The displacement \(x\) of a particle moving in one dimension under the action of a constant force is related to time \(t\) by the equation $$\(t=\sqrt{x}+3,\) where \(x\) is in meters and \(t\) is in seconds. What is the displacement of the particle from \(t=0~\text s\) to \(t = 6~\text s?\)

1. \(0\)
2. \(12~\text m\)
3. \(6~\text m\)
4. \(18~\text m\)

Given below are two statements: 

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

$1.$ $3\mathrm{A}+7\mathrm{B}$
$2.$ $\frac{3\mathrm{A}}{2}+\frac{7\mathrm{B}}{3}$
$3.$ $\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{3}$
$4.$ $\frac{\mathrm{A}}{3}+\frac{\mathrm{B}}{2}$

A particle moves along a path \(ABCD\) as shown in the figure. The magnitude of the displacement of the particle from \(A\) to \(D\) is:

1. $(5+10\sqrt{2})$m
2. \(10\) m
3. $15\sqrt{2}$ m
4. \(15\) m

Which one is not an example of a point object?

A drunkard walking in a narrow lane takes \(5\) steps forward and \(3\) steps backward, followed again by \(5\) steps forward and \(3\) steps backward, and so on. Each step is \(1\) m long and requires \(1\) s. There is a pit on the road \(13\) m away from the starting point. The drunkard will fall into the pit after:
1. \(37\) s
2. \(31\) s
3. \(29\) s
4. \(33\) s