A particle moves along a parabolic path \(y =9x^2\) in such a way that the \(x\) component of the velocity remains constant and has a value of \(\frac{1}{3}~\text{m/s}\)${\mathrm{}}^{}$. It can be deduced that the acceleration of the particle will be:
1. \(\frac{1}{3}\hat j~\text{m/s}^2\)
2. \(3\hat j~\text{m/s}^2\)
3. \(\frac{2}{3}\hat j~\text{m/s}^2\)
4. \(2\hat j~\text{m/s}^2\)

The relation between time and distance is $\mathrm{t}={\mathrm{\alpha x}}^2+\mathrm{\beta x}$, where $\mathrm{\alpha }$ and $\mathrm{\beta }$ are constants. The retardation is

1.  $2{\mathrm{\alpha v}}^3$

2.  $2{\mathrm{\beta v}}^3$

3.  $2{\mathrm{\alpha \beta v}}^3$

4.  $2{\mathrm{\beta }}^2{\mathrm{v}}^3$

A point moves in a straight line under the retardation ${\mathrm{av}}^2$. If the initial velocity is u, the distance covered in 't' second is-

1.  aut

2.  $\frac{1}{\mathrm{a}}\ln <mspace\; width="1em"></mspace>\left(\mathrm{aut}\right)$

3.  $\frac{1}{\mathrm{a}}\ln <mspace\; width="1em"></mspace>\left(1+\mathrm{aut}\right)$

4.  $\mathrm{a}<mspace\; width="1em"></mspace>\ln <mspace\; width="1em"></mspace>\left(\mathrm{aut}\right)$

Four persons K,L,M and N are initially at the corners of a square of side of length d. If every person starts moving, such that K is always headed towards L,L towards M,M is headed directly towards N and N towards K, then the four persons will meet after 

1. $\frac{\mathrm{d}}{\mathrm{v}}<mspace\; width="1em"></mspace>sec$

2. $\frac{\sqrt{2\mathrm{d}}}{\mathrm{v}}<mspace\; width="1em"></mspace>sec$

3. $\frac{d}{\sqrt{2\mathrm{v}}}<mspace\; width="1em"></mspace>sec$

4. $\frac{\mathrm{d}}{2\mathrm{v}}<mspace\; width="1em"></mspace>sec$

The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement?

1.  

2.  

3.  

4.  

A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be

1.  $110<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-1}$

2.  $55<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-1}$

3.  $550<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-1}$

4.  $660<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-1}$

A particle starts from rest at t=0 and undergoes an acceleration a in ms^-2 with time t in second which is as shown

Which one of the following plot represents velocity v in ms^-1 versus time t in second?

1.  

2.  

3.  

4.  

Two particles \(A\) and \(B\) are moving in a uniform circular motion in concentric circles of radii \(r_A\) and \(r_B\) with speeds \(v_A\) and \(v_B\) respectively. Their time periods of rotation are the same. The ratio of the angular speed of \(A\) to that of \(B\) will be:

The speed of a swimmer in still water is \(20~\text{m/s}.\) The speed of river water is \(10~\text{m/s}\) and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes with respect to the north is given by:

Four persons K,L,M and N are initially at the corners of a square of side of length d. If every person starts moving, such that K is always headed towards L,L towards M,M is headed directly towards N and N towards K, then the four persons will meet after 

1. $\frac{\mathrm{d}}{\mathrm{v}}<mspace\; width="1em"></mspace>sec$

2. $\frac{\sqrt{2\mathrm{d}}}{\mathrm{v}}<mspace\; width="1em"></mspace>sec$

3. $\frac{d}{\sqrt{2\mathrm{v}}}<mspace\; width="1em"></mspace>sec$

4. $\frac{\mathrm{d}}{2\mathrm{v}}<mspace\; width="1em"></mspace>sec$