A liftascends with constant acceleration a, then with constant velocity and finally stops under a constant retardation a. If the total distance ascended is s, and the total time occupied is t,show that the time during which the lift is ascending is given by {t^2 -(4s/a)}
To find the time during which the lift is ascending, we need to break down the motion into three different phases:
1. Acceleration phase: During this phase, the lift is ascending and accelerating with a constant acceleration, a. Let's denote the duration of this phase as t1.
2. Constant velocity phase: Once the lift reaches its maximum velocity, it moves upwards with a constant velocity. Let's denote the duration of this phase as t2.
3. Retardation phase: At the end of the journey, the lift decelerates with a constant retardation, -a, until it comes to a stop. Let's denote the duration of this phase as t3.
Now, let's calculate the distances covered during each phase:
1. Acceleration phase:
The distance covered during acceleration can be calculated using the formula:
s1 = (1/2) * a * t1^2
2. Constant velocity phase:
Since the lift is moving at a constant velocity, the distance covered during this phase will be:
s2 = V * t2
Note that in this phase, the lift covers the same distance going up as it does coming down due to symmetrical motion.
3. Retardation phase:
The distance covered during the retardation phase can also be calculated using the formula:
s3 = (1/2) * (-a) * t3^2
Now, let's find the values of t1, t2, and t3 to express them in terms of t:
t1 + t2 + t3 = t ---> (1)
The distances covered during each phase add up to the total distance, s:
s1 + s2 + s3 = s ---> (2)
Substituting the values of s1, s2, and s3 into equation (2):
(1/2) * a * t1^2 + V * t2 + (1/2) * (-a) * t3^2 = s
Multiplying both sides of the equation by 2 to remove the fractions and rearranging:
a * t1^2 + 2V * t2 - a * t3^2 = 2s
Now, let's solve equation (1) for t2:
t2 = t - t1 - t3
Substituting this value into the equation above:
a * t1^2 + 2V * (t - t1 - t3) - a * t3^2 = 2s
Rearrange the terms:
a * t1^2 - 2V * t1 - a * t3^2 + 2V * t3 = 2s - 2V * t
Rearrange the terms and factor out t:
t * (a * t1 - 2V + a * t3) = 2s - a * t1^2 + a * t3^2
Divide both sides by (a * t1 - 2V + a * t3):
t = (2s - a * t1^2 + a * t3^2) / (a * t1 - 2V + a * t3)
Now, let's simplify the equation by substituting the distance values:
t = (2s - a * t1^2 + a * t3^2) / (a * t1 - 2V - a * t3)
Since V is the constant velocity, we can replace V with (a * t1):
t = (2s - a * t1^2 + a * t3^2) / (a * t1 - 2(a * t1) - a * t3)
Simplifying further:
t = (2s - a * t1^2 + a * t3^2) / (- a * t1 - a * t3)
Now, let's look at the expression inside the numerator:
2s - a * t1^2 + a * t3^2 = 2s - a * t1^2 - a * t3^2
Factor out the negative sign to combine like terms:
2s - (a * t1^2 + a * t3^2)
We can recognize this expression as (t1^2 + t3^2) due to the symmetry of motion during acceleration and retardation phases.
So, the numerator simplifies to:
2s - (t1^2 + t3^2)
Therefore, the time during which the lift is ascending (t1 + t2 + t3) is given by:
t = [(2s - (t1^2 + t3^2)) / (- a * t1 - a * t3)]
Simplifying further:
t = [(2s - (t1^2 + t3^2)) / (- a * (t1 + t3))]
Now, let's recall that t1 + t3 = (t - t2) from equation (1):
t = [(2s - (t1^2 + t3^2)) / (- a * (t - t2))]
Since we know that t2 = s / V (distance traveled at constant velocity divided by velocity), substituting this value:
t = [(2s - (t1^2 + t3^2)) / (- a * (t - (s / V)))]
Using common denominators and simplifying:
t = [(2s - (t1^2 + t3^2)) / (- a * (Vt / V - s / V))]
t = [(2s - (t1^2 + t3^2)) / (- a * (Vt - s) / V)]
Now, let's simplify further by multiplying the numerator and denominator by V:
t = [V * (2s - (t1^2 + t3^2)) / (- a * (Vt - s))]
Finally, let's rearrange and factor the numerator:
t = [V * (2s - t1^2 - t3^2) / (- a * (Vt - s))]
t = [V * (2s - (t1^2 + t3^2)) / (- a * (Vt - s))]
t = [V * (2s - (t1^2 + t3^2)) / (a * (s - Vt))]
Now, let's rearrange the terms inside the brackets:
t = [(2s - (t1^2 + t3^2)) / (a * s - a * Vt)]
Notice that t3^2 = t1^2 since both represent the same time interval but in opposite directions.
t = [(2s - 2t1^2) / (a * s - a * Vt)]
t = [(2(s - t1^2)) / (a * (s - Vt))]
Finally, factor out t1^2 from s - t1^2:
t = [(2(s - t1^2)) / (a * (s - Vt))]
t = [2(s - t1^2) / (a * s - a * Vt)]
t = [2(s - t1^2) / a(s - Vt)]
We can see that the denominator, a(s - Vt), is common to both the numerator and the denominator. By canceling it out:
t = 2(s - t1^2) / a(s - Vt)
t = 2s - 2(t1^2) / a(s - Vt)
Which simplifies to:
t = t^2 - (4s / a)
Therefore, the time during which the lift is ascending is given by t^2 - (4s / a).