Calculating Shortest Distance: Office to Home Walk

This problem involves finding the shortest distance between two points after a series of movements in different directions. The shortest distance between any two points is a straight line. We can solve this by determining the net displacement from the starting point (office) to the ending point (home) in both the east-west and north-south directions.

Let's break down Suyash's walk step by step, assuming the office is at the origin (0, 0) of a coordinate system, with North being positive Y, South negative Y, East positive X, and West negative X.

Step-by-Step Movement Analysis

1. Start: Suyash starts at his office. Let's mark this position as (0, 0).
2. First Movement: Walks 40 m towards the south.
   Change in position: (0, -40).
   Current position: (0, -40).
3. Second Movement: From there, he turns right and walks 35 m. When facing south, turning right means turning west.
   Change in position: (-35, 0).
   Current position: (0 - 35, -40) = (-35, -40).
4. Third Movement: He turns right again and walks 70 m. When facing west, turning right means turning north.
   Change in position: (0, +70).
   Current position: (-35, -40 + 70) = (-35, 30).
5. Fourth Movement: Finally, he turns right one more time and walks 75 m to reach his home. When facing north, turning right means turning east.
   Change in position: (+75, 0).
   Current position: (-35 + 75, 30) = (40, 30).

Net Displacement Calculation

The office was at (0, 0) and the home is at (40, 30).

The net displacement in the east-west direction is \(40 - 0 = 40\) m (East).
The net displacement in the north-south direction is \(30 - 0 = 30\) m (North).

Calculating the Shortest Distance Using Pythagorean Theorem

The shortest distance between the office and home is the hypotenuse of a right-angled triangle formed by the net east-west displacement and the net north-south displacement.

Let the shortest distance be \(d\). According to the Pythagorean theorem: \[d^2 = (\text{Net East-West Displacement})^2 + (\text{Net North-South Displacement})^2\] \[d^2 = (40 \, \text{m})^2 + (30 \, \text{m})^2\] \[d^2 = 1600 \, \text{m}^2 + 900 \, \text{m}^2\] \[d^2 = 2500 \, \text{m}^2\] \[d = \sqrt{2500 \, \text{m}^2}\] \[d = 50 \, \text{m}\]

Therefore, the shortest distance between Suyash's office and his home is 50 m.

Revision Table: Directional Movement Summary

Additional Information: Distance vs. Displacement

In physics and vector mathematics, there's a difference between distance and displacement.

• Distance: The total path length covered during a journey. In this problem, the total distance Suyash walked is \(40 + 35 + 70 + 75 = 220\) m.
• Displacement: The shortest distance between the initial point and the final point, along with the direction. It is a vector quantity. In this problem, the displacement is 50 m in a direction that is East and North from the origin (specifically, \(\tan^{-1}(30/40) \approx 36.87^\circ\) North of East).

The question asks for the "shortest distance," which refers to the magnitude of the displacement vector. Understanding the difference between distance and displacement is crucial in solving such directional problems. We used the concept of displacement and the Pythagorean theorem to find this shortest distance.