The displacement of a particle is given by $y<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>a<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>bt<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>c{t}^2<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>d{t}^4$. The initial velocity and acceleration are respectively-

1.  b, -4d

2.  -b, 2c

3.  b, 2c

4.  2c, -4d 

If the velocity of a particle is v=At+Bt${}^2$, where A and B are constants, then the distance travelled by it between 1s and 2s is :

1. 3A+7B

2. $\frac{3}{2}A+\frac{7}{3}B$

3. $\frac{A}{2}+\frac{B}{3}$

4. $\frac{3}{2}A+4B$

The position of a particle with respect to time \(t\) along the \({x}\)-axis is given by \(x=9t^{2}-t^{3}\) where \(x\) is in metres and \(t\) in seconds. What will be the position of this particle when it achieves maximum speed along the \(+{x} \text-\text{direction}?\)
1. \(32~\text m\)
2. \(54~\text m\)
3. \(81~\text m\)
4. \(24~\text m\)

The motion of a particle along a straight line is described by an equation $\mathrm{x}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>8<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>12\mathrm{t}-<mspace\; width="1em"></mspace>{\mathrm{t}}^3$ where x is in metre and t is in second. The retardation of the particle when its velocity becomes zero is:

1.  6 ${\mathrm{ms}}^{-2}$

2.  12 ${\mathrm{ms}}^{-2}$

3.  24 ${\mathrm{ms}}^{-2}$

4.  zero