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?!!!