A car starting from rest accelerates at rate$\alpha $through a distance $x$ then continues at constant speed for time t and then decelerates at a rate of $\dfrac{\alpha }{2}$ to come to rest. If the total distance travelled is $15x$. Then
A.$x=\dfrac{1}{4}\alpha {{t}^{2}}$
B.$x=\dfrac{1}{2}\alpha {{t}^{2}}$
C.$x=\dfrac{1}{8}\alpha {{t}^{2}}$
D.$x=\dfrac{1}{72}\alpha {{t}^{2}}$

Hint: While dealing with problems based on one dimensional Motion of an object in a straight line, we usually come across three kinematic variables namely velocity$\left( v \right)$ , position$\left( s \right)$ and time$\left( t \right)$ and to find one when others are already provided , we use the three equations of motion that are basically equations that describe the behavior of a physical system in terms of the given variables and each could be characterized by an entanglement of two.
The three equations of motion are:
$v=u+gt$
$h=ut+\dfrac{1}{2}g{{t}^{2}}$
${{v}^{2}}-{{u}^{2}}=2gh$
Where $u=Initial\text{ }velocity$
$v=Final\text{ }velocity$
$g=acceleration\text{ }due\text{ }to\text{ }gravity$
$h=height$
$t=time$
It is with the help of these three equations of motion that we are going to solve and give appropriate solutions to the given question.

Complete answer:
The total distance covered by the car = $15x$
$\therefore $ $15x=x+{{x}_{1}}+{{x}_{2}}$
Where, x is the displacement covered during the first phase of travel
${{x}_{1}}$ is the displacement covered during the second phase of travel
${{x}_{2}}$is the displacement covered during the third phase of travel
Let$'{{t}_{0}}'$, $'{{t}_{1}}'$,$'{{t}_{2}}'$be the time duration for covering x, ${{x}_{1}}$,${{x}_{2}}$ displacements respectively.
For phase 1:
 $\Rightarrow x=ut+\dfrac{1}{2}\alpha t_{0}^{2}$
$\Rightarrow x=\dfrac{1}{2}\alpha t_{0}^{2}$
$\left\{ \because u=0 \right\}$
$\Rightarrow v=u+\alpha {{t}_{0}}$
$\Rightarrow v=\alpha {{t}_{0}}$
Where ‘v’ stands for the final velocity in this phase.
For phase 2:
${{x}_{1}}=v{{t}_{1}}$
${{x}_{1}}=(\alpha {{t}_{0}}){{t}_{1}}$
For phase 3:
$\Rightarrow v_{3}^{2}-u_{3}^{2}=2\left( \dfrac{\alpha }{2}{{x}_{3}} \right)$
$\Rightarrow 0-{{(\alpha {{t}_{0}})}^{2}}=2\left( \dfrac{\alpha }{2}{{x}_{3}} \right)$
$\Rightarrow {{x}_{3}}=\alpha t_{0}^{2}$
$\Rightarrow 15x=\dfrac{1}{2}\alpha t_{0}^{2}+(\alpha {{t}_{0}}){{t}_{1}}+\alpha t_{0}^{2}$
$\Rightarrow 15x=x+(\alpha {{t}_{0}}){{t}_{1}}+2x$
$\Rightarrow 12x=\alpha {{t}_{0}}{{t}_{1}}$
From phase 1:
$\Rightarrow x=\dfrac{1}{2}\alpha t_{0}^{2}$
 $\Rightarrow \dfrac{12x}{x}=\dfrac{(\alpha {{t}_{0}}){{t}_{1}}}{\dfrac{1}{2}\alpha t_{0}^{2}}$
$\Rightarrow {{t}_{0}}=\dfrac{{{t}_{1}}}{6}$
After substituting the value of ${{t}_{0}}$
$\Rightarrow 12x=\left( \alpha \dfrac{{{t}_{1}}}{6} \right)\times {{t}_{1}}$
\[\Rightarrow x=\dfrac{\alpha t_{1}^{2}}{72}\]

Therefore, the correct option would be (D) \[x=\dfrac{1}{72}\alpha t_{{}}^{2}\]

Note:
 It is advised to be accurate and be attentive while performing substitution of different values in several equations. And also, the usage of the three equations of motion should also be done carefully and also values must be given appropriately especially the values of initial velocity and final velocity otherwise the student might resultantly get an incorrect solution.