To solve this problem, we can break it down into two parts: the first half where the train is accelerating, and the second half where it is decelerating.
(a) Calculating the travel time: Let's denote the time taken for acceleration as t1 and the time taken for deceleration as t2. We know that the distance covered during each half is half of the total distance, which is 1100 m/2 = 550 m.
For the first half: Using the equation of motion, we have: d = v₀t + (1/2)at²
Since the initial velocity (v₀) is 0 (train starts from rest) and the acceleration (a) is 1.2 m/s², we can rewrite the equation as: 550 = 0 + (1/2)(1.2)t₁²
Simplifying the equation: 1.2t₁² = 550 × 2 t₁² = 1100 t₁ = √1100
For the second half: We have the same distance (550 m), but the acceleration is now -1.2 m/s² (deceleration). So the equation becomes: 550 = 0 + (1/2)(-1.2)t₂²
Simplifying: -1.2t₂² = 550 × 2 t₂² = -1100 t₂ = √-1100
However, the square root of a negative value is not a real number, which indicates that there is no real solution for t₂. This means that the train doesn't come to a complete stop during deceleration. Instead, it reaches its maximum speed and then continues at a constant speed until the end of the distance.
Since t₂ doesn't have a real solution, we can conclude that the train's travel time (a) is equal to t₁.
(b) Calculating the maximum speed: We can find the maximum speed by using the equation of motion:
v = v₀ + at
In this case, since the train starts from rest and we already know the acceleration (1.2 m/s²), we can substitute the values into the equation:
v = 0 + 1.2t₁
Now, substituting t₁ = √1100 (as calculated earlier):
v = 1.2 × √1100
Evaluating this expression will give you the maximum speed of the train.
Note: Keep in mind that when taking the square root, there will be both positive and negative solutions. Since time cannot be negative, we consider the positive square root values.