How Do Different Acceleration Rates Affect Travel Time in Mine Shafts?

[Merlan114]

Ok so the question goes like this: The winding cages in mine shafts are used to move workers in and out of the mines. These cages move much faster than any commercial ele- vators. In one South African mine, speeds of up to 65.0 km/h are at- tained. The mine has a depth of 2072 m. Suppose two cages start their downward journey at the same moment. The first cage quickly attains the maximum speed (an unrealistic situation), then proceeds to de- scend uniformly at that speed all the way to the bottom. The second cage starts at rest and then increases its speed with a constant accelera- tion of magnitude 4.00 × 10–2 m/s2. How long will the trip take for each cage? Which cage will reach the bottom of the mine shaft first?

What I did was use this formula for cage 1 Δt=[2Δx/(vi+vf)] and got 229.5s:
Δt=2(2072m)/[0+(65km/h)(1h/3600s)(1000m/1km)]

For the second cage I used the formula Δt=Δv/a and got 451.38
Δt= [(65km/h)(1h/3600s)(1000m/1km)]/4.00x10^-2 m/s^2

When I look on the answer sheet it said:
1st cage
115 s
They used Δt=Δx/v1
and for second cage they got 322s and they used the formula Δt=√2Δx/a

What did I do wrong? I figured there are multiple way in doing this, but why could I use the formulas I did the first time. Can someone explain to me which formula is right and why it is right and the other is wrong?!

 

[szimmy]

Your formula is right for the first cart, but you have 0 for Vi, the Vi is the same as Vf.
I see why you did this though, it says the carts both start at rest, but it says the first cage instantaneously reaches 65km/h and stays there. This implies that the change in time (and concurrently the distance traveled) is negligible.
The formula comes from the kinematic equation, x = Vi*t + .5 * a * t^2
There is no acceleration, so .5*a*t^2 = 0
This leaves x = Vi*t
Aka T = x/Vi

The formula that you used for part 2 is the time it takes for the cart to reach its maximum velocity. This is not necessarily the time it takes to travel the distance.
Your initial velocity is 0, and the acceleration and displacement x are given. So the equation to use would be x = Vi*t + 1/2 * a * t^2
Since you know Vi = 0, I'm sure you can do some basic algebra to get the equation they got.