The velocity vector at a point A varies with time as `vecv=ahati+bthatj` , where a and b are positive constants. The equation of trajectory of the point would be

Options

A. `y=1/2(bx^2)/a^2`
B. `y=1/2(ax^2)/b^2`
C. `y=1/2(ax)/b`
D. `y=1/2(bx)/a`

Answer

To find the equation of trajectory given the velocity vector \(\vec{v} = a \hat{i} + b t \hat{j}\), we can follow these steps:

### Step 1: Understand the relationship between velocity and position
The velocity vector \(\vec{v}\) can be expressed as the derivative of the position vector \(\vec{r}\) with respect to time \(t\):
\[ \vec{v} = \frac{d\vec{r}}{dt} \]
This means:
\[ \vec{v} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} \]

### Step 2: Set up the equations for \(dx/dt\) and \(dy/dt\)
From the given velocity vector:
\[ \frac{dx}{dt} = a \quad \text{(1)} \]
\[ \frac{dy}{dt} = b t \quad \text{(2)} \]

### Step 3: Integrate the equations to find \(x\) and \(y\)
Integrate equation (1) with respect to \(t\):
\[ x = \int a \, dt = at + C_1 \]
Assuming \(C_1 = 0\) for simplicity, we have:
\[ x = at \quad \text{(3)} \]

Now, integrate equation (2):
\[ y = \int b t \, dt = \frac{b t^2}{2} + C_2 \]
Assuming \(C_2 = 0\) for simplicity, we have:
\[ y = \frac{b t^2}{2} \quad \text{(4)} \]

### Step 4: Solve for \(t\) in terms of \(x\)
From equation (3):
\[ t = \frac{x}{a} \quad \text{(5)} \]

### Step 5: Substitute \(t\) into the equation for \(y\)
Substituting equation (5) into equation (4):
\[ y = \frac{b}{2} \left(\frac{x}{a}\right)^2 \]
This simplifies to:
\[ y = \frac{b}{2a^2} x^2 \]

### Final Result
Thus, the equation of trajectory is:
\[ y = \frac{b}{2a^2} x^2 \]