If the velocity of a particle is given by $v = (180 - 16x)^{1/2} \text{ m/s}$,then its acceleration will be ....... $\text{m/s}^2$.

The distance travelled by a particle is directly proportional to $t^{1/2}$,where $t =$ time elapsed. What is the nature of motion?

$A$ particle is moving in a straight line with initial velocity $u$ and uniform acceleration $a$. If the sum of the distance travelled in the $t^{\text{th}}$ and $(t+1)^{\text{th}}$ seconds is $100 \text{ cm}$,then its velocity after $t$ seconds,in $\text{cm/s}$,is:

$A$ particle starts moving rectilinearly at time $t = 0$ such that its velocity $v$ changes with time $t$ according to the equation $v = t^2 - t$,where $t$ is in seconds and $v$ is in $m/s$. The time interval for which the particle retards is

$A$ particle is moving along the $X$-axis with velocity $v = e^{-\beta x}$. At time $t = 0$,the particle is located at $x = 0$. The displacement of the particle as a function of time is

$A$ body moving with uniform acceleration travels a distance of $25 \ m$ in the fourth second and $37 \ m$ in the sixth second. The distance covered by the body in the next two seconds is: (in $m$)