A car, starting from rest, accelerates at a rate of $\beta$ through a distance S, then continues at constant speed for time t and then decelerates at a rate of $\dfrac{\beta}{2}$ to come to rest. If the total distance travelled is $10\ S$, then ratio $\dfrac{S}{t^2}$ is

Let the maximum speed reached by the car be $v$.

For the first part of the motion, the car accelerates from rest at $\beta$ over a distance $S$. Using the equation of kinematics:

$v^2 = 0 + 2\beta S \Rightarrow v = \sqrt{2\beta S}$

For the second part of the motion, the car travels at a constant speed $v$ for time $t$. The distance covered is:

$S_2 = vt$

For the third part of the motion, the car decelerates at $\dfrac{\beta}{2}$ to come to rest. The distance covered $S_3$ is given by:

$0 = v^2 - 2\left(\dfrac{\beta}{2}\right)S_3 \Rightarrow S_3 = \dfrac{v^2}{\beta}$

Substituting $v^2 = 2\beta S$ gives:

$S_3 = \dfrac{2\beta S}{\beta} = 2S$

The total distance travelled is given as $10S$. Therefore:

$S + S_2 + S_3 = 10S$

$S + vt + 2S = 10S$

$vt = 7S$

Substituting $v = \sqrt{2\beta S}$:

$\sqrt{2\beta S} \cdot t = 7S$

Squaring both sides:

$2\beta S \cdot t^2 = 49S^2$

$\dfrac{S}{t^2} = \dfrac{2\beta}{49}$

Answer: $\dfrac{2\beta}{49}$