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:

If three coordinates of a particle change according to the equations \(x = 3 t^{2}, y = 2 t , z= 4\), then the magnitude of the velocity of the particle at time \(t=1\) second will be:
1. \(2\sqrt{11}~\text{unit}\)
2. \(\sqrt{34}~\text{unit}\)
3. \(40~\text{unit}\)
4. \(2\sqrt{10}~\text{unit}\)

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:

In \(1.0~\text{s}\), a particle goes from point \(A\) to point \(B\), moving in a semicircle of radius \(1.0~\text{m}\) (see figure). The magnitude of the average velocity is:
                    

A particle moves along the positive branch of the curve \(y= \frac{x^{2}}{2}\) where \(x= \frac{t^{2}}{2}\), & \(x\) and \(y\) are measured in metres and in seconds respectively. At \(t= 2~\text{s}\), the velocity of the particle will be:
1. \(\left(\right. 2 \hat{i} - 4 \hat{j})~\text{m/s}\) 
2. \(\left(\right. 4 \hat{i} + 2 \hat{j}\left.\right)\text{m/s}\)
3. \(\left(\right. 2 \hat{i} + 4 \hat{j}\left.\right) \text{m/s}\)
4. \(\left(\right. 4 \hat{i} - 2 \hat{j}\left.\right) \text{m/s}\)

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. \(\frac{\overrightarrow{r_{1}} - \overrightarrow{r_{2}}}{\left|\overrightarrow{r_{1}} -  \overrightarrow{r_{2}}\right|} =   \frac{\overrightarrow{v_{2}} -  \overrightarrow{v_{1}}}{\left|\overrightarrow{v_{2}} -   \overrightarrow{v_{1}}\right|}\)$\frac{\left|\stackrel{}{}\right|}{}$

The position of a particle is given by; \(\overrightarrow{r}=(3.0t\hat{i}-2.0t^{2}\hat{j}+4.0\hat{k})\) m $\stackrel{}{\mathrm{}}$
 where \(t\) is in seconds and the coefficients have the proper units for \(r\) to be in meters. The magnitude and direction of \(\overrightarrow{v}(t)\) at \(t=1.0\) s are:

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. \(\frac{\sqrt{3}}{2}\)

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: