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

### Step 1: Identify the components of the velocity vector
The velocity vector can be separated into its components:
- \(v_x = a\) (the x-component)
- \(v_y = b t\) (the y-component)

### Step 2: Relate velocity to position
The components of velocity can be expressed in terms of position:
- \(v_x = \frac{dx}{dt} = a\)
- \(v_y = \frac{dy}{dt} = b t\)

### Step 3: Integrate the x-component
From \(v_x = a\), we can integrate to find the position in the x-direction:
\[ dx = a \, dt \]
Integrating both sides gives:
\[ x = a t + C_1 \]
Assuming the initial condition \(x(0) = 0\), we find \(C_1 = 0\):
\[ x = a t \]

### Step 4: Integrate the y-component
From \(v_y = b t\), we can integrate to find the position in the y-direction:
\[ dy = b t \, dt \]
Integrating both sides gives:
\[ y = \frac{b}{2} t^2 + C_2 \]
Assuming the initial condition \(y(0) = 0\), we find \(C_2 = 0\):
\[ y = \frac{b}{2} t^2 \]

### Step 5: Eliminate time to find the trajectory equation
Now we have:
- \(x = a t\)
- \(y = \frac{b}{2} t^2\)

From the first equation, we can express \(t\) in terms of \(x\):
\[ t = \frac{x}{a} \]

Substituting this into the equation for \(y\):
\[ y = \frac{b}{2} \left(\frac{x}{a}\right)^2 \]
This simplifies to:
\[ y = \frac{b}{2a^2} x^2 \]

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