Question

Hi I'm kinda stuck on this question and I don't know how to go about it.

For a smooth ("low jerk") ride, an elevator is programmed to start from rest and accelerate according to: a(t)=a(max)/2[1-cos(2*pi*t/T)] from intervals 0 <= t <=T and a(t)= -a(max)/2[1-cos(2*pi*t/T)] from intervals T <= t <=2T where a(max) is the maximum acceleration and 2T is the total time for this trip.

a) Draw sketches of a(t) and the jerk as functions of time.

b)What is the elevator's maximum speed?

c) Find an approximate expression for the seep at short times near the start of the ride, t<<T.

d)What is the time required for a trip of distance D?

Answer 1

We have following acceleration function:

a(t) = (a_max /2) (1 - cos (2πt/T)) for 0<=t<=T

a(t) = -(a_max /2) (1 - cos (2πt/T)) for T<=t<=2T

Part a)

Jerk is defined as  

J = da/dt

For 0<=t<=T, we will have

J(t) =  (a_max /2) (2π/T* sin (2πt/T)) = (π/T) * a_max * sin (2πt/T)

For T<=t<=2T, we will have

J(t) = -(a_max /2) (2π/T* sin (2πt/T)) = -(π/T) * a_max * sin (2πt/T)

Part b):

As Elevator starts from rest at t=0, we will define velocity for 0<=t<=T as:

$v(t)) =\int_{0}^{t}a(t)dt$

$v(t)) =\int_{0}^{t}(amax/2)*(1-cos2\pi t/T)dt$

v(t) = (a_max /2) *[ t - (T/2π)sin (2πt/T)] ------------- Expression 1.

at t = T,

v = (a_max /2) *T ; as sin2π = 0

After t>T, acceleration will be negative and velocity will start decreasing and it will become ZERO(0) at t =2T;

so, maximum velocity will be at t=T and v_T = (a_max /2) *T

Part c)

When t is near to 0; speed will be given at t

v(t) = (a_max /2) *[ t - (T/2π)sin (2πt/T)]

We can find Displacement as :

$x(t)) =\int_{0}^{t}v(t)dt$

$x(t)) =\int_{0}^{t}(amax/2)*(t-(T/2\pi)sin(2\pi t/T))dt$

x(t) = (a_max /2) * ( t²/2 - (T/2π)² cos (2πt/T) + (T/2π)²) ----------------- Expression 2

For approximation at t near to zero,

x(t) = (a_max /2) * t²/2

Part d)

From expression 2, we can find Displacement at t = T,

x(T)= (a_max /2) * ( (T²/2) - (T/2π)² + (T/2π)²) = (a_max /2) * (T²/2)

similarly, elevator will travel another distance (a_max /2) * (T²/2) from t = T to t = 2T

x(2T) = (a_max /2) * (T²/2) *2 = (a_max /2) *T²

If we denote total distance as "D"

D =  (a_max /2) *T²

T = SQRT (2*D /a_max)