Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be `0^@`lt`theta`lt`180^@`. During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions:
An object is projected from origin in x-y plane in which velocity changes according to relation `vecv = a hati + bx hatj`. Path of particle is

Options

A. Hyperbolic
B. Circular
C. Elliptical
D. parabolic

Explanation

To solve the problem of determining the path of a particle projected from the origin in the x-y plane with a velocity that changes according to the relation \(\vec{v} = a \hat{i} + b x \hat{j}\), we can follow these steps:

### Step 1: Identify the components of the velocity vector
The given velocity vector can be broken down into its components:
- Horizontal component: \( V_x = a \)
- Vertical component: \( V_y = b x \)

### Step 2: Relate velocity to position
We know that velocity is the derivative of position with respect to time. Therefore, we can write:
- \( V_x = \frac{dx}{dt} \)
- \( V_y = \frac{dy}{dt} \)

### Step 3: Write the equations for motion
From the horizontal component:
\[ V_x = a = \frac{dx}{dt} \]
This can be rearranged to:
\[ dx = a \, dt \]
Integrating both sides gives:
\[ x = a t + C_1 \]
Since the object is projected from the origin, \(C_1 = 0\), so:
\[ x = a t \quad \text{(Equation 1)} \]

### Step 4: Substitute \(x\) into the vertical component
Now, using the vertical component:
\[ V_y = b x = \frac{dy}{dt} \]
Substituting \(x\) from Equation 1:
\[ V_y = b(a t) = \frac{dy}{dt} \]
This leads to:
\[ dy = b a t \, dt \]
Integrating both sides gives:
\[ y = \frac{b a}{2} t^2 + C_2 \]
Again, since the object starts from the origin, \(C_2 = 0\), so:
\[ y = \frac{b a}{2} t^2 \quad \text{(Equation 2)} \]

### Step 5: Eliminate time to find the relationship between \(x\) and \(y\)
From Equation 1, we can express \(t\) in terms of \(x\):
\[ t = \frac{x}{a} \]
Substituting this into Equation 2:
\[ y = \frac{b a}{2} \left(\frac{x}{a}\right)^2 \]
This simplifies to:
\[ y = \frac{b}{2a} x^2 \]

### Conclusion: Identify the path
The equation \(y = \frac{b}{2a} x^2\) is a quadratic equation in \(x\), which describes a parabolic path. Therefore, the path of the particle is a parabola.

### Final Answer
The path of the particle is a parabolic path.

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