Starting from rest, a body moves at first witha constant accelaration a. Then it moves with a uniform velocity and finally with a constant retardation a before coming to rest. If the displacement is s, and time taen is t, prove that the body was in motion with constant velocity for a time interval of root[(t²-4s)/a].

Dear student 

In the question particle has three motion
1) Accelerated motion
2) Uniform motion
3) Decelerated motion

1) Accelerated motion : Let s₁ be the distance covered by the particle in t₁ time-interval. Since particle starts from rest, u=0 and a is the acceleration so from equation of motion 
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s_1=\frac{1}{2}a{t}_1^2$              .................(1)
and the velocity of the particle at the end of the time interval is given by

$v=u+at$
$\Rightarrow v=a{t}_1$                      ..................(2)

2) Uniform motion : Let ​s₂ be the distance covered by the particle in t₂ time-interval with uniform velocity v so

$s_2=v{t}_2$
$\Rightarrow s_2=a{t}_1{t}_2$                 ...................(3)

3) Decelerated motion : ​Let ​s₃ be the distance covered by the particle in t₃ time-interval with deceleration a. The velocity of the particle at the start of time interval ​t₃ is v and final velocity is 0 so
$$\begin{gathered} 0=v-a{t}_3 \\ \Rightarrow v=a{t}_3 \\ \Rightarrow a{t}_1=a{t}_3 \end{gathered}$$
$\Rightarrow t_1=t_3$                  ..................(4)
and ​s₃ is given by 
    $s_3=v{t}_3-\frac{1}{2}a{t}_3^2$
    $s_3=a{t}_1{t}_3-\frac{1}{2}a{t}_3^2$
but using equation (4)
    $s_3=a{t}_1{t}_1-\frac{1}{2}a{t}_1^2$
$\Rightarrow s_3=\frac{1}{2}a{t}_1^2$         ..............(5)

Now if total distance travelled by particle is s and total time taken is t then 
$$\begin{gathered} t_1+t_2+t_3=t \\ \Rightarrow 2t_1+t_2=t \\ \Rightarrow t_1=\frac{t-t_2}{2} \end{gathered}$$
and 
$$\begin{gathered} s=\frac{1}{2}a{t}_1^2+a{t}_1{t}_2+\frac{1}{2}a{t}_1^2 \\ \Rightarrow s=a{t}_1^2+a{t}_1{t}_2 \\ \Rightarrow s=a{\left(\frac{t-t_2}{2}\right)}^2+a{t}_2\left(\frac{t-t_2}{2}\right) \\ \Rightarrow s=\frac{a}{4}\left(t^2+t_2^2-2t{t}_2\right)+\frac{a}{2}\left(t{t}_2-t_2^2\right) \\ \Rightarrow 4s=a{t}^2+a{t}_2^2-2at{t}_2+2at{t}_2-2a{t}_2^2 \\ \Rightarrow 4s=a{t}^2-a{t}_2^2 \\ \Rightarrow t_2^2=t^2-\frac{4s}{a} \\ \Rightarrow t_2=\sqrt{t^2-\frac{4s}{a}} \end{gathered}$$