Question

(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located 806 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph x, ν, and a versus t for the interval from one start-up to the next.

Answer 1

Answer:

The correct answer is 32.9 m/s

Explanation:

To solve this, we list out the known and the unknown variables as follows

Maximum allowable acceleration = 1.34 m/s²

Distance between sttions = 806 m

Therefore from the equation of motion

v = ut + 0.5·×at²

Where v = final velocity

u = initial velocity

S = distance covered

t  = time

a = acceleration

Also v² = u² + 2·a·S

where u is the initial velocity, which we can take as u = 0, then

v² = 2·1.34·S = 2.68S m²/s²  then

Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z,  so u² = 2.68Z

since u² = 2.68S from the previous calculation, then for v = 0

2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m

and v² = 2.68S m²/s² = 1080.04 m²/s²

v = 32.9 m/s

The maximum speed a subway train can attain between stations is 32.9 m/s