Shweta starts walking from her office and walks 150 m towards the south, then she turns right and walks 80 m, and then she turns left and walks 60 m. She finally turns left and walks 280 m to reach a bank. What is the shortest distance between her office and the bank?
290 m
The problem asks us to find the shortest distance between Shweta's starting point (office) and her ending point (bank) after a series of movements. This involves finding the net displacement, which is the straight-line distance from the start to the end point, regardless of the path taken.
Let's break down Shweta's walk step by step:
We can represent the movements in terms of East-West and North-South directions to find the total change in position from the office.
Southward movements:
Total displacement towards South = $150 \text{ m} + 60 \text{ m} = 210 \text{ m South}$
Eastward/Westward movements:
Total displacement towards East = $280 \text{ m East} - 80 \text{ m West} = 200 \text{ m East}$ (since East > West)
So, the final position (bank) is 210 m South and 200 m East from the starting position (office).
The office, the bank, and a point representing the net displacement towards South and East form a right-angled triangle. The shortest distance between the office and the bank is the hypotenuse of this triangle. We can use the Pythagorean theorem to find this distance.
Let:
According to the Pythagorean theorem: $Hypotenuse^2 = Base^2 + Perpendicular^2$
So, $d^2 = (200 \text{ m})^2 + (210 \text{ m})^2$
Calculating the squares:
$d^2 = 40000 + 44100$
$d^2 = 84100$
To find $d$, we take the square root of both sides:
$d = \sqrt{84100}$
$d = 290 \text{ m}$
The shortest distance between Shweta's office and the bank is 290 m.
| Step | Direction | Distance (m) | Net North-South (m) | Net East-West (m) |
|---|---|---|---|---|
| Start (Office) | - | 0 | 0 | 0 |
| 1 | South | 150 | -150 (South) | 0 |
| 2 | Right (West) | 80 | -150 (South) | -80 (West) |
| 3 | Left (South) | 60 | -150 + (-60) = -210 (South) | -80 (West) |
| 4 | Left (East) | 280 | -210 (South) | -80 + 280 = 200 (East) |
| End (Bank) | - | - | -210 (South) | 200 (East) |
The net displacement is 210 m South and 200 m East. Using Pythagorean theorem: $\sqrt{(-210)^2 + (200)^2} = \sqrt{44100 + 40000} = \sqrt{84100} = 290$ m.
The shortest distance between her office and the bank is 290 m.
| Concept | Definition | Key Characteristics | How it relates to this problem |
|---|---|---|---|
| Distance | Total length of the path covered during motion. | Scalar quantity, always positive, depends on path. | Total distance walked by Shweta is 150+80+60+280 = 570 m. (Not asked in the question) |
| Displacement | Shortest distance between the initial and final positions. | Vector quantity (has magnitude and direction), can be zero or negative, independent of path. | The shortest distance between office (start) and bank (end) is the magnitude of the displacement vector. |
Understanding directions and turns is crucial for solving navigation problems like this. Standard directions are North, South, East, West.
In this problem:
This detailed breakdown of movements helps confirm the directional components used in the displacement calculation.
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