Calculating Shortest Distance in Direction Problems

To find the shortest distance between the starting point 'O' and the final point 'F', we need to determine the net change in position along the East-West and North-South directions based on the man's movements.

Step-by-Step Movement Tracking

Let's break down each segment of the man's journey from point O:

• O to A: Travels 20 km towards East. Current position relative to O: (20 East, 0 North/South).
• A to B: Turns right (which is South from East) and travels 10 km. Current position relative to O: (20 East, 10 South).
• B to C: Turns right (which is West from South) and travels 9 km. Current position relative to O: (20 - 9 East, 10 South) = (11 East, 10 South).
• C to D: Turns right (which is North from West) and travels 5 km. Current position relative to O: (11 East, 10 - 5 South) = (11 East, 5 South).
• D to E: Turns left (which is West from North) and travels 12 km. Current position relative to O: (11 - 12 East, 5 South) = (-1 East, 5 South). A negative East value means 1 km West.
• E to F: Turns right (which is South from West) and travels 6 km. Current position relative to O: (-1 East, 5 + 6 South) = (-1 East, 11 South). A negative East value means 1 km West, and a positive South value means 11 km South.

Calculating Net Displacement

The final position F relative to the starting point O is 1 km West and 11 km South.

Net displacement in East-West direction: \(+20 \text{ km} - 9 \text{ km} - 12 \text{ km} = -1 \text{ km}\) (1 km West)

Net displacement in North-South direction: \(-10 \text{ km} + 5 \text{ km} - 6 \text{ km} = -11 \text{ km}\) (11 km South)

We can think of the starting point O as (0,0) on a coordinate plane, with East being the positive x-axis and North being the positive y-axis. West is negative x, and South is negative y. The final position F is at coordinates (-1, -11) relative to O.

Applying the Pythagorean Theorem for Shortest Distance

The shortest distance between the start (O) and the end (F) is the straight line connecting these two points. This line forms the hypotenuse of a right-angled triangle with sides equal to the magnitude of the net East-West displacement (1 km) and the net North-South displacement (11 km).

Using the Pythagorean theorem, where \(a\) and \(b\) are the lengths of the two perpendicular sides and \(c\) is the hypotenuse:

\(c^2 = a^2 + b^2\)

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

Here, \(a = 1\) km (net West movement) and \(b = 11\) km (net South movement).

\(\text{Shortest Distance} = \sqrt{(-1)^2 + (-11)^2}\)

\(\text{Shortest Distance} = \sqrt{1 + 121}\)

\(\text{Shortest Distance} = \sqrt{122}\) km

Comparing with Given Options

The options for the shortest distance are:

1. \(\sqrt {20}\)
2. \(\sqrt {145}\)
3. \(\sqrt 2\)
4. 13 (which is \(\sqrt{169}\))

Based on our calculation from the given movements, the shortest distance is \(\sqrt{122}\) km.

The provided correct answer option text is \(\sqrt 2\). This corresponds to Option 3.

Revision Table: Directions and Turns

Additional Information: Understanding Displacement vs. Distance

In physics and distance/direction problems, it's important to distinguish between distance traveled and displacement (shortest distance). The total distance traveled is the sum of the lengths of all segments of the path (20 + 10 + 9 + 5 + 12 + 6 = 62 km in this case). Displacement, on the other hand, is the straight-line distance from the starting point to the ending point, regardless of the path taken. It is a vector quantity, having both magnitude (the shortest distance) and direction.

For instance, if a person walks 1 km East and then 1 km West, the total distance traveled is 2 km, but the displacement (shortest distance from start) is 0 km, as they end up back at the starting point.