Vivek's Journey: Calculating Straight Distance

This problem requires us to determine the straight-line distance between Mr. Vivek's starting point and his final position after a series of movements. To do this, we will break down each movement into its horizontal (East-West) and vertical (North-South) components. We will use trigonometry for movements involving angles and then sum up all the components to find the net displacement.

Displacement Components Analysis

Let's consider the starting point as the origin (0,0) on a coordinate plane, where East is the positive x-axis and North is the positive y-axis.

1. First Movement: 6 metres North-east at 60 degrees to East

• This movement is 6 meters long, making an angle of 60 degrees with the East (positive x-axis).
• Horizontal (East) component (\(X_1\)): \(6 \cos(60^\circ) = 6 \times \frac{1}{2} = 3\) meters.
• Vertical (North) component (\(Y_1\)): \(6 \sin(60^\circ) = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\) meters.

2. Second Movement: 6 metres South-east at 60 degrees to East

• This movement is 6 meters long, making an angle of 60 degrees below the East (positive x-axis). We can consider this as \( -60^\circ \) or \(300^\circ \) from the positive x-axis.
• Horizontal (East) component (\(X_2\)): \(6 \cos(60^\circ) = 6 \times \frac{1}{2} = 3\) meters.
• Vertical (South) component (\(Y_2\)): \(6 \sin(-60^\circ) = 6 \times (-\frac{\sqrt{3}}{2}) = -3\sqrt{3}\) meters.

3. Third Movement: 2 metres South

• This movement is purely vertical and downwards.
• Horizontal component (\(X_3\)): \(0\) meters.
• Vertical (South) component (\(Y_3\)): \(-2\) meters.

4. Fourth Movement: 4 metres West

• This movement is purely horizontal and to the left.
• Horizontal (West) component (\(X_4\)): \(-4\) meters.
• Vertical component (\(Y_4\)): \(0\) meters.

Total Displacement Calculation

Now, let's sum up all the horizontal and vertical components to find the net displacement.

Total Horizontal Displacement (\(\Delta X\)):

• \(\Delta X = X_1 + X_2 + X_3 + X_4\)
• \(\Delta X = 3 + 3 + 0 + (-4)\)
• \(\Delta X = 6 - 4 = 2\) meters (East)

Total Vertical Displacement (\(\Delta Y\)):

• \(\Delta Y = Y_1 + Y_2 + Y_3 + Y_4\)
• \(\Delta Y = 3\sqrt{3} + (-3\sqrt{3}) + (-2) + 0\)
• \(\Delta Y = 0 - 2 = -2\) meters (South)

So, Mr. Vivek's final position relative to his starting point is 2 meters East and 2 meters South.

Straight Distance Calculation

The straight distance between the starting point and the final point is the magnitude of the resultant displacement vector. We can find this using the Pythagorean theorem, as the net horizontal and vertical displacements form the sides of a right-angled triangle.

Straight Distance \(D = \sqrt{(\Delta X)^2 + (\Delta Y)^2}\)

Substitute the values:

\(D = \sqrt{(2)^2 + (-2)^2}\)

\(D = \sqrt{4 + 4}\)

\(D = \sqrt{8}\)

To simplify \(\sqrt{8}\):

\(D = \sqrt{4 \times 2}\)

\(D = 2\sqrt{2}\) meters

Summary of Displacement

The straight distance between the starting point and the point he finally reached is \(2\sqrt{2}\) metres.