A particle is moving with veocity $\overrightarrow{\mathrm{v}}=\mathrm{k}(\mathrm{y}\hat{\mathrm{i}}+\mathrm{x}\hat{\mathrm{j}})$; where k is constant. The general equation for the path is:

1. $\mathrm{y}={\mathrm{x}}^2+\mathrm{constant}$

2. ${\mathrm{y}}^2={\mathrm{x}}^2+\mathrm{constant}$

3. $\mathrm{y}=\mathrm{x}+\mathrm{constant}$

4. xy=constant

          
1. \(3.14~\text{m/s}\)
2. \(2.0~\text{m/s}\)
3. \(1.0~\text{m/s}\)
4. zero

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 a time \(t\) is given by:

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. $\frac{11}{5}$($\hat{i}$+$\hat{j}$)

The coordinates of a moving particle at a time t, are given by, x = 5sin10t, y = 5cos10t. The speed of the particle is:

(1)  25

(2)  50

(3)  10

(4)  $50\sqrt{2}$

A car turns at a constant speed on a circular track of radius \(100~\text m,\) taking \(62.8~\text s\) 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. \(2\sqrt{10}~\text{unit}\)

Two particles move from \(A\) to \(C\) and \(A\) to \(D\) on a circle of radius \(R\) and the diameter \(AB.\) If the time taken by both particles is 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 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.