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 \(\mathrm{X}\) to \(\mathrm{Y}\) with a uniform speed \(\mathrm{v_u}\) and returns to \(\mathrm{X}\) with a uniform speed \(\mathrm{v_d}.\) The average speed for this round trip is:

1. $\frac{2{\mathrm{v}}_{\mathrm{d}}{\mathrm{v}}_{\mathrm{u}}}{{\mathrm{v}}_{\mathrm{d}}+{\mathrm{v}}_{\mathrm{u}}}$

2. $\sqrt{{\mathrm{v}}_{\mathrm{u}}{\mathrm{v}}_{\mathrm{d}}}$

3. $\frac{{\mathrm{v}}_{\mathrm{d}}{\mathrm{v}}_{\mathrm{u}}}{{\mathrm{v}}_{\mathrm{d}}+{\mathrm{v}}_{\mathrm{u}}}$

4. $\frac{{\mathrm{v}}_{\mathrm{u}}+{\mathrm{v}}_{\mathrm{d}}}{2}$

The figure gives the \((\mathrm{x-t})\) 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:

The coordinate of an object is given as a function of time by $\mathrm{x}=7\mathrm{t}-3{\mathrm{t}}^2$, 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 speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is:  

1. 4.0 m/s

2. 5.0 m/s

3. 5.5 m/s

4. 4.8 m/s

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 \(\mathrm{v}=\mathrm{At}+\mathrm{Bt^{2}},\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants, then the distance travelled by it between \(1\) s and \(2\) s is:

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

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

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

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