Velocity of a particle varies with time as `v=athati+2ht^(2)hatj`. If the particle starts from point `(0,c)`, the trajectory of the particle is

Options

A. `y=(bx^(3//2))/a+c`
B. `y=(4sqrt(2)b)/3(x/a)^(3//2)+c`
C. `y=(4sqrt(2)b)/3(x/a)^(3//2)-c`
D. `y=(bx^(3//2))/a-c`

Answer

To find the trajectory of the particle given its velocity as \( \mathbf{v} = a \hat{i} + 2b t^2 \hat{j} \) and that it starts from the point \( (0, c) \), we will follow these steps:

### Step 1: Identify the components of velocity
The velocity vector can be broken down into its components:
- In the x-direction: \( v_x = a \)
- In the y-direction: \( v_y = 2b t^2 \)

### Step 2: Write the equations for displacement
Using the relationship between velocity and displacement, we can express the displacements in the x and y directions:
- For the x-direction:
\[ \frac{dx}{dt} = a \implies dx = a \, dt \]
- For the y-direction:
\[ \frac{dy}{dt} = 2b t^2 \implies dy = 2b t^2 \, dt \]

### Step 3: Integrate to find the position functions
Now, we will integrate both equations to find the position as a function of time.

1. **Integrate for x:**
\[ x = \int a \, dt = at + C_x \]
Since the particle starts at \( x = 0 \) when \( t = 0 \):
\[ 0 = a(0) + C_x \implies C_x = 0 \implies x = at \]

2. **Integrate for y:**
\[ y = \int 2b t^2 \, dt = \frac{2b t^3}{3} + C_y \]
Since the particle starts at \( y = c \) when \( t = 0 \):
\[ c = \frac{2b(0)^3}{3} + C_y \implies C_y = c \implies y = \frac{2b t^3}{3} + c \]

### Step 4: Eliminate time to find the trajectory
Now we have:
- \( x = at \)
- \( y = \frac{2b t^3}{3} + c \)

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

Substituting this expression for \( t \) into the equation for \( y \):
\[ y = \frac{2b}{3} \left(\frac{x}{a}\right)^3 + c \]
\[ y = \frac{2b}{3a^3} x^3 + c \]

### Final Result
The trajectory of the particle is given by:
\[ y = \frac{2b}{3a^3} x^3 + c \]