Given below are two statements: 

A car moves with a speed of \(60\) km/h for \(1\) hour in the east direction and with the same speed for \(30\) min in the south direction. The displacement of the car from the initial position is:
1. \(60\) km
2. \(30 \sqrt{2}\)  km
3. \(30 \sqrt{5}\) km
4. \(60 \sqrt{2}\) km

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

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

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 metres and t is in seconds. What is the displacement of the particle from t = 0 s to t = 6 s?

1.  0

2.  12 m

3.  6 m

4.  18 m