A particle moves in the x-y plane according to rule $\mathrm{x}=\mathrm{asin\omega t}$ and $\mathrm{y}=\mathrm{acos\omega t}$. The particle follows:

The magnitude of vector \(\vec{A}\) is constant but it is changing its direction continuously. The angle between \(\vec {A}\) and \(\frac{d \vec{A}}{dt}\) is:
1. \(180^\circ\)
2. \(120^\circ\)
3. \(90^\circ\)
4. \(0^\circ\)

The angle turned by a body undergoing circular motion depends on the time as given by the equation, \(\theta = \theta_{0} + \theta_{1} t + \theta_{2} t^{2}\). It can be deduced that the angular acceleration of the body is? 
1. \(\theta_1\)
2. \(\theta_2\)
3. \(2\theta_1\)
4. \(2\theta_2\)

A particle moves in a circle of radius \(5\) cm with constant speed and time period \(0.2\pi\) s. The acceleration of the particle is:

If the equation for the displacement of a particle moving on a circular path is given by \(\theta = 2t^3 + 0.5\) where \(\theta\) is in radians and \(t\) in seconds, then the angular velocity of the particle after \(2\) sec from its start is:
1. \(8\) rad/sec
2. \(12\) rad/sec
3. \(24\) rad/sec
4. \(36\) rad/sec

A car moves on a circular path such that its speed is given by \(v= Kt\), where \(K\) = constant and \(t\) is time. Also given: radius of the circular path is \(r\). The net acceleration of the car at time \(t\) will be:
1. \(\sqrt{K^{2} +\left(\frac{K^{2} t^{2}}{r}\right)^{2}}\)
2. \(2K\)
3. \(K\)
4. \(\sqrt{K^{2}   +   K^{2} t^{2}}\)

Two particles \(A\) and \(B\) are moving in a uniform circular motion in concentric circles of radii \(r_A\) and \(r_B\) with speeds \(v_A\) and \(v_B\) respectively. Their time periods of rotation are the same. The ratio of the angular speed of \(A\) to that of \(B\) will be:

A stone tied to the end of a \(1\) m long string is whirled in a horizontal circle at a constant speed. If the stone makes \(22\) revolutions in \(44\) seconds, what is the magnitude and direction of acceleration of the stone?