Given below are two statements: 

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

If a body travels some distance in a given time interval, then for that time interval, its:

A car moves from \(X\) to \(Y\) with a uniform speed \(v_u\) and returns to \(X\) with a uniform speed \(v_d.\) The average speed for this round trip is:

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 \(1,\) \(2\) and \(3,\) respectively are:
                       

The coordinate of an object is given as a function of time by \(x = 7 t - 3 t^{2}\), where \(x\) is in metres and \(t\) is in seconds. Its average velocity over the interval \(t=0\) to \(t=4\) is will be:
1. \(5\) m/s
2. \(-5\) m/s
3. \(11\) m/s
​​​​​​​4. \(-11\) m/s

A particle moving in a straight line covers half the distance with a speed of \(3~\text{m/s}\). The other half of the distance is covered in two equal time intervals with speeds of \(4.5~\text{m/s}\) and \(7.5~\text{m/s}\) respectively. The average speed of the particle during this motion is:

The displacement \((x)\) of a point moving in a straight line is given by; \(x=8t^2-4t.\) Then the velocity of the particle is zero at:

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