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

1. $\frac{1}{5}$(13$\hat{i}$+14$\hat{j}$)

2. $\frac{7}{3}$($\hat{i}$+$\hat{j}$)

3. 2($\hat{i}$+$\hat{j}$)

4. $\frac{11}{5}$($\hat{i}$+$\hat{j}$) 

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

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 :

1. 0,0

2. 0, 10 m/s

3. 10 m/s, 10 m/s

4. 10 m/s, 0

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

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

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

A particle moves along the positive branch of the curve y = $\frac{{\mathrm{x}}^2}{2}$ where x = $\frac{{\mathrm{t}}^2}{2}$, & x and y are measured in metres and in seconds respectively. At = 2s, the velocity of the particle will be

1. $(2\hat{\mathrm{i}}-4\hat{\mathrm{j}} )\mathrm{m}/\mathrm{s}$

2. $(4\hat{\mathrm{i}}+2\hat{\mathrm{j}} )\mathrm{m}/\mathrm{s}$

3. $(2\hat{\mathrm{i}}+4\hat{\mathrm{j}} )\mathrm{m}/\mathrm{s}$

4. $(4\hat{\mathrm{i}}-2\hat{\mathrm{j}} )\mathrm{m}/\mathrm{s}$

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:
                 

1. 3.14 m/s

2. 2.0 m/s

3. 1.0 m/s

4. Zero

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

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

1. 30$°$

2. 60$°$

3. 45$°$

4. 75$°$

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

1.  4 m/s at ${53}^o$ with x-axis

2.  4 m/s at ${37}^o$ with x-axis

3.  5 m/s at ${53}^o$ with y-axis

4.  5 m/s at ${53}^o$ with x-axis