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 $\mathrm{x}=3{\mathrm{t}}^2,$ $\mathrm{y}=2\mathrm{t},$ $\mathrm{z}=4$, then the magnitude of the velocity of the particle at time \(t=1\) second will be:
1. $2\sqrt{11}$ unit 
2. $\sqrt{34}$ unit
3. \(40\) unit
4. $2\sqrt{10}$ unit

The coordinates of a moving particle at any time ‘t’ are given by x = αt³ and y = βt³. The speed of the particle at time ‘t’ is given by:

1. $\sqrt{{\alpha }^2+{\beta }^2}$

2. $3$ $t\sqrt{{\alpha }^2+{\beta }^2}$

3. $3$ $t^2\sqrt{{\alpha }^2+{\beta }^2}$

4. $t^2\sqrt{{\alpha }^2+{\beta }^2}$

In 1.0 s, a particle goes from point A to point B, moving in a semicircle of radius 1.0 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 \(\vec{v_1}\)   and  \(\vec{v_2}\) . At the initial moment their position vector are  \(\vec{r_1}\) and  \(\vec{r_2}\)  respectively. The condition for particles A and B for their collision to happen will be:

1.  $\overrightarrow{{\mathrm{r}}_{1 }}.\overrightarrow{{\mathrm{v}}_1} =\overrightarrow{{\mathrm{r}}_{2 }}.\overrightarrow{{\mathrm{v}}_2}$

2.  $\overrightarrow{{\mathrm{r}}_1} \mathrm{x} \overrightarrow{{\mathrm{v}}_1} =\overrightarrow{{\mathrm{r}}_2} \mathrm{x} \overrightarrow{{\mathrm{v}}_2}$

3. $\overrightarrow{{\mathrm{r}}_1}- \stackrel{\rightharpoonup }{{\mathrm{r}}_2} =\overrightarrow{{\mathrm{v}}_1}- \overrightarrow{{\mathrm{v}}_2}$

4. $\frac{\overrightarrow{{\mathrm{r}}_1}- \overrightarrow{{\mathrm{r}}_2}}{\left|\overrightarrow{{\mathrm{r}}_1}- \overrightarrow{{\mathrm{r}}_2}\right|}= \frac{\overrightarrow{{\mathrm{v}}_2}- \overrightarrow{{\mathrm{v}}_1}}{\left|\overrightarrow{{\mathrm{v}}_2}- \overrightarrow{{\mathrm{v}}_1}\right|}$

The position of a particle is given by; \(\vec{r}=(3.0t\hat{i}-2.0t^{2}\hat{j}+4.0\hat{k})\) 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 \(\vec{v}(t)\) 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: