The velocity acquired by a body moving with uniform acceleration is $30 \ m/s$ in $2 \ s$ and $60 \ m/s$ in $4 \ s$. The initial velocity is $............. \ m/s$.

$A$ car passes three points $A$, $B$ and $C$ at $3 \text{ m/s}$, $6 \text{ m/s}$ and $9 \text{ m/s}$ respectively in a straight line with uniform acceleration. If distance $AB = 60 \text{ m}$, then find the distance $BC$. (in $m$)

The position of a particle moving along the $x$ axis is given by the equation $x = (10 + 6t - 3t^2) \ m$. The distance travelled by the particle from $t = 1 \ s$ to $t = 4 \ s$ is: (in $m$)

The acceleration of a particle is increasing linearly with time $t$ as $a = 6t$. The particle starts from the origin with an initial velocity $u = 10 \ m/s$. The distance travelled by the particle after $t = 2 \ s$ will be: (in $m$)

$A$ small block slides down on a smooth inclined plane,starting from rest at time $t=0$. Let $S_{n}$ be the distance travelled by the block in the interval $t=n-1$ to $t=n$. Then,the ratio $\frac{S_{n}}{S_{n+1}}$ is

$A$ body moves on a frictionless plane starting from rest. If $S_{n}$ is the distance moved between $t=n-1$ and $t=n$,and $S_{n-1}$ is the distance moved between $t=n-2$ and $t=n-1$,then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1-\frac{2}{x}\right)$ for $n=10$. The value of $x$ is: