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

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 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. \(2\hat j~\text{m/s}^2\)

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.  

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

Two particles \(\mathrm{A}\) and \(\mathrm{B}\), move with constant velocities \(\overrightarrow{{v}_1}\) and \(\overrightarrow{{v}_2}\) respectively. At the initial moment, their position vectors are \(\overrightarrow{{r}_1}\) and \(\overrightarrow{{r}_2}\) respectively. The condition for particles \(\mathrm{A}\) and \(\mathrm{B}\) for their collision will be:
1. \(\vec{r_1} \cdot \vec{v_1}=\vec{r_2} \cdot \vec{v_2}\)
2. ​​​​​\(\dfrac{\vec{r_1}-\vec{r_2}}{\left|\vec{r_1}-\vec{r_2}\right|}=\dfrac{\vec{v_2}-\vec{v_1}}{\left|\vec{v_2}-\vec{v_1}\right|}\)${\mathrm{}}_{}$
3. \(\vec{r_1} \times \vec{v_1}=\vec{r_2} \times \vec{v_2}\)${\mathrm{<mspace\; width="1em"></mspace>\; <mspace\; width="1em"></mspace>\; <mspace\; width="1em"></mspace>}}_{}$
4. \(\vec{r_1}-\vec{r_2}=\vec{v_1}-\vec{v_2}\) 

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: