A car travels due east on a level road for `30 km`. It then turns due north at an intersection and travels `40 km` before stopping. Find the resultant displacement of the car.

Explanation

To find the resultant displacement of the car, we can follow these steps:

### Step 1: Understand the motion of the car
The car first travels due east for 30 km and then turns north and travels 40 km. We will represent these movements as vectors.

### Step 2: Represent the movements as vectors
- The eastward movement can be represented as a vector:
\[ \vec{A} = 30 \, \text{km} \, \hat{i} \]
- The northward movement can be represented as another vector:
\[ \vec{B} = 40 \, \text{km} \, \hat{j} \]

### Step 3: Calculate the resultant displacement vector
The resultant displacement vector \(\vec{R}\) can be found by adding the two vectors:
\[ \vec{R} = \vec{A} + \vec{B} = 30 \, \hat{i} + 40 \, \hat{j} \]

### Step 4: Calculate the magnitude of the resultant displacement
To find the magnitude of the resultant displacement, we use the Pythagorean theorem:
\[ |\vec{R}| = \sqrt{(30)^2 + (40)^2} \]
Calculating the squares:
\[ |\vec{R}| = \sqrt{900 + 1600} = \sqrt{2500} \]
Thus,
\[ |\vec{R}| = 50 \, \text{km} \]

### Step 5: Determine the direction of the resultant displacement
To find the angle \(\theta\) that the resultant vector makes with the east direction, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{40}{30} = \frac{4}{3} \]
Calculating \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \]
This angle is measured from the east towards the north.

### Final Result
The resultant displacement of the car is:
- Magnitude: \(50 \, \text{km}\)
- Direction: \(53.13^\circ\) north of east.

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