In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: (a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate 1.1 m/s2 of until it reaches 95 km/h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at-2.0 m/s2 Assume it stops at each intermediate station for 22 s.

a) t = 746 s

b) t = 666 s

Explanation:

a)

Total time will be the sum of the partial times between stations plus the time stopped at the stations.Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       

We can find the distance traveled while the train was decelerating as follows:

       

Finally, we need to know the time traveled at constant speed.So, we need to find first the distance traveled at the constant speed of 26.4m/s.This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       

Total time between two stations is simply the sum of the three times we have just found:t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:tm = t*5 = 131.6 * 5 = 658.2 s (9)Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.We can find the distance traveled at constant speed, rewriting (6) as follows:x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       

Total time between two stations is simply the sum of the three times we have just found:t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:tm = t*3 = 207.4 * 3 = 622.2 s (14)Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

Electoral current force mechanical advantage if the wheel and axle is 2 n and the output force is 20 n