If the angle `(theta)` between velocity vector and the acceleration vector of a moving particle is `90^(@) lt theta lt 180^(@)`, then the path of the particle is

Options

A. Straight path with retardation
B. Straight path with acceleration
C. Curvilinear path with acceleration
D. Curvilinear path with retardation

Answer

To solve the question, we need to analyze the motion of a particle when the angle \( \theta \) between the velocity vector and the acceleration vector is between \( 90^\circ \) and \( 180^\circ \).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- The velocity vector \( \vec{v} \) indicates the direction of motion of the particle.
- The acceleration vector \( \vec{a} \) indicates how the velocity is changing over time.

2. **Angle Between Vectors**:
- Given that \( 90^\circ < \theta < 180^\circ \), this means that the acceleration vector is directed opposite to the velocity vector. The particle is slowing down.

3. **Velocity Components**:
- We can decompose the velocity vector into components. If we assume the angle \( \phi = \theta - 90^\circ \), then:
- The component of velocity along the x-axis is \( v \cos(\phi) \).
- The component of velocity along the y-axis is \( v \sin(\phi) \).

4. **Equations of Motion**:
- Assuming the x-axis is along the direction of the velocity, we can write the equations of motion:
- For the x-direction:
\[ x = v \cos(\phi) t + \frac{1}{2} a_x t^2 \]
- For the y-direction:
\[ y = v \sin(\phi) t + \frac{1}{2} a_y t^2 \]

5. **Substituting Time**:
- From the x-direction equation, we can express time \( t \) in terms of \( x \):
\[ t = \frac{x}{v \cos(\phi)} \]

6. **Substituting into the y-equation**:
- Substitute \( t \) into the y-equation:
\[ y = v \sin(\phi) \left(\frac{x}{v \cos(\phi)}\right) + \frac{1}{2} a_y \left(\frac{x}{v \cos(\phi)}\right)^2 \]
- Simplifying this gives:
\[ y = x \tan(\phi) - \frac{a_y x^2}{2 v^2 \cos^2(\phi)} \]

7. **Identifying the Path**:
- The equation derived resembles the standard form of a parabola \( y = ax^2 + bx + c \).
- Since the coefficient of \( x^2 \) is negative, it indicates that the parabola opens downwards.

8. **Conclusion**:
- Therefore, when the angle \( \theta \) between the velocity and acceleration vectors is between \( 90^\circ \) and \( 180^\circ \), the path of the particle is a **curvilinear path with retardation**.

### Final Answer:
The path of the particle is a **curvilinear path with retardation**.