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A particle is moving in a plane with vel...

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 :-

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))`

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The correct Answer is:
To find the trajectory of the particle given its velocity vector, we will integrate the velocity components with respect to time. ### Step-by-Step Solution: 1. **Identify the velocity components**: The velocity of the particle is given as: \[ \vec{u} = u_0 \hat{i} + a \omega \cos(\omega t) \hat{j} \] This means: - The x-component of velocity, \( u_x = u_0 \) - The y-component of velocity, \( u_y = a \omega \cos(\omega t) \) 2. **Integrate the x-component**: To find the position in the x-direction, we integrate the x-component of velocity: \[ x(t) = \int u_x \, dt = \int u_0 \, dt = u_0 t + C_x \] Since the particle is at the origin when \( t = 0 \) (i.e., \( x(0) = 0 \)), we find: \[ C_x = 0 \quad \Rightarrow \quad x(t) = u_0 t \] 3. **Integrate the y-component**: Now, we integrate the y-component of velocity: \[ y(t) = \int u_y \, dt = \int a \omega \cos(\omega t) \, dt \] The integral of \( \cos(\omega t) \) is: \[ y(t) = a \omega \left( \frac{\sin(\omega t)}{\omega} \right) + C_y = a \sin(\omega t) + C_y \] Again, since the particle is at the origin when \( t = 0 \) (i.e., \( y(0) = 0 \)), we find: \[ C_y = 0 \quad \Rightarrow \quad y(t) = a \sin(\omega t) \] 4. **Substitute \( t \) from \( x(t) \)**: From the equation \( x(t) = u_0 t \), we can express \( t \) in terms of \( x \): \[ t = \frac{x}{u_0} \] Now substitute this expression for \( t \) into the equation for \( y(t) \): \[ y = a \sin\left(\omega \frac{x}{u_0}\right) \] 5. **Final trajectory equation**: The trajectory of the particle is given by: \[ y = a \sin\left(\frac{\omega}{u_0} x\right) \] ### Conclusion: The trajectory of the particle is: \[ y = a \sin\left(\frac{\omega}{u_0} x\right) \]

To find the trajectory of the particle given its velocity vector, we will integrate the velocity components with respect to time. ### Step-by-Step Solution: 1. **Identify the velocity components**: The velocity of the particle is given as: \[ \vec{u} = u_0 \hat{i} + a \omega \cos(\omega t) \hat{j} ...
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