A man walks 1km due east and then 1 km due north His displacement?

1. Analyze the question

We are asked to find the displacement of a man who moves in two perpendicular directions: 1 km East and 1 km North. Displacement is a vector quantity, meaning it has both magnitude (how far) and direction (where).

2. Explain the concept

• Displacement: This is the shortest distance between the starting point and the finishing point. Since the man moves East and then North, his path forms a right-angled triangle.
• Pythagoras Theorem: To find the magnitude of the displacement ($R$), we use $R = \sqrt{x^2 + y^2}$, where $x$ and $y$ are the distances moved.
• Direction: We use trigonometry, specifically $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$, to find the angle.

3. Show the solution

Step 1: Calculate the magnitude The man moves 1 km East ($x = 1$) and 1 km North ($y = 1$).

Step 2: Calculate the direction The man is moving from the East towards the North. We find the angle ($\theta$) from the East line:

Since the man is moving North relative to the East line, the direction is 45° North of East (or N 45° E).

4. Identify the answer

Comparing our result ($\sqrt{2}$ km, N 45° E) with the given options, it matches Option A.

5. Exam Tip

Whenever you see a movement problem involving "East" and "North" (or any two perpendicular directions) with equal distances, the resultant magnitude will always be the distance multiplied by $\sqrt{2}$, and the angle will always be 45°. This is a classic property of an isosceles right-angled triangle!