A body starts moving from rest on a horizontal ground such that the position vector of the body with respect to its starting point is given by $\mathrm{r}$ $=$ $2\mathrm{t}\hat{\mathrm{i}}$ $+$ $3{\mathrm{t}}^2\hat{\mathrm{j}}$. The equation of the trajectory of the body is:

1. y = 1.5 x

2. y = 0.75 ${\mathrm{x}}^2$

3. y = 1.5 ${\mathrm{x}}^2$

4. y = 0.45 ${\mathrm{x}}^2$

A particle starts moving with constant acceleration with initial velocity (\(\hat{\mathrm{i}}+5\hat{\mathrm{j}}\)$\stackrel{}{\mathrm{}}$) m/s. After \(4\) seconds, its velocity becomes (\(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}\)) m/s. The magnitude of its displacement in 4 seconds is:
1. \(5\) m
2. \(10\) m
3. \(15\) m
4. \(20\) m

If a particle is moving in a circular orbit with constant speed, then:

A projectile is projected with initial kinetic energy K. If it has kinetic energy 0.25K at its highest point, then the angle of projection is:

1. 30°

2. 45°

3. 60°

4. 75°

If the position of a particle varies according to the equations \(x= 3t^2\)${\mathrm{}}^{}$, \(y =2t\), and \(z= 4t+4\), then which of the following is incorrect?

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:

In a uniform circular motion, if the speed of the particle is 2 m/s and radius of the circle is 2 m, then the values of centripetal and tangential acceleration are, respectively:

1. 2 $\mathrm{m}/{\mathrm{s}}^2$, 2 $\mathrm{m}/{\mathrm{s}}^2$

2. 2 $\mathrm{m}/{\mathrm{s}}^2$, 1 $\mathrm{m}/{\mathrm{s}}^2$

3. 0, 2 $\mathrm{m}/{\mathrm{s}}^2$

4. 2 $\mathrm{m}/{\mathrm{s}}^2$, 0

A person, who can swim with speed u relative to water, wants to cross a river (of width d and water is flowing with speed v). The minimum time in which the person can do so is:

1.  $\frac{\mathrm{d}}{\mathrm{v}}$

2.  $\frac{\mathrm{d}}{\mathrm{u}}$

3.  $\frac{\mathrm{d}}{\sqrt{{\mathrm{v}}^2 + {\mathrm{u}}^2}}$

4.  $\frac{\mathrm{d}}{\sqrt{{\mathrm{v}}^2 - {\mathrm{u}}^2}}$

The position vector of a particle $\overrightarrow{\mathrm{r}}$ as a function of time t (in seconds) is $\overrightarrow{\mathrm{r}}$ $=$ $\left(3\mathrm{t}\right)\hat{\mathrm{i}}$ $+$ $\left(2{\mathrm{t}}^2\right)\hat{\mathrm{j}}$ $\mathrm{m}$. The initial acceleration of the particle is:

1. 2 $\mathrm{m}/{\mathrm{s}}^2$

2. 3 $\mathrm{m}/{\mathrm{s}}^2$

3. 4 $\mathrm{m}/{\mathrm{s}}^2$

4. Zero

Coordinates of a particle as a function of time t are x = 2t,
y = 4t. It can be inferred that the path of the particle will be: