The position vector of a particle is \(\vec{r}= a \sin\omega t \hat{i} + a\cos \omega t \hat{j}\). The velocity of the particle is:

1.	parallel to the position vector.
2.	at \(60^{\circ}\) with position vector.
3.	parallel to the acceleration vector.
4.	perpendicular to the position vector.

A projectile is projected from the ground with the velocity \(v_{0}\) at an angle \(\theta\) with the horizontal. What is the vertical component of the velocity of the projectile when its vertical displacement is equal to half of the maximum height attained?
1. \(\sqrt{3} v_{0}\cos\theta\)
2. \(\frac{v_{0}}{\sqrt{2}} \sin\theta\)
3. \(\frac{v_{0}}{\sqrt{2}} \cos \theta\)
4. \(\sqrt{5} v_{0}\)

A particle starts moving on a circular path from rest, such that its tangential acceleration varies with time as \(a_t=kt\). Distance traveled by particle on the circular path in time \(t\) is:
1. \( \frac{kt^3}{3} \)
2. \(\frac{kt^2}{6} \)
3. \(\frac{kt^3}{6} \)
4. \(\frac{k t^2}{2}\)

The speed of water in a river is \(4\) km/h and a man can swim at \(5\) km/h. The minimum time taken by the man to cross the river of width \(200\) m is:
1. \(\frac{1}{5}~\text{h}\)
2. \(\frac{1}{25}~\text{h}\)
3. \(\frac{1}{15}~\text{h}\)
4. \(\frac{1}{20}~\text{h}\)

A particle is moving on a circular path of radius \(R.\) When the particle moves from point \(A\) to \(B\) (angle \( \theta\)), the ratio of the distance to that of the magnitude of the displacement will be:

Two particles move from \(A\) to \(C\) and \(A\) to \(D\) on a circle of radius \(R\) and diameter \(AB\). If the time taken by both particles are the same, then the ratio of magnitudes of their average velocities is:
                                
1. \(2\)
2. \(2\sqrt{3}\)
3. \(\sqrt{3}\)
4. \(\dfrac{\sqrt{3}}{2}\)

A particle moves on the curve ${\mathrm{}}^{}$\(x^2 = 2y\)$\mathrm{}$. The angle of its velocity vector with the \(x\)-axis at the point \(\left(1, \frac{1}{2}\right )\) will be:

A particle is moving along a curve. Select the correct statement.

A car is moving along east at \(10\) m/s and a bus is moving along north at \(10\) m/s. The velocity of the car with respect to the bus is along:

A particle starts moving from the origin in the XY plane and its velocity after time \(t\) is given by \(\overrightarrow{{v}}=4 \hat{{i}}+2 {t} \hat{{j}}\). The trajectory of the particle is correctly shown in the figure:

1.		2.
3.		4.