The ratio of velocity of two objects A and B is `1:3`. It the position-time graph of object A is inclined to time axis at `30^(@)`, then the position-time graph of object B is inclined to time axis at

Options

A. `0^(@)`
B. `90^(@)`
C. `30^(@)`
D. `60^(@)`

Answer

To solve the problem, we need to analyze the given information about the velocities of two objects A and B and their position-time graphs.

### Step-by-Step Solution:

1. **Understanding the Ratio of Velocities**:
We are given that the ratio of the velocities of objects A and B is \( V_A : V_B = 1 : 3 \). This means:
\[ V_A = k \quad \text{and} \quad V_B = 3k \]
for some constant \( k \).

2. **Analyzing Object A's Position-Time Graph**:
The position-time graph of object A is inclined at an angle of \( 30^\circ \) to the time axis. The slope of a position-time graph gives the velocity of the object. Therefore, we can express the velocity of object A in terms of the angle:
\[ V_A = \text{slope} = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

3. **Finding the Velocity of Object B**:
Since we have \( V_A = \frac{1}{\sqrt{3}} \), we can find \( V_B \) using the ratio:
\[ V_B = 3 \cdot V_A = 3 \cdot \frac{1}{\sqrt{3}} = \sqrt{3} \]

4. **Determining the Angle for Object B's Position-Time Graph**:
The slope of the position-time graph for object B will also give us its velocity:
\[ V_B = \tan(\theta_B) \]
where \( \theta_B \) is the angle of inclination of object B's graph. We know:
\[ V_B = \sqrt{3} \]
Therefore:
\[ \tan(\theta_B) = \sqrt{3} \]

5. **Finding the Angle \( \theta_B \)**:
The angle \( \theta_B \) for which \( \tan(\theta_B) = \sqrt{3} \) is:
\[ \theta_B = 60^\circ \]

6. **Conclusion**:
The position-time graph of object B is inclined to the time axis at \( 60^\circ \).

### Final Answer:
The position-time graph of object B is inclined to the time axis at \( 60^\circ \).

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