Understanding the Problem: Finding Shortest Distance to Starting Point

This question asks for the minimum distance required to travel from a final position back to the original starting position. This shortest distance is also known as the magnitude of the displacement from the final point to the initial point.

Analysing Y's Movement

Let's break down Y's journey step-by-step:

1. Y starts at a point, let's call it the starting point (A).
2. Y walks 6 m west from point A to a new point, say point B. The distance AB is 6 m.
3. From point B, Y turns right. Since Y was walking west, turning right means turning towards North.
4. Y walks 8 m north from point B to a final point, say point C. The distance BC is 8 m.

So, the movement can be visualized as a path from A to B (West) and then from B to C (North).

Visualizing as a Right-Angled Triangle

The path segment AB is along the West direction, and the path segment BC is along the North direction. The West and North directions are perpendicular to each other. Therefore, the angle formed at point B (angle ABC) is 90 degrees. This means points A, B, and C form a right-angled triangle, with the right angle at B.

• Starting Point: A
• Intermediate Point: B
• Final Point: C
• Side AB = 6 m (West)
• Side BC = 8 m (North)
• Angle ABC = 90°

Calculating the Shortest Distance Back

The shortest distance from the final point (C) back to the starting point (A) is the straight-line distance between C and A. In our right-angled triangle ABC, this straight line is the hypotenuse AC.

We can use the Pythagorean theorem to find the length of the hypotenuse AC. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

Mathematically, the Pythagorean theorem is expressed as:

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

Where 'a' and 'b' are the lengths of the two shorter sides (legs), and 'c' is the length of the hypotenuse.

In our case:

• a = AB = 6 m
• b = BC = 8 m
• c = AC (the shortest distance we need to find)

Applying the Pythagorean theorem:

\(\text{AC}^2 = \text{AB}^2 + \text{BC}^2\)

Substitute the given values:

\(\text{AC}^2 = (6 \text{ m})^2 + (8 \text{ m})^2\)

\(\text{AC}^2 = 36 \text{ m}^2 + 64 \text{ m}^2\)

\(\text{AC}^2 = 100 \text{ m}^2\)

To find AC, take the square root of both sides:

\(\text{AC} = \sqrt{100 \text{ m}^2}\)

\(\text{AC} = 10 \text{ m}\)

Thus, the shortest distance Y needs to travel to go back to his starting point is 10 m.

Revision Table: Movement Summary and Shortest Distance

Additional Information: Distance vs. Displacement

It's important to understand the difference between distance and displacement in problems like this:

• Distance: This is the total length of the path traveled. In this case, Y traveled 6 m west and then 8 m north, so the total distance covered is 6 m + 8 m = 14 m. Distance is a scalar quantity (it only has magnitude).
• Displacement: This is the shortest distance between the starting point and the final point, along with direction. It is the change in position. In this case, the displacement is the vector from A to C. Its magnitude is the length of the straight line AC, which we calculated as 10 m. Displacement is a vector quantity (it has both magnitude and direction). The question asks for the "shortest distance to travel back to his starting point," which is the magnitude of the displacement from the final point (C) to the starting point (A), equivalent to the magnitude of the displacement from A to C.

The problem specifically asks for the "shortest distance" to return, which refers to the straight-line distance between the final location and the starting location, not the total path length.