Verified answer

Explanation:

To find the minimum time for the train to travel between the stations, we need to consider the acceleration and deceleration phases.

Given:

Acceleration: 20 cm/s²

Deceleration: 100 cm/s²

Distance between stations: 2 km = 200,000 cm (since 1 km = 100,000 cm)

Let's analyze the different phases of the train's motion:

1. Acceleration phase:

During this phase, the train starts from rest and accelerates until it reaches its maximum velocity. The time taken during acceleration can be calculated using the formula:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Since the train starts from rest (u = 0) and the final velocity is unknown, we can rewrite the formula as:

v = at

Substituting the values:

a = 20 cm/s²

d = distance covered during acceleration (unknown)

t₁ = time taken during acceleration (unknown)

Using the formula for distance covered during constant acceleration:

d = ut + 1/2 at²

Since the initial velocity u = 0:

d = 1/2 at²

200,000 cm = 1/2 (20 cm/s²) t₁²

400,000 = 20 t₁²

t₁² = 20,000

t₁ ≈ 141.4 s

2. Deceleration phase:

During this phase, the train decelerates from its maximum velocity to come to a stop at the second station. Similar to the acceleration phase, we can calculate the time taken during deceleration using the formula:

v = u + at

Here, the initial velocity u is the maximum velocity achieved during acceleration, and the final velocity v is 0 since the train comes to a stop.

0 = u + (-a) t

Substituting the values:

a = 100 cm/s²

d = distance covered during deceleration (unknown)

t₂ = time taken during deceleration (unknown)

Using the formula for distance covered during constant acceleration:

d = ut + 1/2 at²

Substituting the values:

d = ut₂ + 1/2 a t₂²

200,000 cm = u t₂ + 1/2 (-100 cm/s²) t₂²

200,000 cm = u t₂ - 50 t₂²

Since u = at₁ (maximum velocity is achieved at the end of acceleration)

200,000 cm = (20 cm/s²)(141.4 s) t₂ - 50 t₂²

200,000 cm = 2828 cm t₂ - 50 t₂²

4,000 cm = 80 t₂ - t₂²

This equation is a quadratic equation. We can solve it to find the time t₂.

Using a quadratic solver, we find two solutions for t₂: t₂ ≈ 48.29 s and t₂ ≈ 52.58 s.

3. Total time:

The total time for the train to travel between the stations is the sum of the time taken during acceleration (t₁), the time taken during deceleration (t₂), and the time spent at the maximum velocity (tₘ).

t_total = t₁ + t₂ + tₘ

Since the train comes to a stop at the second station, the time spent at the maximum velocity is 0.

t_total = t₁ + t₂

t_total = 141.4 s + 48.29 s (taking the smaller value of t₂)

t_total ≈ 189.69 s

Therefore, the minimum time for the train to travel between the stations 2 km apart is approximately 189.69 seconds.

To find the minimum time for the train to travel between two stations, we need to calculate the time taken for both acceleration and deceleration.

Given:

Acceleration (a) = 20 cm/s²

Deceleration (d) = -100 cm/s² (negative sign indicates deceleration)

Distance (s) = 2 km = 200,000 cm (since 1 km = 100,000 cm)

First, let's calculate the time taken for acceleration.

We can use the equation: v = u + at, where

v = final velocity

u = initial velocity (which is 0 in this case)

a = acceleration

t = time

To find the final velocity, we can use the equation: v² = u² + 2as.

For the acceleration phase:

u = 0 (initial velocity)

v = ? (final velocity)

a = 20 cm/s² (acceleration)

s = 200,000 cm (distance)

v² = u² + 2as

v² = 0 + 2 * 20 * 200,000

v² = 800,0000

v ≈ 8944.27 cm/s (approximately)

Now, using the equation v = u + at, we can calculate the time taken for acceleration:

8944.27 = 0 + 20t

t ≈ 447.21 seconds (approximately)

Next, let's calculate the time taken for deceleration.

Using the same equation v = u + at, but with the deceleration value and the final velocity as 0, we have:

0 = v + (-100)t

Substituting v ≈ 8944.27 cm/s:

0 = 8944.27 - 100t

100t = 8944.27

t ≈ 89.44 seconds (approximately)

Since the time taken for acceleration and deceleration is the same (89.44 seconds), the total minimum time taken for the train to travel between the two stations is the sum of both phases:

Total time = 2 * 89.44 = 178.88 seconds (approximately)

Therefore, the minimum time for the train to travel between the two stations is approximately 178.88 seconds.