Consider the motion of the tip of the second hand of a clock. In one minute (assuming \(R\) to be the length of the second hand), its:

Three girls skating on a circular ice ground of radius \(200\) m start from a point \(P\) on the edge of the ground and reach a point \(Q\) diametrically opposite to \(P\) following different paths as shown in the figure. The correct relationship among the magnitude of the displacement vector for three girls will be:

1. \(A > B > C\)
2. \(C > A > B\)
3. \(B > A > C\)
4. \(A = B = C\)

A cat is situated at point \(A\) (\(0,3,4\)) and a rat is situated at point \(B\) (\(5,3,-8\)$\mathrm{}$). The cat is free to move but the rat is always at rest. The minimum distance travelled by the cat to catch the rat is:
1. \(5\) unit
2. \(12\) unit
3. \(13\) unit
4. \(17\) unit

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:

A vector $\overrightarrow{a}$ is turned without a change in its length through a small angle $d\theta .$ The value of $\left|\Delta \overrightarrow{a}\right|$ and $\Delta a$ are, respectively:

A particle is moving such that its position coordinates \((x,y)\) are \((2\) m, \(3\) m) at time \(t=0,\)  \((6\) m, \(7\) m) at time \(t=2\) s and   \((13\) m, \(14\) m) at time \(t=5\) s. Average velocity vector \((v_{avg})\) from \(t=0\) to \(t=5\) s is:

A car turns at a constant speed on a circular track of radius \(100\) m, taking \(62.8\) s for every circular lap. The average velocity and average speed for each circular lap, respectively, is:

The coordinates of a moving particle at any time \(t\) are given by \(x= \alpha t^3\) and \(y = \beta t^3\). The speed of the particle at time \(t\) is given by:

Two particles \(A\) and \(B\), move with constant velocities \(\overrightarrow{v_1}\) and \(\overrightarrow{v_2}\). At the initial moment their position vector are \(\overrightarrow {r_1}\) and \(\overrightarrow {r_2}\) respectively. The condition for particles \(A\) and \(B\) for their collision to happen will be:
1. \(\overrightarrow{r_{1  }} . \overrightarrow{v_{1}}   = \overrightarrow{r_{2  }} . \overrightarrow{v_{2}}\)
2. \(\overrightarrow{r_{1}} \times\overrightarrow{v_{1}}   = \overrightarrow{r_{2}} \times \overrightarrow {v_{2}}\)
3. \(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}=\overrightarrow{v_{1}} - \overrightarrow{v_{2}}\)
4. \(\frac{\overrightarrow{r_{1}} - \overrightarrow{r_{2}}}{\left|\overrightarrow{r_{1}} -  \overrightarrow{r_{2}}\right|} =   \frac{\overrightarrow{v_{2}} -  \overrightarrow{v_{1}}}{\left|\overrightarrow{v_{2}} -   \overrightarrow{v_{1}}\right|}\)

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. \(\frac{\sqrt{3}}{2}\)