Mechanics question (requiring SUVAT formulas)

Question

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

$v = u + at$

$s = \frac{1}{2}(u+v)t$

$s = ut + \frac{1}{2}at^2$

$v^2 = u^2 + 2as$

Answer 1

$$a_1 = \phantom{-}0.6 {\text{ m}}/{\text{s}^2}$$ $$a_2 = -0.8 {\text{ m}}/{\text{s}^2}$$

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.

Answer 2

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.

Answer 3

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 $$s_1+s_2=\frac{v_{max}^2}{2}*\frac{0.6+0.8}{0.6*0.8}z$$