A plane is travelling eastward at a speed of `500 kmh^(-1)`. But a `90 km h^(-1)` wind is blowing southward. What is the direction and speed of the plane relative to the ground ?

Answer

To find the speed and direction of the plane relative to the ground, we can follow these steps:

### Step 1: Define the vectors
- The velocity of the plane traveling eastward can be represented as:
\[ \vec{V_p} = 500 \, \text{km/h} \, \hat{i} \]
- The velocity of the wind blowing southward can be represented as:
\[ \vec{V_w} = -90 \, \text{km/h} \, \hat{j} \]

### Step 2: Calculate the resultant velocity
To find the resultant velocity of the plane relative to the ground, we need to add the two vectors:
\[ \vec{V_{pg}} = \vec{V_p} + \vec{V_w} = 500 \, \hat{i} - 90 \, \hat{j} \]

### Step 3: Find the magnitude of the resultant velocity
The magnitude of the resultant velocity can be calculated using the Pythagorean theorem:
\[ |\vec{V_{pg}}| = \sqrt{(500)^2 + (-90)^2} \]
Calculating this gives:
\[ |\vec{V_{pg}}| = \sqrt{250000 + 8100} = \sqrt{258100} \approx 508 \, \text{km/h} \]

### Step 4: Find the direction of the resultant velocity
To find the direction of the resultant velocity, we can use the tangent function:
\[ \tan(\alpha) = \frac{|\text{southward component}|}{|\text{eastward component}|} = \frac{90}{500} \]
Calculating the angle:
\[ \alpha = \tan^{-1}\left(\frac{90}{500}\right) \approx 10.2^\circ \]

### Step 5: State the direction
The direction of the plane relative to the ground is:
\[ 10.2^\circ \text{ south of east} \]

### Final Answer
The speed of the plane relative to the ground is approximately **508 km/h**, and the direction is **10.2 degrees south of east**.

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