An autonomous security robot at an industrial facility begins its inspection patrol from c… — State PSC Exam

An autonomous security robot at an industrial facility begins its inspection patrol from central gate $O$ facing East. It moves $14\text{ m}$ due East to reach checkpoint $A$ . From checkpoint $A$ , it turns $135^\circ$ anti-clockwise and travels $10\sqrt{2}\text{ m}$ to checkpoint $B$ . At checkpoint $B$ , it turns $90^\circ$ clockwise and travels $5\sqrt{2}\text{ m}$ to checkpoint $C$ . Next, it turns $90^\circ$ anti-clockwise and travels $4\sqrt{2}\text{ m}$ to reach checkpoint $D$ . Finally, it turns to face due South and moves $7\text{ m}$ to reach its final parking dock $E$ . What is the shortest straight-line distance (in meters) between central gate $O$ and final parking dock $E$ ?

Answer: 13 m

Answer

The shortest straight-line distance between central gate

O

and final parking dock

E

13\text{ m}

By resolving each directional move into its respective orthogonal

(x, y)

components, the cumulative net position of the robot from origin

O(0, 0)

is found to be

(5, 12)

. Applying the distance formula gives

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

Step-by-Step Solution

Establish Cartesian Coordinates

Central gate

O = (0, 0)

, with

+x

as East,

-x

as West,

+y

as North, and

-y

as South.

Cartesian representation allows precise vector addition for multi-directional paths.

Calculate displacement to checkpoint A

Displacement is

(14, 0)

, so

A = (14, 0)

The robot travels

14\text{ m}

directly East along the positive x-axis.

Calculate displacement to checkpoint B

Displacement vector is

(-10, 10)

, so

B = (4, 10)

Turning

135^\circ

anti-clockwise from East points North-West (

135^\circ

). Distance

10\sqrt{2}\text{ m}

gives

\Delta x = -10\sqrt{2}\cdot\frac{1}{\sqrt{2}} = -10

and

\Delta y = 10\sqrt{2}\cdot\frac{1}{\sqrt{2}} = 10

Calculate displacement to checkpoint C

Displacement vector is

(5, 5)

, so

C = (9, 15)

Turning

90^\circ

clockwise from North-West points North-East (

45^\circ

). Distance

5\sqrt{2}\text{ m}

gives

\Delta x = 5

and

\Delta y = 5

Calculate displacement to checkpoint D

Displacement vector is

(-4, 4)

, so

D = (5, 19)

Turning

90^\circ

anti-clockwise from North-East points North-West (

135^\circ

). Distance

4\sqrt{2}\text{ m}

gives

\Delta x = -4

and

\Delta y = 4

Calculate displacement to final dock E

Displacement vector is

(0, -7)

, so

E = (5, 12)

Moving

7\text{ m}

due South reduces the y-coordinate by

7

Compute shortest straight-line distance OE

Distance

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

Apply the Pythagorean theorem to net displacement components

x = 5

and

y = 12

Key Concept

Vector Addition of Angular Displacements and Pythagorean Theorem

Estimated Time:2m 0s

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