A body lying initially at point (3,7) starts moving with a constant acceleration of `4 hati`. Its position after 3s is given by the coordinates

Options

A. (7,3)
B. (7,18)
C. (21,7)
D. (3,7)

Explanation

To solve the problem step-by-step, we will follow the principles of motion under constant acceleration.

### Step 1: Identify Initial Conditions
The body starts at the initial position given by the coordinates (3, 7). This means:
- Initial position \( x_0 = 3 \)
- Initial position \( y_0 = 7 \)

### Step 2: Understand the Acceleration
The body has a constant acceleration of \( 4 \hat{i} \). This means:
- The acceleration in the x-direction \( a_x = 4 \, \text{m/s}^2 \)
- There is no acceleration in the y-direction, so \( a_y = 0 \).

### Step 3: Determine Initial Velocity
Since the body starts moving from rest, the initial velocity is:
- Initial velocity \( u_x = 0 \, \text{m/s} \) (in the x-direction)
- Initial velocity \( u_y = 0 \, \text{m/s} \) (in the y-direction)

### Step 4: Use the Second Equation of Motion
To find the displacement in the x-direction after a time \( t = 3 \, \text{s} \), we will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
- \( u = 0 \)
- \( a = 4 \, \text{m/s}^2 \)
- \( t = 3 \, \text{s} \)

Calculating the displacement:
\[ s_x = 0 \cdot 3 + \frac{1}{2} \cdot 4 \cdot (3^2) \]
\[ s_x = 0 + \frac{1}{2} \cdot 4 \cdot 9 \]
\[ s_x = 2 \cdot 9 = 18 \, \text{m} \]

### Step 5: Calculate the Final Position
Now, we can find the final position in the x-direction:
\[ x_f = x_0 + s_x = 3 + 18 = 21 \]

Since there is no change in the y-direction:
\[ y_f = y_0 = 7 \]

### Step 6: Write the Final Coordinates
The final coordinates of the body after 3 seconds are:
\[ (21, 7) \]

### Final Answer
The position of the body after 3 seconds is given by the coordinates \( (21, 7) \).

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