Someone plz help me with this problem: An elevator ascends from rest with an acceleration of 0.6 ms^2 before slowing down with a decelleration of 0.8 ms^2 for the next stop. The total time taken is 10 seconds. Find the distance between the stops. The answer is 17.1 but I dont understand how to get it . Dont answer if you dont know or I will report you.

I'll be using the following variables and values

I will also be using the two kinematics equations

where Vi and Vf are the initial and final velocities respectively, 'a' is the acceleration, t is time, and x is displacement.

To start, let's focus on just the accelerating portion of the journey.

The elevator starts at rest, so VS1 = 0. It reaches some current unknown final velocity, but we can calculate that as such

Vf = Vi + at  

VF1 = VS1 + a1*t

VF1 = 0 + 0.6*t

VF1 = 0.6t

The elevator reaches a top speed of 0.6t meters per second after traveling and accelerating t seconds.

Let's see how much the elevator is displaced after those t seconds (again only focus on the journey where the elevator is speeding up for now)

x = Vi*t + 0.5*at^2

d1 = VS1*t + 0.5*a1*t^2

d1 = 0*t + 0.5*0.6*t^2

d1 = 0.3t^2

The elevator travels 0.3t^2 meters when it is speeding up. We'll use this later.

Now onto the portion where the elevator is slowing down.  

The starting speed for this portion is VS2 = VF1 = 0.6t because the transition from the last bit of the accelerating portion to the decelerating portion has no gaps. Let's assume it's a smooth transition without any sudden bumps to make the ride uncomfortable.  

Let's find the displacement for this portion.

x = Vi*t + 0.5*at^2

d2 = VS2*(10-t) + 0.5*a2*(10-t)^2 note the 10-t terms

d2 = 0.6t*(10-t) + 0.5*(-0.8)(100-20t+t^2)

d2 = 6t-0.6t^2 - 40+8t-0.4t^2

d2 = -t^2 + 14t - 40

This is the distance it travels from its top speed, decelerating all the way down to rest.  

Add the two distances

d = total distance

d = d1+d2

d = ( d1 ) + ( d2 )

d = ( 0.3t^2 ) + ( -t^2 + 14t - 40 )

d = 0.3t^2 - t^2 + 14t - 40

d = -0.7t^2 + 14t - 40

If we knew the value of t (as defined at the very top of this solution), then we could find the total distance traveled.

Let's find that value of t.

Recall the following three items

So,

Vf = Vi + at

VF2 = VS2 + a2*(10-t) note the 10-t portion

0 = 0.6t + (-0.8)(10-t)

0 = 0.6t - 8 + 0.8t

0 = 1.4t - 8

8 = 1.4t

1.4t = 8

t = 8/1.4

t = 5.714 approximately

It takes about 5.714 seconds for the elevator to go from rest to its top speed, before the elevator starts the deceleration process.

We can now find the value of d

d = -0.7t^2 + 14t - 40

d = -0.7(5.714)^2 + 14(5.714) - 40

d = 17.1411428

d = 17.1 is the approximate total distance traveled (in meters)

step-by-step explanation: