Question

A train starts from rest at station A and accelerates at 0.5 m/s 2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m/s2 until it is brought to rest at station B. Determine the distance between the stations.

Answer 1

d = 28,350 m.

Step-by-step explanation:

• We can divide the problem in 3 parts, according to the type of movement that train had:

1. A distance traveled while the train was speeding up.
2. A distance traveled at constant speed.
3. A distance traveled while the train was slowing down.

• For the first part, we can get the distance traveled using the kinematic equation for the displacement, starting from rest, as follows:

`x_(1) = (1)/(2) * a * t^(2) = (1)/(2) *0.5m/s2* (60s)^(2) = 900 m. (1)`

• In order to get the final speed at this distance, we can simply apply the definition of acceleration, in this way:

`v_(f1) = a* t = 0.5 m/s2 * 60 s = 30 m/s (2)`

• Since the second part is traveled at a constant speed, we can find the displacement applying the definition of average velocity, as follows:

`x_(2) = v_(f1) * t = 30 m/s* 60 s* 15 = 27,000 m (3)`

• Finally, as we know the value of the deceleration, and that the final velocity is zero from the givens, we can apply the following kinematic equation to get the displacement during the third stage:

`v_(f3)^(2) - v_(o3) ^(2) = 2* a_(3) * x_(3) (4)`

• Replacing by the values and solving for x₃, we get:

`x_(3) = (v_(o3)^(2))/(2*a_(3) ) = ((30m/s)^(2))/(2*1m/s2) = 450 m (5)`

• The total distance d is just the sum of x₁, x₂ and x₃, from (1), (3) and (5), as follows:
• d = x₁ + x₂ + x₃ = 900 m + 27,000 m + 450 m = 28,350 m.