Mechanics question (requiring SUVAT formulas) An elevator ascends from rest with an acceleration of 0.6 m/s^2, before slowing down with a deceleration of 0.8 m/s^2 for the next stop. The total time taken is 10 seconds. Find the distance between the stops.
I have tried this problem over multiple spells over and over again using the four suvat formulas  and some basics Mechanics concepts I have been taught (I have just begun the course). The calculations have wounded up too complicated to mention here. Can someone help?
Edit: The "suvat formulas" (as they called it in my book) are as follows:
$v$ = final velocity, $u$ = initial velocity, $t$ = time, $s$ = displacement, $a$ = acceleration

$v = u + at$
$s = \frac{1}{2}(u+v)t$
$s = ut + \frac{1}{2}at^2$
$v^2 = u^2 + 2as$

 A: The elevator starts from rest.
Suppose the time spent increasing the speed is $t_{1}$ and the time spent in decreasing the speed and coming to rest is $t_{2}$.
It is given $t_{1} + t_{2}=10$.
Let the maximum speed attained is $v_{max}$
By the 1st equation of motion,applied during the acceleration period, we get
$0 +0.6*t_1=v_{max}$
Similarly,$v_{max}=0.8*t_2,$.
i.e.$$3*t_{1}=4t_{2}$$.
Thus, the maximum speed attained is $v_{max}= 0+0.6*10/7=6/7.$
Now, we use the third equation of motion, $v^2=u^2+2*a*s$.
This gives $s_1$, the time spent increasing the speed as $$s_1=\frac{v_{max}^2}{2*0.6}$$.
Similarly, the distance  travelled while decrasing the speed to 0 is
$$s_2=\frac{v_{max}}{2*0.6*0.8}$$.
Hence, we have the total distance $$s_1+s_2=\frac{v_{max}^2}{2}*\frac{0.6+0.8}{0.6*0.8}z$$
A: $$a_1 = \phantom{-}0.6 {\text{ m}}/{\text{s}^2}$$
$$a_2 = -0.8 {\text{ m}}/{\text{s}^2}$$
The basic equations of motion that we will use are:
$$\begin{aligned}
v & = a t + v_0  \\
x & = \frac{1}{2} a t^2 + v_0 t + x_0
\end{aligned}$$
Applying the first equation for the accelearation phase:
$$v^\star = a_1 t_1$$
On deceleration $$0=a_2 t_2 + v^\star$$
Combining these equations:
$$ a_1 t_1 + a_2 t_2 =0.$$
The total time is given as 10 seconds.
So now we have two equations in two unknowns:
\begin{aligned}
a_1 t_1 &+ a_2 t_2 &= \phantom{0}0\\
t_1 &+ \phantom{a_2}t_2 &=10
\end{aligned}
So $$\begin{aligned} t_1 &= -\frac{10 a_2 }{a_1-a_2} =\frac{8}{1.4}  \\
t_2 &= \phantom{-}\frac{10a_1}{a_1-a_2} = \frac{6}{1.4}\end{aligned}$$
The distance is
$$d= \frac{1}{2} \left( a_1 t_1^2 -a_2 t_2^2\right) \approx 17.1429 \text{ m}$$
We've applied our distance equation for the acceleration phase with $v_0=0$ starting at $x_0=0$.  For the decelaration phase, we've "run the equation backwards."  The distance covered accelerating from $0$ m/s to $v^\star$ is the distance covered decelarating from $v^\star$ to $0$ m/s.  The total distance is the sum of the distances over each phase.
A: Let $T$ denote the time at which the elevator stops accelerating and starts decelerating. Then, since the final velocity is 0, find the time:
$$0.6(m/sec^2) T-0.8(m/sec^2)(10sec-T)=0.$$
Solve for $T$, then use the fact that $v=0.6t$ for $t<T$ and $v=0.6T-0.8t$ for $t\geq T$ to get the distance travelled.
