$A$ 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 $t = 2 \, s$?

$A$ particle starts from the origin at $t=0$ with a velocity $5 \hat{i} \text{ m/s}$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3 \hat{i} + 2 \hat{j}) \text{ m/s}^2$. If the $x$-coordinate of the particle at that instant is $84 \text{ m}$,then the speed of the particle at this time is $\sqrt{\alpha} \text{ m/s}$. The value of $\alpha$ is . . . . . . .

Four persons $K, L, M$ and $N$ are initially at the corners of a square of side length $d$. If every person starts moving with speed $v$ such that $K$ is always headed towards $L$,$L$ towards $M$,$M$ towards $N$,and $N$ towards $K$,then the four persons will meet after

The hour hand of a clock is $6\,cm$ long. The magnitude of the displacement of the tip of the hour hand between $1:00\,PM$ and $5:00\,PM$ is:

At $t = 0$,a projectile is fired from a point $O$ (taken as origin) on the ground with a speed of $50 \, m/s$ at an angle of $53^{\circ}$ with the horizontal. It passes through two points $A$ and $B$,each at a height of $75 \, m$ above the horizontal,as shown in the figure. Find the distance (in meters) of the particle from the origin at $t = 2 \, s$.

The initial position of an object at rest is given by $3 \hat{i}-8 \hat{j}$. It moves with constant acceleration and reaches the position $2 \hat{i}+4 \hat{j}$ after $4 \, s$. What is its acceleration?