A vacuum-propelled capsule for a high-speed tube transportation system of the future is being designed for operation between two stations A and B, which are 11.6 km apart. If the acceleration and deceleration are to have a limiting magnitude of 0.64g and if velocities are to be limited to 430 km/h, determine the minimum time t for the capsule to make the 11.6-km trip. B 11.6 km

Question

A vacuum-propelled capsule for a high-speed tube transportation system of the future is being designed for operation between two stations A and B, which are 11.6 km apart. If the acceleration and deceleration are to have a limiting magnitude of 0.64g and if velocities are to be limited to 430 km/h, determine the minimum time t for the capsule to make the 11.6-km trip. B 11.6 km

Answer

Determining the Minimum Travel Time for the Capsule

The problem involves optimizing the travel time for the capsule considering the limitations on acceleration, deceleration, and maximum velocity. Here’s how we can solve it:

1. Define the Travel Phases:

The trip can be divided into three phases:

  • Acceleration Phase: The capsule accelerates from rest to its maximum velocity (430 km/h).
  • Constant Velocity Phase: The capsule maintains its maximum velocity for a certain distance.
  • Deceleration Phase: The capsule decelerates from its maximum velocity to rest at station B.

2. Calculate Time Spent in Each Phase:

a) Acceleration Phase:

  • Convert maximum velocity to m/s: 430 km/h * (1000 m/km) * (1 h/3600 s) = 119.44 m/s
  • Use the equation v = u + at, where v is final velocity (119.44 m/s), u is initial velocity (0 m/s), a is acceleration (0.64g * 9.81 m/s²), and t is time.
  • Solve for t: t = (v – u) / a = 119.44 m/s / (0.64 * 9.81 m/s²) ≈ 18.8 s

b) Constant Velocity Phase:

  • Let x be the distance traveled at constant velocity.
  • The total distance is 11.6 km = 11600 m.
  • Allocate the remaining distance after accounting for acceleration and deceleration to the constant velocity phase: x = 11600 m – 2 * (0.5 * 18.8 s * 0.64 * 9.81 m/s²) ≈ 11223.4 m.
  • Time spent at constant velocity (t_c): t_c = x / v = 11223.4 m / 119.44 m/s ≈ 94.06 s

c) Deceleration Phase:

  • Time spent decelerating is equal to the time spent accelerating (t_d = 18.8 s).

3. Calculate Total Minimum Travel Time:

  • Minimum total travel time (t_min): t_min = t_a + t_c + t_d = 18.8 s + 94.06 s + 18.8 s ≈ 131.7 s

Therefore, the minimum time for the capsule to travel the 11.6 km distance between stations A and B is approximately 131.7 seconds.

Additional Notes:

  • This solution assumes that the acceleration and deceleration can be achieved instantaneously, which may not be realistic in practice. Realistically, both phases would take some finite time.
  • Optimizing the distance traveled at constant velocity for a shorter total travel time could involve a more complex analysis considering the limitations on acceleration and deceleration.

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