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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))`

Text Solution

AI Generated Solution

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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A particle is moving in a plane with velocity given by u = u_(0)i+ (aomega cos omegat)j , where i and are j are unit vectors along x and y axes respectively. If particle is at the origin at t = 0 . (a) Calculate the trajectroy of the particle. (b) Find the distance form the origin at time 3pi//2omega .

A particle is moving in a plane with velocity vec(v) = u_(0)hat(i) + k omega cos omega t hat(j) . If the particle is at origin at t = 0 , (a) determine the trajectory of the particle. (b) Find its distance from the origin at t = 3pi//2 omega .

Knowledge Check

  • The angle which the vector vec(A)=2hat(i)+3hat(j) makes with the y-axis, where hat(i) and hat(j) are unit vectors along x- and y-axis, respectively, is

    A
    `cos^(-1)(3//5)`
    B
    `cos^(-1)(2//3)`
    C
    `tan^(-1)(2//3)
    D
    `sin^(-1)(2//3)`
  • The unit vector along hat(i)+hat(j) is

    A
    `hat(k)`
    B
    `hat(i)+hat(j)`
    C
    `(hat(i)+hat(j))/(sqrt(2))`
    D
    `(hat(i)+hat(j))/(2)`
  • If hat(i), hat(j) and hat(k) are unit vectors along x,y and z-axis respectively, the angle theta between the vector hat(i) + hat(j) + hat(k)" ""and vector" hat(j) is given by

    A
    `theta = cos^(-1) ((1)/(sqrt3))`
    B
    `theta = sin^(-1) ((1)/(sqrt3))`
    C
    `theta = cos^(-1) ((sqrt3)/(2))`
    D
    `theta = sin^(-1) ((sqrt3)/(2))`
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