Question

A subway train is designed with a maximum acceleration of +0.20 m/s^2 which allows for both passenger safety and comfort. a. If subway stations are 1.2 km apart, what is the maximum velocity that can be obtained between stations? b. How long does it take to travel between two stations? (Do not include time spent stopped at a station.) c. The train stops for a total of 45 s at each station. What is the overall average velocity of the train from station to station?

Answer 1

a) v = 480 m / s , b) t = 109.5 s , c) v = 7.77 m / s

Step-by-step explanation:

Let's use the kinematic equations

v² = v₀² + 2 a x

As part of rest v₀ = 0

v = 2 ax

Let's reduce the magnitudes to the SI system

x = 1.2 km (1000 m / 1 km) = 1200 m

v = 2 0.20 1200

v = 480 m / s

b) x = v₀ t + ½ a t²

x = 0 + 1/2 to t²

t = √ 2 x / a

t = √ (2 1200 / 0.2)

t = 109.5 s

c) the average speed is the distance traveled between the time interval

.x = 1200 m

Total time is travel time plus time that is stopped

t = 109.5 + 45

t = 154.5 s

v = x / t

v = 1200 / 154.5

v = 7.77 m / s