gives

(20, 0)

.

5

Determine the final distance and direction from the origin.

The shortest distance is

20\text{ m}

, and the direction is East.

The final coordinate is

(20, 0)

. The distance from

(0,0)

is

\sqrt{20^2 + 0^2} = 20\text{ m}

. Since it lies on the positive x-axis, the direction is East.

Key Concept

Vector displacement and cardinal alignment using sun-shadow rules.

Question 19Question →

Four tourists (Alex, Blake, Casey, and Drew) start exploring a flat city from the exact same central plaza. Match each tourist's movement sequence with their final direction relative to the starting point.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Alex is Exactly East, Blake is South-East, Casey is Exactly West, and Drew is North-East of the starting point.

By placing the starting point at the origin (0,0) of a Cartesian plane, we can translate every movement into (x,y) coordinate changes. Alex ends at (40, 0), indicating a positive x-axis position (East). Blake ends at (30, -30), indicating a positive x and negative y position (South-East). Casey ends at (-60, 0), indicating a negative x-axis position (West). Drew ends at (40, 40), indicating a positive x and positive y position (North-East).

Step-by-Step Solution

1

Map Alex's path on a coordinate plane starting at (0,0).

Moves

40\text{ m}

North to

(0, 40)

. A

135^\circ

clockwise turn from North faces South-East. Moving

40\sqrt{2}\text{ m}

SE results in

\Delta x = 40

and

\Delta y = -40

. Final position:

(40, 0)

.

Vector decomposition (using Pythagoras/trigonometry on

45^\circ

right triangles) accurately tracks diagonal displacement.

2

Map Blake's path starting at (0,0).

Moves West

30\text{ m}

to

(-30, 0)

. Turning

90^\circ

left (faces South) and moving

30\text{ m}

gives

(-30, -30)

. Turning left again (faces East) and moving

60\text{ m}

yields

(-30 + 60, -30) = (30, -30)

.

Step-by-step orthogonal tracking determines the final Cartesian coordinates.

3

Map Casey's path starting at (0,0).

Moves South

20\text{ m}

to

(0, -20)

. Turning

90^\circ

right (faces West) and moving

40\text{ m}

gives

(-40, -20)

. A

45^\circ

clockwise turn from West faces North-West. Moving

20\sqrt{2}\text{ m}

NW gives

\Delta x = -20

,

\Delta y = +20

. Final position:

(-40 - 20, -20 + 20) = (-60, 0)

.

Translating relative turns (right, clockwise) into cardinal headings ensures correct vector application.

4

Map Drew's path starting at (0,0).

Moves North

10\text{ m}

to

(0, 10)

. Turning

90^\circ

right (faces East) and moving

40\text{ m}

gives

(40, 10)

. Turning

90^\circ

left (faces North) and moving

30\text{ m}

yields

(40, 10 + 30) = (40, 40)

.

Summing the

x

and

y

vectors gives the exact final location in the first quadrant.

Key Concept

Direction and Distance Vector Mapping

Question 20Question →

An autonomous delivery drone is programmed to navigate from a central warehouse to a remote drop-off location. The drone executes the following sequence of movements:

1. It takes off and flies $10\text{ km}$ due North.
2. It turns $90^\circ$ to its right and flies $20\text{ km}$ .
3. It then rotates $135^\circ$ anti-clockwise and flies $10\sqrt{2}\text{ km}$ .
4. Finally, it turns $135^\circ$ clockwise and flies $5\text{ km}$ to reach the destination.

What is the shortest direct distance between the warehouse and the drop-off destination?

15\text{ km}

20\text{ km}

35\text{ km}

25\text{ km}

Show answer & explanation

Answer: $25\text{ km}$

Answer

The shortest direct distance between the warehouse and the drop-off destination is

25\text{ km}

.

The correct answer is found by tracking the drone's position on a 2D plane. By applying the vector components of each movement, the net displacement is

15\text{ km}

East and

20\text{ km}

North. The shortest direct distance is the hypotenuse of these components:

\sqrt{15^2 + 20^2} = 25\text{ km}

.

Step-by-Step Solution

1

Determine the drone's position after the first two movements.

Starting at origin

(0,0)

, the drone flies

10\text{ km}

North to

(0, 10)

. Turning

90^\circ

right faces it East; flying

20\text{ km}

brings it to

(20, 10)

.

Tracking coordinates on a plane allows for precise calculation of vector displacements.

2

Calculate the coordinates after the

135^\circ

anti-clockwise rotation.

Facing East, a

135^\circ

anti-clockwise turn points the drone North-West. A distance of

10\sqrt{2}\text{ km}

NW means moving

10\text{ km}

West and

10\text{ km}

North. The new position is

(20-10, 10+10) = (10, 20)

.

The North-West vector components of

10\sqrt{2}

are split equally into

(-10, +10)

on a standard coordinate grid.

3

Calculate the final position after the

135^\circ

clockwise rotation.

Facing North-West, a

135^\circ

clockwise turn points the drone East. Flying

5\text{ km}

East adds

+5

to the x-coordinate. The final position is

(10+5, 20) = (15, 20)

.

Establishing the final drop-off location coordinates relative to the warehouse to allow for distance calculation.

4

Compute the shortest direct distance using the final coordinates.

Using the Pythagorean theorem: Distance

= \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ km}

.

The shortest direct distance corresponds to the hypotenuse of the right-angled triangle formed by the net horizontal and vertical displacements.

Key Concept

Vector displacement, angular rotation tracking, and Pythagorean theorem application.

Direction and Distance Test

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept