A car moves 40 m due east and turns towards north and moves 30 m then tursn `45^(@)` east of norht and moves `20sqrt(2)m`. The net displacment of car is (east is taken positive x-axis, North as positive y-axis)

Options

A. `50hati+60hatj`
B. `60hati+50hatj`
C. `30hati+40hatj`
D. `40hatj+30hatj`

Explanation

To find the net displacement of the car, we will break down its movement into vector components and then sum them up. Here’s a step-by-step solution:

### Step 1: Identify the movements and their vectors
1. The car moves **40 m due east**. In vector form, this is represented as:
\[ \vec{A} = 40 \hat{i} \]
where \(\hat{i}\) represents the east direction (positive x-axis).

2. Next, the car moves **30 m north**. In vector form, this is:
\[ \vec{B} = 30 \hat{j} \]
where \(\hat{j}\) represents the north direction (positive y-axis).

3. Finally, the car turns **45 degrees east of north** and moves **20√2 m**. We need to break this movement into its x and y components:
- The x-component (east) is given by:
\[ \vec{C_x} = 20\sqrt{2} \cos(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \]
- The y-component (north) is given by:
\[ \vec{C_y} = 20\sqrt{2} \sin(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \]
Therefore, the vector for this movement is:
\[ \vec{C} = 20 \hat{i} + 20 \hat{j} \]

### Step 2: Sum the vectors
Now we will sum all the vectors:
\[ \vec{R} = \vec{A} + \vec{B} + \vec{C} \]
Substituting the values:
\[ \vec{R} = (40 \hat{i}) + (30 \hat{j}) + (20 \hat{i} + 20 \hat{j}) \]
Combining like terms:
\[ \vec{R} = (40 + 20) \hat{i} + (30 + 20) \hat{j} = 60 \hat{i} + 50 \hat{j} \]

### Step 3: Final Result
The net displacement of the car is:
\[ \vec{R} = 60 \hat{i} + 50 \hat{j} \]