Respuesta :
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.
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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.
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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.
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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.
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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.
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Let's find that value of t.
Recall the following three items
- Vf = Vi + at
- VS2 = VF1 = 0.6t
- VF2 = 0
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.
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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)
