Mathematics, 28.09.2020 14:01 jjjjjj6643

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.

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Answer from: reaperdraws

I'll be using the following variables and values

t = duration of time from rest to max speed (before it starts to decelerate). This means the other portion of the trip takes 10-t seconds a1 = 0.6 = acceleration when elevator is speeding up a2 = -0.8 = acceleration when elevator is slowing down VS1 = start velocity #1 = 0 = elevator is at rest VF1 = final velocity #1 = final velocity when acceleration process stops VS2 = start velocity #2 = starting velocity when deceleration kicks in VF2 = final velocity #2 = final velocity when deceleration is done = 0 d1 = distance from rest to max speed (acceleration) d2 = distance from max speed to 0 (deceleration)

I will also be using the two kinematics equations

Vf = Vi + at x = Vi*t + 0.5*at^2

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 )