The driver of a bus moving with a velocity of $72 \ km/h$ observes a boy walking across the road at a distance of $50 \ m$ in front of the bus and decelerates the bus at $5 \ m/s^2$ by applying brakes and is just able to avoid an accident. The reaction time of the driver is (in $s$)

$A$ car moves in the positive $Y$-direction with velocity $v$ proportional to the distance travelled $y$ as $v(y) \propto y^\beta$, where $\beta$ is a positive constant. The car covers a distance $L$ with average velocity $\langle v \rangle$ proportional to $L$ as $\langle v \rangle \propto L^{1/3}$. The constant $\beta$ is given as:

$A$ man moves in an open field such that after moving $10 \, m$ on a straight line,he makes a sharp turn of $60^{\circ}$ to his left. The total displacement just at the start of the $8^{\text{th}}$ turn is equal to ........ $m$.

$A$ 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 $1\; s$. Plot the $x-t$ graph of his motion. How long (in $s$) does the drunkard take to fall into a pit $13\; m$ away from the start?

The relation between time $t$ and distance $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. The relation between acceleration $a$ and velocity $v$ is:

At time $t=0$, a particle leaves the origin and moves in the positive direction of the $X$-axis. If the velocity of the particle varies as $v(t)=v_0(1-t/t_0)$, where $|v_0|=10 \ m/s$ and $t_0=10 \ s$, then the distance covered by the particle during the first $20 \ s$ is: (in $m$)