Solving the Direction and Distance Problem

The question asks us to find the straight-line distance of Srinija from her starting point after a series of movements in different directions. To solve this, we need to track her net displacement in both the east-west and north-south directions.

Analyzing Srinija's Movements

Srinija's movements are:

• 9 km towards the west
• 11 km towards the south
• 13 km towards the east
• 4 km towards the west

Calculating Net Displacement

Let's consider the movements along the East-West axis and the North-South axis separately. We can assign directions as positive or negative. For example, let East be positive and West be negative. Let North be positive and South be negative.

Net Movement in East-West Direction:

Initial movement: 9 km West (-9 km)

Next movement: 13 km East (+13 km)

Final movement: 4 km West (-4 km)

Total displacement in the East-West direction = $(-9) + (+13) + (-4)$ km

Total East-West displacement = $-9 + 13 - 4$ km

Total East-West displacement = $13 - (9 + 4)$ km

Total East-West displacement = $13 - 13$ km

Total East-West displacement = $0$ km

This means there is no net displacement in the East-West direction relative to the starting point.

Net Movement in North-South Direction:

Only one movement in the North-South direction: 11 km South (-11 km)

Total displacement in the North-South direction = $-11$ km

This means Srinija's final position is 11 km south of her starting point in the North-South direction.

Determining Final Distance from Initial Position

Srinija's final position is 0 km East/West and 11 km South from her initial position. We can visualize this as a right-angled triangle where the net East-West displacement is one side and the net North-South displacement is the other side. However, in this case, the net East-West displacement is 0.

The straight-line distance from the initial position is the magnitude of the net displacement vector. If the net East-West displacement is $\Delta x$ and the net North-South displacement is $\Delta y$, the distance $D$ is given by the formula:

\begin{equation*} D = \sqrt{(\Delta x)^2 + (\Delta y)^2} \end{equation*}

In this case, $\Delta x = 0$ km and $\Delta y = -11$ km (or simply 11 km magnitude in the South direction).

\begin{equation*} D = \sqrt{(0)^2 + (-11)^2} \end{equation*}

\begin{equation*} D = \sqrt{0 + 121} \end{equation*}

\begin{equation*} D = \sqrt{121} \end{equation*}

\begin{equation*} D = 11 \text{ km} \end{equation*}

So, Srinija is 11 km away from her initial position.

Summary of Displacements

Net East-West displacement = $-9 + 13 - 4 = 0$ km

Net North-South displacement = $-11$ km

Distance from start = $\sqrt{(0)^2 + (-11)^2} = 11$ km.

The final answer is 11 km.

Revision Table: Key Concepts in Direction and Distance

Additional Information: Direction and Distance Calculations

Direction and distance problems often involve visualizing movements on a 2D plane. We typically use a coordinate system where:

• East is along the positive x-axis.
• West is along the negative x-axis.
• North is along the positive y-axis.
• South is along the negative y-axis.

Each movement segment can be represented as a vector. The final position is the sum of all these displacement vectors. The distance from the initial position is the magnitude of the resultant vector.

In this problem:

• 9 km West is a vector $(-9, 0)$.
• 11 km South is a vector $(0, -11)$.
• 13 km East is a vector $(13, 0)$.
• 4 km West is a vector $(-4, 0)$.

The total displacement vector is the sum:

$(-9, 0) + (0, -11) + (13, 0) + (-4, 0) = (-9 + 0 + 13 - 4, 0 - 11 + 0 + 0) = (0, -11)$.

The final position is at coordinates $(0, -11)$ assuming the start is at $(0,0)$.

The distance from the initial position $(0,0)$ to the final position $(0, -11)$ is indeed $\sqrt{(0-0)^2 + (-11-0)^2} = \sqrt{0^2 + (-11)^2} = \sqrt{121} = 11$ km.

This vector approach confirms the result obtained by calculating net displacements along axes separately.