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:
1. $4.0~\text{m/s}$
2. $5.0~\text{m/s}$
3. $5.5~\text{m/s}$
4. $4.8~\text{m/s}$

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