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

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`.

Text Solution

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Given that `u = u_(0)I +(a omega cos omegat)j`,
Hence, velocity along `x`-axis `u_(x) = u_(0) …(1)`
Velocity along `y`-axis `u_(y) = a omega cos omegat …(2)`
We know that `v=(ds)/(dt)` or `s - int v.dt`
So from equations (1) and (2), we have
Displacement at time `t` in horizontal direction
`x =int u_(0) dt =u_(0).t ...(3)`
Displacement in vertical direction
`y = int a omega cos omegat = dt a sin omegat ...(4)`
Eliminating `t` form equaitons (3) and (4) we get
`y =a sin (omegax//u_(0)) ...(5)`
Equations (5) gives the trajectroy of the particle.
(b) At time, `t = 3pi//2 omega`
`x = u_(0) (3pi//2omega)` and `y = a sin 3pi//2 =-a`
`:.` Distance of the particle form teh origin
`=sqrt((x^(2)+y^(2)))=sqrt([((3piu_(0))/(2omega))^(2)+a^(2)])`
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