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:

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:

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

The figure gives the \((x\text-t)\) plot of a particle in a one-dimensional motion. Three different equal intervals of time are shown. The signs of average velocity for each of the intervals \(1,\) \(2\) & \(3,\) respectively are: