Car A has an acceleration of `2m//s^2` due east and car B,`4 m//s^2.` due north. What is the acceleration of car B with respect to car A?

Explanation

To find the acceleration of car B with respect to car A, we can follow these steps:

### Step 1: Define the accelerations of both cars
- Car A has an acceleration of \( \vec{a_A} = 2 \, \text{m/s}^2 \) due east, which can be represented as \( \vec{a_A} = 2 \hat{i} \).
- Car B has an acceleration of \( \vec{a_B} = 4 \, \text{m/s}^2 \) due north, represented as \( \vec{a_B} = 4 \hat{j} \).

### Step 2: Write the formula for relative acceleration
The acceleration of car B with respect to car A is given by:
\[ \vec{a_{BA}} = \vec{a_B} - \vec{a_A} \]

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ \vec{a_{BA}} = (4 \hat{j}) - (2 \hat{i}) = -2 \hat{i} + 4 \hat{j} \]

### Step 4: Calculate the magnitude of the relative acceleration
To find the magnitude of \( \vec{a_{BA}} \):
\[ |\vec{a_{BA}}| = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \, \text{m/s}^2 \]

### Step 5: Calculate the angle of the relative acceleration
To find the angle \( \alpha \) that the acceleration makes with the east direction (the x-axis), we can use the tangent function:
\[ \tan \alpha = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{4}{-2} \]
This simplifies to:
\[ \tan \alpha = -2 \]
Thus,
\[ \alpha = \tan^{-1}(-2) \]

### Final Result
The acceleration of car B with respect to car A is:
- Magnitude: \( 2\sqrt{5} \, \text{m/s}^2 \)
- Direction: \( \tan^{-1}(-2) \) radians from the east direction towards the north.

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