What is nature of trajectory of a particle having a uniformly accelerated motion in a plane?

Answer

To determine the nature of the trajectory of a particle undergoing uniformly accelerated motion in a plane, we can follow these steps:

### Step 1: Understand the Motion
In uniformly accelerated motion, the acceleration of the particle is constant in both magnitude and direction. We can represent the motion in a two-dimensional coordinate system (x-y plane).

### Step 2: Set Up the Coordinate System
Assume the x-axis is horizontal and the y-axis is vertical. Let the acceleration vector be along the y-axis. The initial velocity of the particle makes an angle θ with the x-axis.

### Step 3: Break Down the Initial Velocity
The initial velocity \( \mathbf{u} \) can be resolved into two components:
- \( u_x = u \cos \theta \) (horizontal component)
- \( u_y = u \sin \theta \) (vertical component)

### Step 4: Analyze Motion Along the x-axis
Since there is no acceleration in the x-direction, the velocity in the x-direction remains constant:
- \( v_x = u_x = u \cos \theta \)

The displacement in the x-direction after time \( t \) is given by:
\[ x = u_x t \]
Thus, we can express time \( t \) in terms of \( x \):
\[ t = \frac{x}{u \cos \theta} \]

### Step 5: Analyze Motion Along the y-axis
The motion in the y-direction is uniformly accelerated due to the constant acceleration \( \mathbf{a} \). The displacement in the y-direction is given by:
\[ y = u_y t + \frac{1}{2} a t^2 \]
Substituting \( u_y \) and \( t \) into this equation:
\[ y = (u \sin \theta) \left(\frac{x}{u \cos \theta}\right) + \frac{1}{2} a \left(\frac{x}{u \cos \theta}\right)^2 \]

### Step 6: Simplify the Equation
Substituting and simplifying:
\[ y = \frac{u \sin \theta}{u \cos \theta} x + \frac{1}{2} a \frac{x^2}{(u \cos \theta)^2} \]
This can be rewritten as:
\[ y = x \tan \theta + \frac{a}{2u^2 \cos^2 \theta} x^2 \]

### Step 7: Identify the Nature of the Trajectory
The equation derived is of the form:
\[ y = mx + bx^2 \]
where \( m = \tan \theta \) and \( b = \frac{a}{2u^2 \cos^2 \theta} \). This is a quadratic equation in \( x \), which represents a parabola.

### Conclusion
Thus, the trajectory of a particle undergoing uniformly accelerated motion in a plane is parabolic.

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