Understanding Radha's Movement and Displacement

This problem asks us to find Radha's final distance from her starting point after a series of movements in different directions. This type of problem involves calculating the total displacement, which is a vector quantity representing the straight-line distance and direction from the start to the end point.

To solve this, we can break down each movement into horizontal (East-West) and vertical (North-South) components. We'll use a coordinate system where East is positive x, West is negative x, North is positive y, and South is negative y.

Breaking Down Each Movement Leg

Let's analyze each part of Radha's walk:

1. 9 m towards the South-East: South-East is exactly between South and East. This means the angle with the East (positive x) axis is -45 degrees (or 45 degrees towards South). The components are:
   • East component (x): $9 \times \cos(45^\circ) = 9 \times \frac{1}{\sqrt{2}}$
   • South component (y): $9 \times \sin(45^\circ)$, but in the negative y direction, so $-9 \times \frac{1}{\sqrt{2}}$
2. 15 m towards the West: West is purely in the negative x direction.
   • East component (x): $-15$
   • South component (y): $0$
3. 9 m towards the North-West: North-West is exactly between North and West. This means the angle with the West (negative x) axis is 45 degrees towards North. Or, measured from the positive x-axis, the angle is 135 degrees. The components are:
   • West component (x): $9 \times \cos(45^\circ)$, but in the negative x direction, so $-9 \times \frac{1}{\sqrt{2}}$
   • North component (y): $9 \times \sin(45^\circ)$, in the positive y direction, so $9 \times \frac{1}{\sqrt{2}}$
4. 6 m towards the East: East is purely in the positive x direction.
   • East component (x): $6$
   • South component (y): $0$

Calculating Total Displacement Components

Now, we sum up the x-components and y-components from all four movements to find the total displacement vector (x, y) from the starting point (0,0).

Total x-component ($\Delta x$):

$\Delta x = \left(9 \times \frac{1}{\sqrt{2}}\right) + (-15) + \left(-9 \times \frac{1}{\sqrt{2}}\right) + 6$

$\Delta x = \frac{9}{\sqrt{2}} - 15 - \frac{9}{\sqrt{2}} + 6$

The terms $\frac{9}{\sqrt{2}}$ and $-\frac{9}{\sqrt{2}}$ cancel each other out.

$\Delta x = -15 + 6 = -9$ m

Total y-component ($\Delta y$):

$\Delta y = \left(-9 \times \frac{1}{\sqrt{2}}\right) + 0 + \left(9 \times \frac{1}{\sqrt{2}}\right) + 0$

$\Delta y = -\frac{9}{\sqrt{2}} + \frac{9}{\sqrt{2}}$

The terms $-\frac{9}{\sqrt{2}}$ and $\frac{9}{\sqrt{2}}$ cancel each other out.

$\Delta y = 0$ m

The total displacement vector is $(-9, 0)$. This means Radha's final position is 9 meters to the West of her starting point and at the same North-South level as her starting point.

Calculating Final Distance from Starting Point

The distance from the starting point (0,0) to the final position $(-9, 0)$ is the magnitude of the displacement vector. Using the distance formula (which is also the magnitude of the vector):

Distance $= \sqrt{(\Delta x)^2 + (\Delta y)^2}$

Distance $= \sqrt{(-9)^2 + (0)^2}$

Distance $= \sqrt{81 + 0}$

Distance $= \sqrt{81}$

Distance $= 9$ m

Radha is standing 9 meters away from her starting point.

The final position relative to the start is $(-9, 0)$, meaning 9 meters West and 0 meters North/South. The distance from the start is the magnitude of this vector, which is 9 meters.

Conclusion on Radha's Distance

After all her movements, Radha is 9 meters away from where she started.

Revision Table: Key Concepts for Distance and Displacement

Additional Information on Vector Displacement

Understanding vectors is crucial for solving problems involving movement in multiple directions. A vector has both magnitude (size) and direction. When movements are along diagonal directions like South-East or North-West, we use trigonometry (sine and cosine) to find their components along the standard East-West and North-South axes.

For directions like South-East, North-West, North-East, and South-West (assuming they mean exactly halfway between the cardinal directions), the angle with the axes is 45 degrees. The components are found using $d \cos(45^\circ)$ and $d \sin(45^\circ)$, where $d$ is the distance moved. Remember to assign the correct positive or negative sign based on the quadrant the direction is in (e.g., South-East is +x, -y; North-West is -x, +y).

The total displacement vector is the sum of individual displacement vectors. If the total displacement vector is $(\Delta x, \Delta y)$, the final distance from the starting point is its magnitude, calculated as $\sqrt{(\Delta x)^2 + (\Delta y)^2}$.