Answer:
The time of travel between the two stations is 56.6 s.
Explanation:
The equation for the position and velocity of the train will be as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where
x = position of the train at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
First let´s calculate the time of travel for the first half of the distance:
In this case x0 and v0 = 0. Then:
x = 1/2 · a · t²
600 m = 1/2 · 1.50 m/s² · t²
2 · 600 m / 1.50 m/s² = t²
t = 28.3 s
The velocity at the end of this part of the travel will be:
v = v0 + a · t (v0 = 0)
v = a · t
v = 1.50 m/s² · 28.3 s = 42.5 m/s
Now, let´s calculate the time for the second half of the travel. The initial velocity will be the final velocity of the first part of the travel (42.5 m/s).
Using the equation for velocity:
v = v0 + a · t
0 m/s = 42.5 m/s - 1.50 m/s² · t
- 42.5 m/s / - 1.50 m/s² = t
t = 28.3 s
This makes sense because the acceleration is of the same magnitude.
The time of travel between the two stations is (28.3 s + 28.3 s) 56.6 s.
