A particle is moving in a plane with velocity given by `vec(u)=u_(0)hat(i)+(aomega cos omegat)hat(j)`, where `hat(i)` and `hat(j)` are unit vectors along x and y axes respectively. If particle is at the origin at `t=0`. Calculate the trajectory of the particle :-

Options

A. `y= a sin (u_(0)/(omega x))`
B. `y= asin ((omega x)/u_(0))`
C. `y=1/a. sin (u_(0)/(omega x))`
D. `y=1/a. sin((omega x)/u_(0))`

Explanation

To solve the problem, we need to find the trajectory of the particle given its velocity vector:

\[ \vec{u} = u_0 \hat{i} + a \omega \cos(\omega t) \hat{j} \]

where \( \hat{i} \) and \( \hat{j} \) are the unit vectors along the x and y axes, respectively.

### Step 1: Analyze the velocity components
The velocity vector can be broken down into its components:
- The x-component of the velocity is \( u_x = u_0 \).
- The y-component of the velocity is \( u_y = a \omega \cos(\omega t) \).

### Step 2: Relate velocity to position
The relationship between velocity and position is given by:

\[ u_x = \frac{dx}{dt} \quad \text{and} \quad u_y = \frac{dy}{dt} \]

### Step 3: Integrate the x-component
For the x-component:
\[ \frac{dx}{dt} = u_0 \]

Integrating both sides with respect to time \( t \):

\[ dx = u_0 dt \]

Integrating gives:

\[ x = u_0 t + C_x \]

Since the particle is at the origin at \( t = 0 \) (i.e., \( x = 0 \)), we find that \( C_x = 0 \). Thus, we have:

\[ x = u_0 t \quad \text{(Equation 1)} \]

### Step 4: Integrate the y-component
Now, for the y-component:

\[ \frac{dy}{dt} = a \omega \cos(\omega t) \]

Integrating both sides:

\[ dy = a \omega \cos(\omega t) dt \]

Integrating gives:

\[ y = a \omega \int \cos(\omega t) dt = a \omega \left( \frac{\sin(\omega t)}{\omega} \right) + C_y \]

This simplifies to:

\[ y = a \sin(\omega t) + C_y \]

Again, since the particle is at the origin at \( t = 0 \) (i.e., \( y = 0 \)), we find that \( C_y = 0 \). Thus, we have:

\[ y = a \sin(\omega t) \quad \text{(Equation 2)} \]

### Step 5: Eliminate time to find the trajectory
From Equation 1, we can express \( t \) in terms of \( x \):

\[ t = \frac{x}{u_0} \]

Now, substitute this expression for \( t \) into Equation 2:

\[ y = a \sin\left(\omega \frac{x}{u_0}\right) \]

This gives us the trajectory of the particle in the form:

\[ y = a \sin\left(\frac{\omega}{u_0} x\right) \]

### Final Result
The trajectory of the particle is:

\[ y = a \sin\left(\frac{\omega}{u_0} x\right) \]