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 yaxes 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 given by u = u_(0)i+ (ao

1.	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 yaxes 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.
Answer» Solution :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)])`