When a body moves with a constant speed along a circle,

$A$ man wants to reach from $A$ to the opposite corner of the square $C$. The sides of the square are $100\, m$. $A$ central square of $50\, m \times 50\, m$ is filled with sand. Outside this square,he can walk at a speed of $1\, m/s$. In the central square,he can walk only at a speed of $v\, m/s$ $(v < 1)$. What is the smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

Twelve persons are initially at $12$ corners of a regular polygon of $12$ sides of side length $a$. Each person moves with a uniform speed $v$ in such a manner that person $1$ is always directed towards person $2$,person $2$ towards person $3$,person $3$ towards person $4$,and so on. The time after which they meet is

$A$ particle moves from the point $(2.0\hat i + 4.0\hat j) \, m$ at $t = 0$ with an initial velocity $(5.0\hat i + 4.0\hat j) \, m/s$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat i + 4.0\hat j) \, m/s^2$. What is the distance of the particle from the origin at time $t = 2 \, s$?

$A$ body of mass $m$ is projected at an angle of $45^{\circ}$ with the horizontal with an initial velocity $v$. If air resistance is negligible,then the total change in momentum when it strikes the ground is:

$A$ particle moving in the $xy$ plane experiences a velocity-dependent force $\overrightarrow{F} = k(v_y \hat{i} + v_x \hat{j})$,where $v_x$ and $v_y$ are the $x$ and $y$ components of its velocity $\overrightarrow{v}$. If $\overrightarrow{a}$ is the acceleration of the particle,then which of the following statements is true for the particle?