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
